Page 1 :
https://t.me/NEET_StudyMaterial, , Chapter, , 29, , Semiconductor Electronics :, Materials, Devices and Simple Circuits, Solutions, SECTION - A, Objective Type Questions, 1., , The semiconductors are generally, (1) Monovalent, , (2) Divalent, , (3) Trivalent, , (4) Tetravalent, , Sol. Answer (4), Semiconductors are generally tetravalent like silicon and gallium., 2., , The resistivity of a semiconductor depends upon, (1) Size of the atom, , (2) The nature of atoms, , (3) Type of bonds, , (4) Size and types of motion, , Sol. Answer (2), The resistivity of a semiconductor depends mainly on the kind of atoms and the valence electrons they, possess., 3., , The impurity atoms with which pure silicon should be doped to make a p-type semiconductor are those of, (1) Phosphorus, , (2) Antimony, , (3) Boron, , (4) Copper, , Sol. Answer (3), The impurities needed to make holes it should be a trivalent substance, of the third group which happens to, be boron., 4., , In semiconductors, which of the following relations is correct at thermal equilibrium?, (1) ni = ne = nh, , (2) ni2 = nenh, , (3) ni , , ne, nh, , (4) ni = ne + nh, , Sol. Answer (2), n12 = ne × nh, 5., , Due to conservation of charge, , A pure semiconductor has, (1) An infinite resistance at 0°C, (2) A finite resistance which does not depend upon temperature, (3) A finite resistance which increases with temperature, (4) A finite resistance which decreases with temperature, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 2 :
182, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (4), A simple semiconductor has a finite resistance. An increase in temperature increases number of charge, carries and increases conductivity., 6., , The rate of recombination or generation are governed by the law(s) of, (1) Mass conservation, , (2) Electrical neutrality, , (3) Thermodynamics, , (4) Chromodynamics, , Sol. Answer (3), Carriers flow from higher to lower concentration like heat., 7., , An n-type semiconductor is electrically, (1) Positive, , (2) Negative, , (3) May be positive or negative, , (4) Neutral, , Sol. Answer (4), The presence of charge carries does not mean a semiconductors has any net charge., 8., , A solid having uppermost energy band partially filled with electrons is called, (1) An insulator, , (2) A conductor, , (3) A semiconductor, , (4) None of these, , Sol. Answer (2), A solid which has uppermost energy band partially filled with electron is called a conductor., 9., , The energy gap for an insulator may be, (1) 1.1 eV, , (2) 0.02 eV, , (3) 6 eV, , (4) 0.7 eV, , Sol. Answer (3), The energy gap for an insulators is very high around 6 eV., 10. In an intrinsic semiconductor, the density of conduction electrons is 7.07 × 1015 m–3. When it is doped with, indium, the density of holes becomes 5 × 1022 m–3. Find the density of conduction electrons in doped, semiconductor, (2) 1 × 109 m–3, , (1) Zero, , (3) 7 × 1015 m–3, , (4) 5 × 1022 m–3, , Sol. Answer (2), ne . nh = ni2, and ne = nh = ni, , initially, , or ni = 7.07 × 1015 m–3, ne =, , (7.07 × 1015 )2, 5 × 1022, , on ne 1 × 109 m–3, 11. If NA is number density of acceptor atoms added and ND is number density of donor atoms added to a, semiconductor, ne and nh are the number density of electrons and holes in it, then, (1) ne = ND, nh = NA, , (2) ne = NA, nh = ND, , (3) ne + ND = nh + NA, , (4) ne + NA = nh + ND, , Sol. Answer (4), Donor atoms increase number of conduction electron and must be added to available electrons., Similarly for holes and acceptor atoms., The equation is formed according to the law of electrical neutrality., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 3 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 183, , 12. In an unbiased p-n junction which of the following is correct?, (1) p-side is at higher potential than n-side, (2) n-side is at higher potential than p-side, (3) Both p-side and n-side are at the same potential, (4) Any of the above is possible depending upon the carrier density in the two sides, Sol. Answer (2), In the depletion region n-side has positive ions and p-side is with negative ion. Hence n-side has longer, potential., 13. In a full wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the ripple, would be, (1) 25 Hz, , (2) 50 Hz, , (3) 70.7 Hz, , (4) 100 Hz, , Sol. Answer (4), If mains frequency is 50 Hz after full wave rectification the frequency becomes double that of mains., So answer is 100 Hz., 14. In a semiconductor diode, the reverse biased current is due to drift of free electrons and holes caused by, (1) Thermal excitations only, , (2) Impurity atoms only, , (3) Both (1) & (2), , (4) Neither (1) nor (2), , Sol. Answer (1), In case of reverse bias the reverse current is independent of reverse bias voltage but depends only on, temperature of junction., 15. The value of form factor in case of half wave rectifier is, (1) 1.11, , (2) 1.57, , (3) 1.27, , (4) 0.48, , Sol. Answer (2), Form factor =, , rms of out put voltage, average value of out put voltage, , or form factor =, , v max / 2 π, = = 1.57, v max / π 2, , 16. In a semiconductor diode, P-side is earthed and N-side is put at potential of –2 V, the diode shall, (1) Conduct, , (2) Not conduct, , (3) Conduct partially, , (4) Break down, , Sol. Answer (1), P side is put at higher potential than N side hence the diode will conduct., 17. Two identical p-n junctions may be connected in series with a battery in three ways as shown in the adjoining, figure. The potential drop across the p-n junctions are equal in, , P N, , N P, , First circuit, , P N, , P N, , Second circuit, , N P, , N P, , Third circuit, , (1) First and second circuits, , (2) Second and third circuits, , (3) Third and first circuits, , (4) All of these, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 4 :
184, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (2), First is not bias second and third are bias and have same potential drop across diodes., 18. The zener diode is used for, (1) Rectification, , (2) Amplification, , (3) Stabilization, , (4) All of these, , Sol. Answer (3), Zener diode is a reverse biased transistor used for voltage stabilisation., 19. In the diagram shown below, the input is across the terminals A and C and the output is across B and D., Then the output is, , B, A, , C, D, , (1) Zero, , (2) Same as input, , (3) Full wave rectified, , (4), , Half wave rectified, , Sol. Answer (3), The diagram is an example of a full wave rectifying circuit., 20. A junction diode, in which one of the p or n-sections is made very thin, can be used to convert light energy, into electrical energy, then the diode is called, (1) Light emitting diode, , (2) Zener diode, , (3) Solar cell, , (4) Photo diode, , Sol. Answer (3), A diode used to convert light energy to electrical energy is called a photo diode., 21. The material suitable for making a solar cell is, (1) PbS, , (2) GaAs, , (3) CdSe, , (4) Ge, , Sol. Answer (2), Ga As has a band gap close to 1.5 eV which in same as maximum intensity of solar radiation spectrum., 22. In which of the configurations of a transistor, the power gain is highest?, (1) Common base, , (2) Common emitter, , (3) Common collector, , (4) Same in all the three, , Sol. Answer (2), The power gain is highest in the case common emitter type transistor., 23. In a common base amplifier, the phase difference between the input signal voltage and the output voltage, (across collector and base) is, (1) 0, , (2), , , 4, , (3), , , 2, , (4) , , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 5 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 185, , 24. The current gain of a transistor is 50. The input resistance of the transistor, when used in the common emitter, configuration, is 1 k. The peak value of the collector a.c. current for an alternating peak input voltage 0.01V, is, (1) 100 A, , (2) 250 A, , (3) 500 A, , (4) 800 A, , Sol. Answer (3), , ic, 50, ib, Ri ic, 50, Vi, , ic , , 50 .01, 103, , = 500 A, , 25. In a common emitter transistor circuit, the base current is 40 A, then VBE is, , VCC = 10 V, 2 k, 245 k, , C, B, E, , (1) 2 V, , (2) 0.2 V, , (3) 0.8 V, , (4) Zero, , Sol. Answer (2), Applying kirchoff, 10 – 245 × 40 × 10–3 = VBE, , 10V = Vcc, 40 A, 245 k, , 2 k, C, , B, E, , 26. In a transistor the base is very lightly doped as compared to the emitter because by doing so, (1) The flow across the base region is mainly because of electrons, (2) The flow across the base region is mainly because of holes, (3) Recombination is decreased in the base region, (4) Base current is high, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 6 :
186, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (3), Base is lightly doped because if it had more of its charge carries. It would recombine with the majority charge, carrier of the transistor, reducing conductivity., 27. A transistor is operated in CE configuration at Vcc = 2V such that a change in base current from 100 A to, 200 A produces a change in the collector current from 9 mA to 16.5 mA. The value of current gain, is, (1) 45, , (2) 50, , (3) 60, , (4) 75, , Sol. Answer (4), β=, , =, , ∆IC, ∆IB, 7.5 × 10−3, 100 × 10−6, , = 75, 28. The input resistance of a silicon transistor is 1 k. If base current is changed by 100 A, it causes the change in, collector current by 2 mA. This transistor is used as a CE amplifier with a load resistance of, 5 k. What is the ac voltage gain of amplifier?, (1) 10, , (2) 100, , (3) 500, , (4) 200, , Sol. Answer (2), R2, Av = R, i, , β=, , Ri = 1 k; R2 = 3 k, , ∆IC, 2 × 10−3, =, ∆IB 100 × 10−6, , ∆IB = 100 μA, , IC = 2 mA, Av =, , 20 × 5 × 10−2, 1 × 10−2, , = 100, , 29. The relationship between and is given by, (1) = , , (2) , , 1, , , (3) , , , 1 , , (4) , , , 1 , , Sol. Answer (3), Relation between and is given by β =, , α, 1− α, , 30. Input signal to a common emitter amplifier having a voltage gain of 1000 is given by, vi = (0.004 V) sin (t + /2). The corresponding output signal is, (1) (40V) sin (t + /2), , (2) (0.004V) cos (t + /2) (3) (4V) cos (t – /2), , (4) (4V) sin (t – /2), , Sol. Answer (4), Av = 1000, Vi = 0.004 sin(t + 90º), V0 = ?, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 7 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 187, , V0, Av = V, i, V0 = Av × Vi = 1000 × 4 × 10–3 = 4, π⎞, ⎛, V0 = 4 sin ⎜ ωt − ⎟, ⎝, 2⎠, , 31. In a common base transistor circuit, the current gain is 0.98. On changing emitter current by, 5.00 mA, the change in collector current is, (1) 0.196 mA, , (2) 2.45 mA, , (3) 4.9 mA, , (4) 5.1 mA, , Sol. Answer (3), CB - configuration, = 0.98, , =, , ∆IC, ∆IE, , IE = 5 mA, IC = IE = 0.98 × 5 × 10–3, = 4.9 × 10–3 A, = 4.9 mA, 32. For a transistor amplifier power gain and voltage gain are 7.5 and 2.5 respectively. The value of the current, gain will be, (1) 0.33, , (2) 0.66, , (3) 0.99, , (4) 3, , Sol. Answer (4), Power gain = Voltage gain × Current gain, Amplifier power gain = 7.5, and voltage gain is 2.5, Current gain =, , 7.5, =3, 2.5, , 33. The input resistance of a common-emitter amplifier is 2 k and a.c. current gain is 20. If the load resistor used, is 5 k, calculate the transconductance of the transistor used, (1) 0.01 –1, , (2) 0.03 –1, , (3) 0.04 –1, , (4) 0.07 –1, , Sol. Answer (1), Ri = 2 k, = 20, RL = 5 k, , β, 20, = 0.01 Ω−1, gm = R =, 3, 2, ×, 10, i, 34. In a silicon transistor, a change of 7.89 mA is the emitter current produces a change of 7.8 mA in the collector, current. What change in the base current is necessary to produce an equivalent change in the collector current?, (1) 9 mA, , (2) 0.9 mA, , (3) 0.09 mA, , (4) Zero, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 8 :
188, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (3), IE = 7.89 mA, IC = 7.8 mA, IE = IC + IB, IB = 0.09 mA, 35. The adjoining logic symbol is equivalent to, , A, T, B, (1) OR gate, , (2) AND gate, , (3) NOT gate, , (4) NAND gate, , Sol. Answer (2), Let after 1st NAND gate the result is Z., , Z = A⋅B, 2nd NAND gate with common input behave like a NOT, result y = Z = A.B = A.B, 36. Which of the following gates corresponds to the truth table given below?, , (1) NAND, , (2) NOR, , A, , B, , Y, , 1, , 1, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 1, (3) XOR, , (4) OR, , Sol. Answer (1), , y A. B, 37. Figure shows the practical realisation of a logic gate. Identify the logic gate, , VCC = 5V, A, , RB, , RC, , 5V, B, , (1) NAND, , (2) NOR, , (3) XOR, , (4) XNOR, , Sol. Answer (1), The inputs A and B are both high when zero signal is observed only. Hence it is as NAND gate., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 9 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 189, , 38. The combination of gates shown in the circuit is equivalent to, , A, Y, B, (1) OR, , (2) AND, , (3) NAND, , (4) NOR, , (3) XOR gate, , (4) NOR gate, , Sol. Answer (1), NAND and NOR with common input behave line a NOT., Output of 1st = A, Output of 2nd = B, Feed to the NAND gate, , Y A. B A B A B (OR ), A, , Y The gate is, , 39., , B, (1) OR gate, , (2) AND gate, , Sol. Answer (3), It is a symbol for XOR gate., 40. Write down the boolean expression for output Y of a system shown in figure, , A, Y, , B, , (2) ( A B ) ( A B ), , (1) AB A B, , (3) A B AB, , (4) A.B ( A B ), , Sol. Answer (3), , A, , A, , B, , B, , A.B, Y, A, B, , A.B, , Y A. B A.B, 41. Copper has face centred cubic lattice with interatomic spacing 2.5 Å. The value of lattice constant will be about, (1) 0.35 Å, , (2) 3.5 Å, , (3) 7.0 Å, , (4) 1.5 Å, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 10 :
190, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (2), , 5, 2., 45º, , Lattice constant is the side of the cubic crystal., , S=, , 2.5, 2, , ×2, , S, , = 3.5 Å, 42. Distance between body centred atom and a corner atom in sodium is (where a = lattice constant), , a 3, 2, Sol. Answer (1), , (2) a 3, , (1), , (3), , a 3, 4, , (4) a 2, , Body centred atoms are in the centre of the cube. Value will be half the body diagonal ., Body diagonal = a 3, Distance =, , a 3, 2, , 43. Liquid crystal display monitors are made of, (1) Monocrystals, , (2) Single crystals, , (3) Liquid crystals, , (4) Polycrystals, , (3) Inductor, , (4) Capacitor, , Sol. Answer (3), 44. Which of the following cannot be obtained from an IC?, (1) Resistor, , (2) Diode, , Sol. Answer (3), IC consists of many passive components like R and C (not L)., 45. Operational amplifier is a, (1) Digital IC, , (2) Linear IC, , (3) OR gate, , (4) AND gate, , Sol. Answer (2), Operational amplifier is one of the very useful linear IC., , SECTION - B, Objective Type Questions, 1., , In a zener diode, break down occurs in reverse bias due to, (1) Impact ionisation, , (2) Internal field emission, , (3) High doping concentration, , (4) All of these, , Sol. Answer (2), 2., , In an n-p-n transistor working in active mode, the depletion region, (1) Is not formed, (2) At emitter-base junction is wider than that at collector-base junction, (3) At emitter-base junction is thinner than that at collector-base junction, (4) At the two junctions have equals width, , Sol. Answer (3), Emitter base junction is forward biased and collector base junction is reverse biased., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 11 :
Solution of Assignment, , 3., , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 191, , A transistor has a current amplification factor of 60. In a CE amplifier, input resistance is 1 k and output voltage, is 0.01 V. The transconductance is (in SI units), (1) 10–5, , (2) 6 × 10–2, , (3) 6 × 104, , (4) 10, , Sol. Answer (2), β=, , ∆IC, = 60, ∆IB, , gm =, , 4., , Ri = 1 k, V0 = 0.01 V, , β, 60, =, = 6 × 10−2, Ri 1 × 103, , A common emitter amplifier is designed with NPN transistor ( = 0.99). The input impedance is 1 kand load, is 10 k. The voltage gain will be, (1) 9.9, , (2) 99, , (3) 990, , Ri = 1 k, , RL = 10 k, , (4) 9900, , Sol. Answer (3), = 0.99, , ⎡R ⎤, 10 × 103, = 9.9, AV = α ⎢ L ⎥ = 0.99 ×, 1 × 103, ⎣ Ri ⎦, , 5., , Pure Si at 300 K has hole and electron densities are 1.5 × 1016 m–3. Doping it by an impurity increases the, hole density nh to 4.5 × 1022 m–3. Electron density in the doped silicon is, (1) 1.5 × 1016 m–3, , (2) 3.0 × 1022 m–3, , (3) 5 × 109 m–3, , ni = 1.5 × 1016, , nh = 4.5 × 1022, , (4) 3 × 106 m–3, , Sol. Answer (3), ni2 = ne × nh, , ne =, , ni2, ne, , ne = 3 × 106, 6., , A common emitter transistor amplifier has a current gain of 50. If the load resistance is 9 k and the input, resistance is 500 , the voltage gain of the amplifier will be, (1) 900, , (2) 300, , (3) 200, , Ri = 500, , RL = 9 k, , (4) 100, , Sol. Answer (1), = 50, , Av = α ×, 7., , RL 50 × 9 × 103, =, = 900, 500, Ri, , In the figure given, voltage of point A is, , A, –3V, (1) 0 V, , (2) –3 V, , 5, , (3) –2.3 V, , (4) –2.7 V, , Sol. Answer (1), The diode is reverse biased., Since no current flows in circuit no voltage drop occurs in resistor., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 12 :
192, 8., , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , A p-n photodiode is manufactured from a semiconductor with band gap of 3.1 eV. Which of the following, wavelengths can be detected by it?, (1) 4000 Å, , (2) 3900 Å, , (3) 4200 Å, , (4) Both (1) & (2), , Sol. Answer (4), Energy of photons =, For current to flow :, , hc, J, λ, hce, > 3.1 eV, λ, , < 4010 Å, 9., , In the circuit shown, the input waveform is given. Which of the following correctly gives the output waveform across, RL?, , Input, , RL, , Output, (1), , (2), , (3), , (4), , Sol. Answer (2), The circuit diagram is like a full wave rectifier and current through RL flows only in one direction., 10. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter, than 2480 nm is incident on it. The band gap of the semiconductor is, (1) 0.3 eV, , (2) 0.5 eV, , (3) 0.7 eV, , (4) 1.1 eV, , Sol. Answer (2), = 24800Å, Energy correspond to this wavelength =, , hc, , , 11. Zener breakdown takes place if, (1) Doped impurity is low, , (2) Doped impurity is high, , (3) Less impurity in N-part, , (4) Less impurity in p-type, , Sol. Answer (2), Zener breakdown is more easily possible when doped impurity is higher as there is greater conductive tendency in the diode and a larger depletion zone., 12. In a p-n junction solar cell, the value of photo-electromotive force produced by monochromatic light is, proportional to the, (1) Voltage applied at the p-n junction, , (2) Barrier voltage at the p-n junction, , (3) Intensity of light falling on the cell, , (4) Frequency of light falling on the cell, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 13 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 193, , 13. In a transistor the collector current is always less than the emitter current because, (1) Collector side is reverse biased and the emitter side is forward biased, (2) Collector being reverse biased, attracts less electrons, (3) A few electrons are lost in the base and only remaining one’s reach the collector, (4) Collector side is forward biased and emitter side is reverse biased, Sol. Answer (3), 14. The depletion region of p-n junction has a thickness of the order of, (1) 10–12 m, , (2) 10–6 m, , (3) 10–3 m, , (4) 10–2 m, , Sol. Answer (2), 15. In an n-p-n transistor, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, then, the emitter current will be, (1) 9 mA, , (2) 11 mA, , (3) 1 mA, , (4) 0.1 mA, , Sol. Answer (2), ie = ib + i c, ie =, , 100, × ic, 90, , ie = 11.1 mA or 11 mA, 16. Four equal resistors, each of resistance 10 , are connected as shown in the circuit diagram. The equivalent, resistance between A and B is, 10 , 10 , , A, , B, 10 , , (1) 5 , , (2) 10 , , 10 , (3) 20 , , (4) 40 , , Sol. Answer (2), It is a simple wheat stone's bridge and no current will flow through the middle wire because of lack of potential difference across transistor., , Req =, , 20 × 20, = 10 Ω, 20 + 20, , 17. A transistor cannot be used as an, (1) Amplifier, , (2) Oscillator, , (3) Modulator, , (4) Rectifier, , Sol. Answer (4), A transistor cannot be used as a Rectifier., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 14 :
194, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , 18. What is the value of output voltage V0 in the circuit shown in the figure?, 2 k, +, 20 V, –, , (1) 6 V, , 6 k, , V0, , 6V, , (2) 14 V, , (3) 20 V, , (4) 26 V, , Sol. Answer (1), The zener diode maintains a voltage of 6 V across its ends and voltage remains equal to breakdown voltage, as 6 V., 19. What is the power gain in a CE amplifier, where input resistance is 3k and load resistance 24k given = 6?, (1) 180, , (2) 288, , (3) 240, , (4) 480, , Sol. Answer (2), Power gain = 2, , RL, 2 ⎛ 24 ⎞, = (6) ⎜ ⎟ = 288, Ri, ⎝ 3 ⎠, , 20. For inputs (A, B) and output (Y) of the following gate can be expressed as, , A, B, , (1) A B, Sol. Answer (1), , A, B, , Y, , (2) A.B, , (3) A + B, , (4) A B, , A, B, , A.B, , A, , A.B, , Y = A. B + A. B, Y=AB, , B, 21. Calculate the current I in the following circuit, if all the diodes are ideal. All resistances are of 200, , I, (1) Zero, , (2) 1 A, , 200 V, , (3) 2 A, , (4) 4 A, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 15 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 195, , Sol. Answer (2), Only the forward biased transistor allows flow of current., , Hence I =, , V, =1A, R, , 22. A transistor having = 0.99 is used in a common base amplifier. If the load resistance is 4.5k and the dynamic, resistance of the emitter junction is 50 the voltage gain of the amplifier will be, (1) 79.1, , (2) 89.1, , (3) 78.2, , (4) 450, , Sol. Answer (2), = 0.99, Av = α ⋅, , RI = 50 , , RL = 4.5 k, , RL, 4.5, = 0.99 ×, × 103, RI, 50, , Av = 0.0891 × 103 = 89.1, 23. A potential difference of 2.5 V is applied across the faces of a germanium crystal plate. The face area of the crystal, is 1 cm2 and its thickness is 1.0 mm. The free electron concentration in germanium is 2 × 1019 m–3 and the electron, and holes mobilities are 0.33 m2/V s and 0.17 m2/V s respectively. The current across the plate will be, (1) 0.2A, , (2) 0.4A, , (3) 0.6A, , (4) 0.8A, , Sol. Answer (2), l = 1 mm V = 2.5 volt, e = 0.33 m2/vs, n = 0.17 m2/vs, ne = nn = 2 × 10–19 m–3, , i, , eAv, (ne e nn n ), l, , 24. In the following transistor amplifier circuit = 50. VCE of the transistor is, , +10 V, , 2k , C, , 245k , 40 A, , IB, B, E, , (1) 4 V, , (2) 6 V, , (3) 10 V, , (4) 8 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 16 :
196, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (2), = 50, , IC, 50 , I = 40 , B, A, IB, IC = 50 × 40 × 10–6, = 2 × 10–3 A, VCE = VCC – ICRC, = 10 – 2 × 10–3 × 2 × 103, =6V, 25. The given transistor amplifier connection is, , +20 V, , 10k , , 250k , , (1) Common base connection, , (2) Common emitter connection, , (3) Common collector connection, , (4) All of these, , Sol. Answer (2), In this case the emitter is grounded here, it is a common emitter configuration., 26. The circuit shown in the figure contains two diodes each with a forward resistance of 50 and with infinite backward, resistance. If the battery of 6 V is connected in the circuit, then the current through the 100 resistance is, , 50, 50, 6V, (1) Zero, , (2) 0.02 A, , 100, (3) 0.03 A, , (4) 0.036 A, , Sol. Answer (3), Only the forward biased diode will allow the transfer of current., v, 6, =, = 0.04 A, R 150, Current 0.4 0.036 A hence answer is (4)., I=, , 27. Three amplifiers each having voltage gain 10, are connected in series. The resultant gain would be, (1) 10, , (2) 30, , (3) 1000, , (4), , 10, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 17 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 197, , Sol. Answer (3), Each multiplies initial power by 10 times., Hence final voltage compared to initial voltage, V = 10 × 10 × 10 V0, or V = 1000 V0, 28. A semiconductor X is made by doping a germanium crystal with arsenic (Z = 33). A second semiconductor Y is, made by doping germanium with indium (Z = 49). X and Y are used to form a junction as shown in figure and, connected to a battery as shown. Which of following statement is correct?, , X, , Y, , (1) X is p-type, Y is n-type and the junction is forward biased, (2) X is n-type, Y is p-type and the junction is forward biased, (3) X is p-type, Y is n-type and the junction is reverse biased, (4) X is n-type, Y is p-type and the junction is reverse biased, Sol. Answer (4), Group 5 elements like arsenic are used to create, N type semiconductors and indium is used to create P type, semiconductor., 29. The maximum efficiency of a full wave rectifier is, , (1), , 4, 2, , 100 %, , (2), , 8, 2, , 100 %, , (3) 40%, , (4) 80%, , Sol. Answer (2), Maximum efficiency of a full wave rectifier is, , 8, π2, , × 100%, , 30. Logic gate realised from pn junctions shown in figure is, A, Y, B, , (1) OR gate, , (2) AND gate, , (3) NOT gate, , (4) NOR gate, , Sol. Answer (1), It is an OR gate as the current will flow through resistor if any one A or B is of correct input., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 18 :
198, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , 31. The output across the load resistance R is, , R, , Input, a.c, (1) Half wave rectified, , (2) Full wave rectified, , (3) Quarter wave rectified, , (4) ac, , Sol. Answer (2), It is a full wave rectifier., 32. Which of the following pn junction is not used in reverse bias?, (1) LED, , (2) Solar cell, , (3) Zener diode, , (4) Both (1) & (2), , Sol. Answer (4), 33. Which of the following break down of pn junction is reversible?, (1) Avalanche breakdown (2) Zener breakdown, , (3) Dielectric breakdown (4) All of these, , Sol. Answer (2), 34. In the circuit shown, the average power dissipated in the resistor is (assume diode to be ideal), R, , ~, , E0 sin t, (1), , E02, , 2R, Sol. Answer (2), , (2), , E02, 4R, , It's a half wave rectification Erms , Pav =, =, , (3), , E02, R, , (4) Zero, , E0, 2, , Erms 2, R, , E02, 4R, , 35. A crystal has bcc structure and its lattice constant is 3.6 Å. What is the atomic radius?, (1) 3.6 Å, , (2) 1.8 Å, , (3) 1.27 Å, , (4) 1.56 Å, , Sol. Answer (4), 36. All the diodes are ideal. The current flowing in 2resistor connected between the diodes D1 and D2 is, D2, D1, 1, 2, 1, , D3, , 7, , D4 5, , 4, , (1) 1A, , (2) 2A, , 10V 3, (3) 3A, , (4) Zero, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 19 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 199, , Sol. Answer (1), D1 and D2 are forward bias, i=, , 10, 32, , i = 2A, Branch resistors are same, i1 = i2 = 1A, , 37. In a transistor ( = 50), the voltage across 5k load resistance in collector circuit is 5V. The base current is, (1) 0.02mA, , (2) 0.03mA, , (3) 0.08mA, , (4) 0.09mA, , Sol. Answer (1), , IC, 5, = I , IC , 50, 103, , B, , IB , , IC, , , SECTION - C, Previous Years Questions, 1., , In the given figure, a diode D is connected to an external resistance R = 100 and an e.m.f. of, 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be, , D, , 100 , , [Re-AIPMT-2015], , R, , 3.5 V, (1) 35 mA, , (2) 30 mA, , (3) 40 mA, , (4) 20 mA, , Sol. Answer (2), , 0.5 V, , 100 , , I, 3.5 V, I, , (3.5 0.5) V, 30 mA, 100 , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 20 :
200, 2., , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , ⎞, ⎛, The input signal given to a CE amplifier having a voltage gain of 150 is Vi = 2cos ⎜ 15t ⎟ . The corresponding, 3⎠, ⎝, output signal will be, [Re-AIPMT-2015], 4 ⎞, ⎛, (1) 300cos⎜15t ⎟, 3⎠, ⎝, , 2 ⎞, ⎛, (3) 75cos ⎜ 15t , 3 ⎟⎠, ⎝, , ⎞, ⎛, (2) 300cos⎜15t ⎟, 3, ⎝, ⎠, , 5 ⎞, ⎛, (4) 2cos ⎜ 15t , 6 ⎟⎠, ⎝, , Sol. Answer (1), V0 = Vi and phase difference of , , 4 ⎞, , ⎛, V0 = 150 2 cos ⎛⎜ 15t ⎞⎟ = 300 cos ⎜ 15t , 3 ⎟⎠, 3, ⎝, ⎝, ⎠, 3., , Which logic gate is represented by the following combination of logic gates?, , [AIPMT-2015], , Y1, , A, , Y, B, (1) NOR, , Y2, , (2) OR, , (3) NAND, , (4) AND, , Sol. Answer (4), From boolean, –, –, A + B, , 4., , =AB, , If in a p-n junction, a square input signal of 10 V is applied, as shown, , +5 V, RL, –5 V, then the output across RL will be, , (1), , 5V, , [AIPMT-2015], 10 V, , (2), , (3), , – 10 V, , (4), , –5 V, , Sol. Answer (1), Diode will be in forward bias in 0-5 volt so it will conduct., 5., , The given graph represent V-I characteristic for a semiconductor device., , I, A, V, B, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 21 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Which of the following statement is correct ?, , 201, , [AIPMT-2014], , (1) It is V – I characteristic for solar cell where, point A represents open circuit voltage and point B short circuit, current., (2) It is for a solar cell and points A and B represent open circuit voltage and current, respectively., (3) It is for a photodiode and points A and B represent open circuit voltage and current, respectively., (4) It is for a LED and points A and B represent open circuit voltage and short circuit current, respectively., Sol. Answer (1), Voltage at A will be VOC i.e., open circuit voltage i.e., cell supplies no current to circuit, at pont B condition, of short circuit current., 6., , The barrier potential of a p-n junction depends on:, (a) Type of semi conductor material, (b) Amount of doping, (c) Temperature, Which one of the following is correct?, (1) (a) and (b) only, , (2) (b) only, , [AIPMT-2014], (3) (b) and (c) only, , (4) (a), (b) and (c), , Sol. Answer (4), Barier potential depends upon types of semiconductor amount of dopping., 7., , In a n-type semiconductor, which of the following statement is true, , [NEET-2013], , (1) Electrons are minority carriers and pentavalent atoms are dopants., (2) Holes are minority carriers and pentavalent atoms are dopants, (3) Holes are majority carriers and trivalent atoms are dopants, (4) Electrons are majority carriers and trivalent atoms are dopants., Sol. Answer (2), 8., , In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03, mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho, and current gain 20, the voltage gain will be, [NEET-2013], (1) 1.5 G, , (2), , 1, G, 3, , (3), , 5, G, 4, , (4), , 2, G, 3, , Sol. Answer (4), 9., , The output (X) of the logic circuit shown in figure will be, , [NEET-2013], , A•, B•, (1) X A B, , (2) X = A.B, , •X, , (3) X A B, , (4) X A B, , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 22 :
202, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , 10. Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is, [AIPMT (Prelims)-2012], , D1, , 10 , , D2, , 20 , , 5V, (1) 0.25 A, , (2) 0.5 A, , (3) 0.75 A, , (4) Zero, , Sol. Answer (2), Current =, , V, 5, =, = 0.5 A, R 10, , A2 is reverse biased and it is assumed in this questions that current does not pass through it., 11. The figure shows a logic circuit with two inputs A and B and the output C. The voltage wave forms across A, B, and C are given. The logic circuit gate is, [AIPMT (Prelims)-2012], , A, B, C, 0, (1) AND gate, , (2) NAND gate, , t1 t2 t3 t4 t5 t6, (3) OR gate, , (4) NOR gate, , Sol. Answer (3), Using the truth table., 12. In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k is 2 V. If the base, resistance is 1 k and the current amplification of the transistor is 100, the input signal voltage is, [AIPMT (Prelims)-2012], (1) 1 mV, , (2) 10 mV, , (3) 0.1 V, , (4) 1.0 V, , Sol. Answer (2), RC = 2 k, vCE = 2V = V0, , = 100, , Rb = 1 k, Av = β, , Av =, , Rc 100 × 2 × 103, =, = 200, Rb, 1 × 103, , V0, 2, 2, =, = 200 ⇒ Vi =, = 10 mV, 200, V1 Vi, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 23 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 203, , 13. C and Si both have same lattice structure, having 4 bonding electrons in each. However, C is insulator whereas, Si is intrinsic semiconductor. This is because, [AIPMT (Prelims)-2012], (1) The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in, the third., (2) The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit., (3) In case of C the valence band is not completely filled at absolute zero temperature., (4) In case of C the conduction bans is partly filled even at absolute zero temperature, Sol. Answer (1), Fact., 14. Transfer characteristics [output voltage (V0) vs input voltage (Vi)] for a base biased transistor in CE configuration, is as shown in the figure. For using transistor as a switch, it is used, [AIPMT (Prelims)-2012], , V0, , I, , II, , III, , Vi, (1) In region II, , (2) In region I, , (3) In region III, , (4) Both in region (I) and (III), , Sol. Answer (4), , V0, , I, , II, , III, , Vi, To use it as a switch it must be used cut off and in saturation region which are I and III., 15. The input resistance of a silicon transistor is 100 . Base current is changed by 40 A, which results in a, change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load, resistance of 4 k. The voltage gain of the amplifier is, [AIPMT (Mains)-2012], (1) 2000, , (2) 3000, , (3) 4000, , (4) 1000, , Sol. Answer (1), ri = 100 , , β=, , RL = 4 k, , Av = β, , RL, Ri, , IB = 40 A, , ∆IC, 2 × 10−3, 2000, =, =, = 50, −, 6, ∆IB 40 × 10, 40, , Av =, , 50 × 4 × 103, 102, , = 2000, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 24 :
204, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , 16. To get an output Y = 1 in given circuit which of the following input will be correct?, [AIPMT (Mains)-2012], A, B, B, , A, , B, , C, , (1) 1, , 0, , 0, , (2) 1, , 0, , 1, , (3) 1, , 1, , 0, , (4) 0, , 1, , 0, , Sol. Answer (2), y = (A + B)· C, C should be 1, Either of A and B OR both A and B are 1., 17. A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current, from 100 A to 300 A produces a change in the collector current from 10 mA to 20 mA. The current gain is, [AIPMT (Prelims)-2011], (1) 25, , (2) 50, , (3) 75, , (4) 100, , Sol. Answer (2), vc = 2 V, , β=, , IB = (300 – 100) A = 200 A Ic = (20 – 10) mA = 10 mA, , ∆Ic, 10 × 10−3, 10 × 10−3 100, =, =, =, = 50, ∆Ib 200 × 10−6, 200, 2, , 18. If a small amount of antimony is added to germanium crystal, , [AIPMT (Prelims)-2011], , (1) Its resistance is increased, (2) It becomes a p-type semiconductor, (3) The antimony becomes an acceptor atom, (4) There will be more free electrons than holes in the semiconductor, Sol. Answer (4), Antimony in pentavalent atom., 19. In forward biasing of the p-n junction, , [AIPMT (Prelims)-2011], , (1) The positive terminal of the battery is connected to p-side and the depletion region becomes thin, (2) The positive terminal of the battery is connected to p-side and the depletion region becomes thick, (3) The positive terminal of the battery is connected to n-side and the depletion region becomes thin, (4) The positive terminal of the battery is connected to n-side and the depletion region becomes thick, Sol. Answer (1), Fact, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 25 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 205, , 20. Symbolic representation of four logic gates are shown as, (i), , (ii), , (iii), , (iv), , Pick out which ones are for AND, NAND and NOT gates, respectively, (1) (ii), (iv) and (iii), , (2) (ii), (iii) and (iv), , [AIPMT (Prelims)-2011], , (3) (iii), (ii) and (i), , (4) (iii), (ii) and (iv), , Sol. Answer (1), Fact., 21. A zener diode, having breakdown voltage equal to 15 V is used in a voltage regulator circuit shown in figure., The current through the diode is, [AIPMT (Mains)-2011], , +, , 250 , , 20 V, , 15 V, , 1 k, , –, (1) 20 mA, , (2) 5 mA, , (3) 10 mA, , (4) 15 mA, , Sol. Answer (2), Potential through zener diode is maintained 15 V., Current through 1000 =, Current through 250 =, , +, , 15, = 0.015 A, 1000, , 250 , , 20 V, , 15 V, , 1 k, , –, , 5, 1, =, = 0.02 A, 250, 50, , Current through diode = 0.02 – 0.015 = 0.005 A, 22. In the following figure, the diodes which are forward biased are, , +10 V, (a), , R, (b), , R, , [AIPMT (Mains)-2011], , –10 V, , +5 V, –12 V, , R, , (c), , (d), , +5 V, , –5 V, (1) (b) and (d), , (2) (a), (b) and (d), , R, , (3) (c) only, , (4) (c) and (a), , Sol. Answer (4), P should be at higher voltage than N., 23. Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 × 1016 m–3. Doping, by indium increases nh to 4.5 × 1022 m–3. The doped semiconductor is of :, [AIPMT (Mains)-2011], (1) n-type with electron concentration ne = 2.5 × 1023 m–3, (2) p-type having electron concentrations ne = 5 × 109 m–3, (3) n-type with electron concentration ne = 2.5 × 1022 m–3, (4) p-type with electron concentration ne = 2.5 × 1010 m–3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 26 :
206, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (2), , (1.5 1016 )2, ne 5 109 m–3, 4.5 1022, 24. A common emitter amplifier has a voltage gain of 50, an input impedance of 100 and an output impedance, of 200 . The power gain of the amplifier is, [AIPMT (Prelims)-2010], (1) 50, , (2) 500, , (3) 1000, , R1 = 100 , , R2 = 200 , , (4) 1250, , Sol. Answer (4), Av = 50, R2, Av = β R, 1, 2 R1, ⎛, R ⎞, Power gain = Av Av ⎜ Av 1 ⎟ AV R, 2, ⎝ R2 ⎠, , 25. To get an output Y = 1 from the circuit shown below, the input must be:, , A, B, C, A, , B, , C, , (1) 0, , 1, , 0, , (2) 0, , 0, , 1, , (3) 1, , 0, , 1, , (4) 1, , 0, , 0, , [AIPMT (Prelims)-2010], , Y, , Sol. Answer (3), To get output y = 1 C must be 1 and A + B must be 1. Hence answer is option (2)., 26. The device that can act as a complete electronic circuit is:, (1) Zener diode, , (2) Junction diode, , (3) Integrated circuit, , [AIPMT (Prelims)-2010], (4) Junction transistor, , Sol. Answer (3), 27. Which of the following statement is False?, , [AIPMT (Prelims)-2010], , (1) The resistance of intrinisic semiconductor decreases with increase of temperature, (2) Pure Si doped with trivalent impurities gives a p-type semiconductor, (3) Majority carriers in a n-type semiconductors are holes, (4) Minority carriers in a p-type semiconductor are electrons, Sol. Answer (3), Majority carries in N-type semiconductor as electron. Hence option (3) is correct., 28. Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical, conductivity decreases with temperature and has high melting point?, [AIPMT (Prelims)-2010], (1) Covalent bonding, , (2) Metallic bonding, , (3) Van der Waal's bonding, , (4) Ionic bonding, , Sol. Answer (2), The description is of a metal and is produced a special bond called metallic bond., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 27 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 207, , 29. The following figure shows a logic gate circuit with two inputs A and B and the output Y. The voltage waveforms, of A, B and Y are as given, , A, B, A, , Logic gate, circuit, , Y, , 1, 0, , B, , 1, 0, 1, , Y, , 0, , t1, , t2, , t3, , t4, , t5, , t6, , The logic gate is, (1) NOR gate, , [AIPMT (Mains)-2010], (2) OR gate, , (3) AND gate, , (4) NAND gate, , Sol. Answer (4), , A, , B, , Y, , 1, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , 0, , 1, , 30. For transistor action, (a) Base, emitter and collector regions should have similar size and droping concentrations., (b) The base region must be very thin and lightly doped., (c) The emitter-base junction is forward biased and base-collector junction is reverse biased., (d) Both the emitter-base junction as well as the base collector junction are forward biased., Which one of the following pairs of statement is correct?, (1) (d), (a), , (2) (a), (b), , (3) (b), (c), , [AIPMT (Mains)-2010], (4) (c), (d), , Sol. Answer (3), 31. The symbolic representation of four logic gates are given below:, (i), , (ii), , (iii), , (iv), , The logic symbols for OR, NOT and NAND gates are respectively, (1) (iv), (i), (iii), , (2) (iv), (ii), (i), , (3) (i), (iii), (iv), , [AIPMT (Prelims)-2009], (4) (iii), (iv), (ii), , Sol. Answer (2), 32. A p-n photodiode is fabricaed from a semiconductor with a band gap of 2.5 eV. It can detect a signal of, wavelength:, [AIPMT (Prelims)-2009], (1) 4000 nm, , (2) 6000 nm, , (3) 4000 Å, , (4) 6000 Å, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 28 :
208, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (3), , hc, = 2.5 eV, λ, λmax =, , 6.6 × 10−34 × 3 × 108, 2.5 × 1.6 × 10−19, , = 4.95 × 10–7 = 4.950 × 10–10 = 4950 Å, , Hence is less than max., 33. A transistor is operated in common-emitter configuration at Vc = 2 V such that a change in the base current, from 100 A to 200 A produces a change in the collector current from 5 mA to 10 mA. The current gain is:, [AIPMT (Prelims)-2009], (1) 100, , (2) 150, , (3) 50, , (4) 75, , Sol. Answer (3), 34. Sodium has body centred packing. Distance between two nearest atoms is 3.7 Å. The lattice parameter is, [AIPMT (Prelims)-2009], (1) 4.3 Å, , (2) 3.0 Å, , (3) 8.6 Å, , (4) 6.8 Å, , Sol. Answer (1), 35. If the lattice parameter for a crystalline structure is 3.6 Å, then the atomic radius in fcc crystal is, [AIPMT (Prelims)-2008], (1) 1.27 Å, , (2) 1.81 Å, , (3) 2.10 Å, , (4) 2.92 Å, , Sol. Answer (1), 36. The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be, [AIPMT (Prelims)-2008], (1) 100, , (2) 90, , (3) 10, , (4) 1.25, , Sol. Answer (1), Afb=, , A, 1 mA, , Afb= Voltage gain with feed back = 10, A = Voltage gain without feed back, m=, , 10 , , A, 9A, 1, 100, 9, A =A, 10, , A = 100, NOR, , NOT, , 37. The circuit, (1) OR gate, , is equivalent to, (2) AND gate, , [AIPMT (Prelims)-2008], , (3) NAND gate, , (4) NOR gate, , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 29 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 209, , 38. A p-n photodiode is made of a material with a band gap of 2 eV. The minimum frequency of the radiation that, can be absorbed by the material is nearly, [AIPMT (Prelims)-2008], (1) 20 × 1014 Hz, , (2) 10 × 1014 Hz, , (3) 5 × 1014 Hz, , (4) 1 × 1014 Hz, , Sol. Answer (3), 2 × 1.6 × 10–19 = hf, f=, , 2 × 1.6 × 10−19, 6.6 × 10, , −34, , = 4.828 × 1014 Hz 5 × 1014 Hz, , 39. In the energy band diagram of a material shown below, the open circles and filled circles denote holes and, electrons respectively. The material is, [AIPMT (Prelims)-2007], Ei, , Eg, , Ev, (1) An n-type semiconductor, , (2) A p-type semiconductor, , (3) An insulator, , (4) A metal, , Sol. Answer (2), Holes are more in numbers., 40. A common emitter amplifier has a voltage gain of 50, an input impedance of 100 and an output impedance, of 200 . The power gain of the amplifier is, [AIPMT (Prelims)-2007], (1) 100, , (2) 500, , (3) 1000, , (4) 1250, , Sol. Answer (4), 41. In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table, [AIPMT (Prelims)-2007], A, , A, , B, , B, , Y, , (1) A, , B, , Y, , (2) A, , B, , Y, , 0, , 1, , 1, , 0, , 0, , 1, , 0, , 1, , 1, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 0, , 1, , 1, , 0, , (3) A, , B, , Y, , (4) A, , B, , Y, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 1, , 1, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 1, , 1, , 1, , 1, , Sol. Answer (3), A, , A+B, , A+B=A+B, , B, , Hence it is an OR gate., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 30 :
210, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , 42. For a cubic crystal structure which one of the following relations indicating the cell characteristics is correct?, [AIPMT (Prelims)-2007], (1) a = b = c and = = = 90°, , (2) a b c and 90°, , (3) a b c and = = = 90°, , (4) a = b = c and = 90°, , Sol. Answer (1), 43. A transistor is operated in common emitter configuration at constant collector voltage Vc = 1.5 V such that a, change in the base current from 100 A to 150 A produces a change in the collector current from 5 mA to, 10 mA. The current gain (â) is:, [AIPMT (Prelims)-2006], (1) 67, , (2) 75, , (3) 100, , (4) 50, , Sol. Answer (3), 44. A forward biased diode is :, , [AIPMT (Prelims)-2006], , (1) –4 V, , –3 V, , (2) 3 V, , 5V, , (3) –2 V, , +2 V, , (4) 0 V, , –2 V, , Sol. Answer (4), Higher to lower potential is achieved in forward bias in option (1) only., 45. The following figure shows a logic gate circuit with two inputs A and B and the output C. The voltage waveforms, of A, B and C are as shown below, , 1, A, 1, , t, , B, 1, , t, , C, , t, , The logic circuit gate is, (1) AND gate, , [AIPMT (Prelims)-2006], (2) NAND gate, , (3) NOR gate, , (4) OR gate, , Sol. Answer (1), 46. Application of a forward bias to a p-n junction:, , [AIPMT (Prelims)-2005], , (1) Increases the number of donors on the n-side, (2) Increases the electric field in the depletion zone, (3) Increases the potential difference across the depletion zone, (4) Widens the depletion zone, Sol. Answer (1), Increases the diffusion current., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 31 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 47. Zener diode is used for, , 211, , [AIPMT (Prelims)-2005], , (1) Producing oscillations in an oscillator, , (2) Amplification, , (3) Stabilisation, , (4) Rectification, , Sol. Answer (3), Zener diode is used as a voltage stabiliser., 48. Carbon, silicon and germanium atoms have four valence electrons each. Their valence and conduction bands, are separated by energy band gaps represented by (Eg)C, (Eg)Si and (Eg)Ge respectively. Which one of the, following relationships is true in their case?, [AIPMT (Prelims)-2005], (1) (Eg)C > (Eg)Si, , (2) (Eg)C = (Eg)Si, , (3) (Eg)C < (Eg)Ge, , (4) (Eg)C < (Eg)Si, , Sol. Answer (1), Carbon is a dielectric and has higher forbidden energy gap., 49. Choose the only false statement from the following:, , [AIPMT (Prelims)-2005], , (1) Substances with energy gap of the order of 10 eV are insulators, (2) The conductivity of a semiconductor increases with increase in temperature, (3) In conductors the valence and conduction bands may overlap, (4) The resistivity of a semiconductor increases with increase in temperature, Sol. Answer (4), Resistivity of semiconductor decreases with increase in temperature, due to availability of more carrier particles., 50. Copper has face-centered cubic (fcc) lattice with interatomic spacing equal to 2.54 Å. The value of lattice, constant for this lattice is, [AIPMT (Prelims)-2005], (1) 1.27 Å, , (2) 5.08 Å, , (3) 2.54 Å, , (4) 3.59 Å, , Sol. Answer (4), 51. The cations and anions are arranged in alternate form in, (1) Metallic crystal, , (2) Ionic crystal, , (3) Covalent crystal, , (4) Semi-conductor crystal, , Sol. Answer (2), Ionic crystal consisting the cation and anion as basic constituents, 52. In semiconductors at room temperature, (1) The valence band is partially empty and the conduction band is partially filled, (2) The valence band is completely filled and the conduction band is partially filled, (3) The valence band is completely filled, (4) The conduction band is completely empty, Sol. Answer (1), Due to excitation energy provided by room temperature., 53. In good conductors of electricity, the type of bonding that exists is, (1) Metallic, , (2) van der Waal’s, , (3) Ionic, , (4) Covalent, , Sol. Answer (1), Good conductor and conduct electricity both due to their property of metallic bonds., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 32 :
212, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , 54. In an insulator, the forbidden energy gap between the valence band and conduction band is of the order of, (1) 1 MeV, , (2) 0.1 MeV, , (3) 1 eV, , (4) 5 eV, , Sol. Answer (4), In an insulator there is a high energy gap of the order., 55. A p-n junction has a depletion layer thickness of the order of, (1) 10–10 m, , (2) 10–8 m, , (3) 10–6 m, , (4) 10–4 m, , Sol. Answer (3), 56. Pure Si at 300 K has hole and electron densities are 1.5 × 1016 m–3. Doping it by an impurity increases the, hole density nh to 4.5 × 1022 m–3. Electron density in the doped silicon is, (1) 1.5 × 1016 m–3, , (2) 3.0 × 1022 m–3, , (3) 5 × 109 m–3, , (4) 3 × 106 m–3, , Sol. Answer (3), ni2 = ne ⋅ nn, (1.5 × 1016 )2 =, , 2.25 × 1032, 4.5 × 10, , 22, , = 5 × 109 m−3, , 57. In a reverse-biased p-n junction, when the applied bias voltage is equal to the breakdown voltage, then, (1) Current remains constant while voltage increase sharply, (2) Voltage remains constant while current increases sharply, (3) Current and voltage increase, (4) Current and voltage decrease, Sol. Answer (2), 58. In the case of forward biased of p-n junction, which one of the following figures correctly depicts the direction, of flow of carriers?, , Vb, p, , Vb, n, , p, , n, , (2), , (1), , VF, , Vb, , Vb, p, , n, , (3), , VF, , p, , n, , (4), , VF, , VF, , Sol. Answer (2), Holes will move in the direction of current in the P region and electrons will move in the direction opposite, to current., 59. When arsenic is added as an impurity to silicon, the resulting material is, (1) n-type conductor, , (2) n-type semiconductor, , (3) p-type semiconductor, , (4) p-type conductor, , Sol. Answer (2), Arsenic is pentavalent atom., 60. To obtain a p-type germanium semiconductor, it must be doped with, (1) Indium, , (2) Phosphorus, , (3) Arsenic, , (4) Antimony, , Sol. Answer (1), Indium is an acceptor doping agent and it increases number of holes., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 33 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 213, , 61. The cause of the potential barrier in a p-n diode is, (1) Depletion of negative charges near the junction, (2) Concentration of positive charges near the junction, (3) Depletion of positive charges near the junction, (4) Concentration of positive and negative charges near the junction, Sol. Answer (4), The depletion layer causes the potential barriers., 62. A semi-conducting device is connected in a series circuit with a battery and a resistance. A current is found, to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device, may be, (1) A p-type semi-conductor, , (2) An intrinsic semi-conductor, , (3) A p-n junction, , (4) An n-type semi-conductor, , Sol. Answer (3), The P–N junction initially in forward bias and finally in reverse bias is described., 63. A p-n junction diode can be used as, (1) Condenser, , (2) Oscillator, , (3) Amplifier, , (4) Rectifier, , Sol. Answer (4), 64. In a p-type semiconductor, the majority carriers of current are, (1) Protons, , (2) Electrons, , (3) Holes, , (4) Neutrons, , Sol. Answer (3), 65. In forward bias, the width of potential barrier in a p-n junction diode, (1) Remains constant, , (2) Decreases, , (3) Increases, , (4) First (1) then (2), , (2) Protons, , (3) Electrons, , (4) Immobile ions, , (3) Protons, , (4) Missing of electrons, , Sol. Answer (2), 66. Depletion layer consists of, (1) Mobile ions, Sol. Answer (4), 67. In a junction diode, the holes are due to, (1) Extra electrons, , (2) Neutrons, , Sol. Answer (4), Holes are created due to electron vacancy., 68. In a PN junction, (1) High potential at N side and low potential at P side, (2) High potential P side and low potential at N side, (3) P and N both are at same potential, (4) Undetermined, Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 34 :
214, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , 69. For the given circuit of ideal P-N junction diode which is correct, Diode, , V, (1) In forward bias the voltage across R is V, , (2) In reverse bias the voltage across R is V, , (3) In forward bias the voltage across R is 2V, , (4) In reverse bias the voltage across R is 2V, , Sol. Answer (1), In forward bias P – N situation is thaught to have very low or negligible resistance. Hence entire voltage drop, is across resistor., 70. Reverse bias applied to a junction diode, (1) Lowers the potential barrier, , (2) Raises the potential barrier, , (3) Increases the majority carrier current, , (4) Increases the minority carrier current, , Sol. Answer (2), Reverse bias increases depletion layer and hence the potential barrier., 71. Barrier potential of a p-n junction diode does not depend on, (1) Diode design, , (2) Temperature, , (3) Forward bias, , (4) Doping density, , Sol. Answer (1), Barrier potential does not depend on diode design but the other three factors in options., 72. In a p-n junction photo cell, the value of the photo-electromotive force produced by monochromatic light is, proportional to, (1) The barrier voltage at the p-n junction, , (2) The intensity of the light falling on the cell, , (3) The frequency of the light falling on the cell, , (4) The voltage applied at the p-n junction, , Sol. Answer (2), Photo emf depends on intensity., 73. In a zener diode, break down occurs in reverse bias due to, (1) Impact ionisation, , (2) Internal field emission, , (3) High doping concentration, , (4) All of these, , Sol. Answer (2), 74. In a p-n junction, depletion region contains, (1) No charges at all, , (2) Equal number of conduction electrons and holes, , (3) Equal number of donor and acceptor ions, , (4) More conduction holes than electrons, , Sol. Answer (3), Answer is option (3) in order to maintain charge neutrality., 75. In the given circuit, the value of current is, , 30, , 4V, Si, (1) 1 ampere, , (2) 0.1 ampere, , 0.3 V, (3) 0.5 ampere, , (4) Zero, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 35 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 215, , Sol. Answer (2), Voltage drop across silicon P – N junction is forward bias = 0.7 V, , or, , I=, , ∆V, R, , I=, , 4 − 0.7 − 0.3, = 0.1 A, 30, , 76. The diode used in the circuit shown in the figure has a constant voltage drop at 0.5 V at all currents and a, maximum power rating of 100 milli watt. What should be the value of the resistor R, connected in series and, with diode for obtaining maximum current?, , R, , 0.5 V, , 1.5 V, (1) 6.76 , , (2) 20 , , (3) 5 , , (4) 5.6 , , Sol. Answer (3), Max curent through diode, 0.1 watt, i = 0.5 watt .2A, 1 Volt, R = .2 A 5 , , 77. The current in the circuit will be, , 20 , D2, 20 , (1), , 5, A, 40, , (2), , 5, A, 50, , D1, 30 , 5V, , (3), , 5, A, 10, , (4), , 5, A, 20, , Sol. Answer (2), , 20 , D2, 20 , , D1, 30 , 5V, , D1 is reverse biased, no current flows through it., D2 is forward biased so current flows with minimum resistance., i=, , 5, A, 50, , 78. If a full wave rectifier circuit is operating from 50 Hz mains, the fundamental frequency in the ripple will be, (1) 25 Hz, , (2) 50 Hz, , (3) 70.7 Hz, , (4) 100 Hz, , Sol. Answer (4), Frequency of full wave rectifier = 2 × Frequency of mains = 100 Hz., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 36 :
216, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , 79. The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10 V., The D.C. component of the output voltage is, , 10, V, 2, , (1), , (2), , 10, V, , , (3) 10 V, , (4), , 20, V, , , Sol. Answer (2), D.C. component of output voltage =, , Peak voltage, π, , 80. When NPN transistor is used as an amplifier, then, (1) Electrons move from collector to base, , (2) Holes move from collector to base, , (3) Holes move from base to collector, , (4) Electrons move from emitter to base, , Sol. Answer (4), Emitter base junction is forward bias., 81. An oscillator is nothing but an amplifier with, (1) Positive feedback, , (2) Negative feedback, , (3) Voltage gain, , (4) No feedback, , Sol. Answer (1), 82. The correct relationship between the two current gains and in a transistor is, (1) , , , 1 , , (2) , , 1 , , , (3) , , , 1 , , (4) , , , 1 , , Sol. Answer (4), 83. The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter, configuration is 1 k. The peak value of the collector A.C. current for an A.C. input voltage of 0.01 V peak is, (1) 0.25 mA, , (2) 0.01 mA, , (3) 100 A, , (4) 500 A, , Sol. Answer (4), , IC, Vi, = I , IB R, B, i, , =, , IC Ri, Vi, , Vi, Ic = R, i, , , , 50(.01), 103, , = 500 A, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 37 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 217, , IC, 84. For a common emitter circuit if I = 0.98 then current gain for common emitter circuit will be, E, , (1) 49 × 10–2, , (2) 98, , (3) 4.9 × 101, , (4) 25.5, , Sol. Answer (3), IC, = α = 0.98, IE, β=, , =?, , α, 0.98, =, = 49 = 4.9 × 10, 1 − α 0.02, , 85. A n-p-n transistor conducts in active mode when, (1) Both collector and emitter are positive with respect to the base, (2) Collector is positive and emitter is negative with respect to the base, (3) Collector is positive and emitter is at same potential as the base, (4) Both collector and emitter are negative with respect to the base, Sol. Answer (2), 86. A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C, in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be, (1), , f, 2, , f, 4, , (2), , (3) 8f, , (4), , f, 2 2, , Sol. Answer (4), At resonance condition f =, , f1 =, , f1 =, , 1, 2π 8LC, , =, , 1, 4 2π LC, , 1, 2π LC, , =, , f, 2 2, , f, 2 2, , 87. In an n-p-n transistor working in active mode, the depletion region, (1) Is not formed, (2) At emitter-base junction is wider than that at collector-base junction, (3) At emitter-base junction is thinner than that at collector-base junction, (4) At the two junctions have equals width, Sol. Answer (3), 88. A transistor has a current amplification factor of 60. In a CE amplifier, input resistance is 1 k and output, voltage is 0.01 V. The transconductance is (in SI units), (1) 10–5, , (2) 6 × 10–2, , (3) 6 × 104, , (4) 10, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 38 :
218, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (2), = 60, , R1 = 1 k, , V0 = 0.01 V, , gm = ?, , , 60, 2, gm = R 3 6 10, 10, i, , 89. Which of the conditions must be satisfied to operate a transistor amplifier?, (1) Emitter-base and collector-base junctions are forward biased, (2) Emitter-base is forward biased junctions collector-base is reverse biased, (3) Emitter-base and collector-base junctions both are reverse biased, (4) Emitter-base is reverse biased collector-base is forward biased, Sol. Answer (2), 90. In a common emitter configuration base current is 40 A and current gain is 100. The collector current will, be, (1) 4 mA, , (2) 4 A, , (3) 1 mA, , (4) 1 A, , Sol. Answer (1), IB = 40 A = 40 × 10–6 A, , = 100, , IC = ?, , IC = IB, 100 × 40 × 10–6 = 4 × 10–3 = 4 mA, 91. Which one of the following gates correspond to the truth table given below?, A B Y, 1 1 0, 1, 0, 0, , (1) NAND, , 0, 1, 0, , (2) NOR, , 1, 1, 1, , (3) XOR, , (4) OR, , Sol. Answer (1), It is opposite of truth table of and gate hence it is a nand gate., 92. A truth table is given. Which of the following has this type of truth table?, A B Y, 0 0 1, 1, 0, 1, , (1) AND gate, , (2) OR gate, , 0, 1, 1, , 0, 0, 0, , (3) XOR gate, , (4) NOR gate, , Sol. Answer (4), It is opposite of truth table of an OR gate hence it is or NOR gate., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 39 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 219, , 93. This symbol represents, , A, B, (1) AND gate, , y, , (2) NOR gate, , (3) NAND gate, , (4) OR gate, , (3) 1, 1, , 0, (4) 1, , Sol. Answer (3), 94. Which of the following gates will have an output of 1?, , 0, (2) 1, , 0, (1) 1, , Sol. Answer (4), It is a XOR gate, 95. Following diagram performs the logic function of, A, B, , (1) AND gate, , (2) NAND gate, , Y, , (3) OR gate, , (4) XOR gate, , Sol. Answer (1), , A, , A B, , A B = A.B, , B, Hence it is an AND gate., 96. The output of OR gate is 1, (1) If both inputs are zero, , (2) If either or both inputs are 1, , (3) Only if both inputs are 1, , (4) If either input is zero, , Sol. Answer (2), 97. In the figure shown if A = 1 and B = 0 then y will be, , A, y, B, , (1) 0, , (2) 1, , (3) 2, , (4) Any of these, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 40 :
220, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , Sol. Answer (2), , A B, , A, , A B + B, B, , B, , X = 0.1 + 0, , [A = 1 B = 0], , X=1, 98. Symbol given below represents, , A, , y, , B, (1) AND, , (2) OR, , (3) NAND, , (4) NOR, , Sol. Answer (2), , A, B, , A, B, , y, , y = A•B, , y = A B AB, 99. The given symbol represents the, , B, A, (1) AND gate, , (2) NAND gate, , Y, , (3) NOR gate, , (4) XOR gate, , Sol. Answer (4), 100. Output Y of the gate shown in figure, , A, B, , (1) Y = AB, , (2) Y = A + B, , Y, , (3) Y = A B, , (4) Y = A B, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 41 :
Solution of Assignment, , Semiconductor Electronics : Materials, Devices and Simple Circuits, , 221, , Sol. Answer (2), , A, B, , A, B, , A B=A+B=A+B, Y, , Y=A+B, 101. Which of the following is the boolean expression for XOR gate?, (1) y AB AB, , (2) y ( AB )i ( AB ), , (3) y ( A B )i ( A B ) (4) y ( AB ) ( AB ), , Sol. Answer (1), , SECTION - D, Assertion-Reason Type Questions, 1., , A : The conductivity of an intrinsic semiconductor at zero kelvin is zero., R : The bond strength of the semiconductor at zero kelvin is much higher as compared to the bond strength, at room temperature., , Sol. Answer (3), 2., , A : When base region has larger width, the collector current is small., R : At larger width of the base region the rate of electron-hole recombination is more which results in larger, value of base current., , Sol. Answer (1), 3., , A : The conductivity of a pure semiconductor increases on doping., R : Doping causes the reduction in bond strength., , Sol. Answer (3), 4., , A : Semiconductors do not obey Ohm’s law., R : In semiconductors the rate of flow of charge not only depends on the applied electric field but also on the, availability of charge carriers., , Sol. Answer (1), 5., , A : When a pure semiconductor is doped with a pentavalent impurity, the number of conduction electrons is, increased while the number of holes is decreased., R : Some of the holes get recombined with the conduction electrons as the concentration of the conduction, electrons is increased., , Sol. Answer (1), 6., , A : In transistor common emitter mode as an amplifier is preferred over common base mode., R : In common emitter mode, the input signal is connect in series with the voltage applied to the base emitter, junction., , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 42 :
222, 7., , Semiconductor Electronics : Materials, Devices and Simple Circuits, , Solution of Assignment, , A : The energy gap between the valence band and conduction band is greater in silicon than in germanium., R : Thermal energy produces fewer minority carriers in silicon than in germanium., , Sol. Answer (2), 8., , A : Light emitting diode (LED) emits spontaneous radiation., R : LED are forward biased p-n junctions., , Sol. Answer (1), The flow of current through forward wired P – N junction career radial., 9., , A : NAND or NOR gates are called digital building blocks., R : The repeated use of NAND (or NOR) gates can produce all the basic or complicated gates., , Sol. Answer (1), 10. A : In NPN transistors, electrons are current carriers inside as well as outside the transistor circuits., R : In PNP transistors, holes are responsible for current inside the transistors but outside the transistor,, electrons are the current carriers., Sol. Answer (2), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 43 :
Chapter, , 28, , Nuclei, Solutions, SECTION - A, Objective Type Questions, 1., , The nuclear radius as compared to the atomic radius is of the order, (1) 10–3, , (2) 10–5, , (3) 10–7, , (4) 10–9, , Sol. Answer (2), Fact., 2., , Two nuclei which are not identical but have the same number of nucleons represent, (1) Isotones, , (2) Isobars, , (3) Isotopes, , (4) Isotones, , Sol. Answer (2), Definition based. Fact., 3., , In a nuclear fusion reaction, if the energy is released then, (1) BEproducts = BEreactants, , (2) BEreactants > BEproducts, , (3) BEproducts > BEreactants, , (4) Mass of product > Mass of reactant, , Sol. Answer (3), Product is more stable than reactant and hence has more binding energy., 4., , The binding energy per nucleon for a 6C12 nucleus is, (Nuclear mass of 6C12 = 12.00000 a.m.u., Mass of hydrogen nucleus = 1.007825 a.m.u, Mass of neutron = 1.008665 a.m.u), (1) 2.675 MeV, , (2) 7.675 MeV, , (3) 0 MeV, , (4) 3.675 MeV, , Sol. Answer (2), m = 6mP + (12 – 6)mn – mN, mP = mass of proton, mn = mass of neutron, mN = mass of nucleus., or m = 6 × 1.007825 + 6(1.008665) – 12, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 44 :
148, 5., , Nuclei, , Solution of Assignment, , Which of the following pairs of particles cannot exert nuclear force on each other?, (1) Proton and electron, , (2) Neutron and electron, , (3) Electron and neutron (4) All of these, , Sol. Answer (4), Answer is all of these as electrons are not affected by nuclear forces at all., 6., , When two nuclei of mass X and Y respectively fuse to form a nucleus of mass m alongwith the liberation of, some energy, then, (1) X + Y > m, , (2) X – Y = m, , (3) X + Y = m, , (4) X + Y < m, , Sol. Answer (1), Mass reduces and is converted into energy in fusion reactions., 7., , The nuclear force between two nucleons is explained by, (1) Quark exchange theory, , (2) Meson exchange theory, , (3) Photon exchange theory, , (4) Gravitation exchange theory, , Sol. Answer (2), Fact., 8., , If Fpp , Fpn and Fnn are the magnitudes of nuclear force between proton-proton, proton-neutron and neutron-neutron, respectively, then, (1) Fpp = Fpn = Fnn, , (2) Fpp < Fpn = Fnn, , (3) Fpp > Fpn > Fnn, , (4) Fpp < Fpn < Fnn, , Sol. Answer (1), Nuclear force between all nucleons is the same., 9., , The atomic mass of 7N15 is 15.000108 a.m.u. and that of 8O16 is 15.994915 a.m.u. If the mass of a proton is, 1.007825 a.m.u. then the minimum energy provided to remove the least tightly bound proton is, (1) 0.013018 MeV, , (2) 12.13 MeV, , (3) 13.018 MeV, , (4) 12.13 eV, , Sol. Answer (2), Energy + 8O16 7N15 + 1H1, Energy = [(MN + MH) – M0] c2 = [(15.000108+1.007825) – 15.994915] × 931.5 MeV, 10. Nuclear energy is released in fusion reaction, since binding energy per nucleon is, (1) Smaller of fusion products than for fusing nuclei, , (2) Same for fusion products as for fusing nuclei, , (3) Larger for fusion products than for fusing nuclei, , (4) Sometimes larger and sometimes smaller, , Sol. Answer (3), Since binding energy is larger for products than reactants., , 11. A nucleus X undergoes following transformation, , α, X , Y, , Y Z, , (1) X and Y are isotopes (2) X and Z are isobars, , 2β, , then, , (3) X and Y are isobars (4) X and Z are isotopes, , Sol. Answer (4), α, , x ⎯⎯→ y, , , y = xzm−−24, 2β, , y ⎯⎯→ z, , , z = xzm−−24+ 2, , Hence, x and z are isotopes., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 45 :
Solution of Assignment, , Nuclei, , 149, , 12. Consider the nuclear reaction X200 A110 + B90, If the binding energy per nucleon for X, A and B is 7.4 MeV, 8.2 MeV and 8.2 MeV respectively, then the, amount of the energy released is, (1) 200 MeV, , (2) 160 MeV, , (3) 110 MeV, , (4) 90 MeV, , (3) Isotopes of carbon, , (4) Isotopes of nitrogen, , (3) 2.750, , (4) 3.375, , Sol. Answer (2), Initial BE = 7.4 × 200 MeV, Final BE = 8.2 × 110 + 8.2 × 90 MeV, Energy release = Final – Initial, = (8.2 – 7.4) × 200 = 0.8 × 200, = 160 MeV, 13. The nuclei 6A13 and 7B14 can be described as, (1) Isotones, , (2) Isobars, , Sol. Answer (1), Number of neutrons in A = 13 – 6 = 7, Number of neutrons in B = 14 – 7 = 7, Hence, A and B are isotones., 14. Ratio of nuclear radii of, (1) 1.40, , 135, , Cs to 40Ca is, (2) 1.50, , Sol. Answer (2), r A1/3, 1/3, , rCs ⎛ 135 ⎞, =⎜, ⎟, rCa ⎝ 40 ⎠, , 1/3, , rCs ⎛ 27 ⎞, =⎜ ⎟, rCa ⎝ 8 ⎠, rCs 3, =, rCa 2, , 15. A nucleus with Z = 92 emits the following in a sequence ,–,–, , , , , , –, –, +, , +, . The Z, of the resulting nucleus is, (1) 74, , (2) 76, , (3) 78, , (4) 82, , Sol. Answer (3), Z = 92, If it goes through 8 alpha decays and two + decays, Hence, net decrease = 8 × 2 + 2 × 1, = 18 protons, Net increase is due to – decays = 4 × 1, Hence, final Z = 92 – 18 + 4 = 78, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 46 :
150, , Nuclei, , Solution of Assignment, , 16. In nuclear reactions we have the conservation of, (1) Mass only, , (2) Energy only, , (3) Momentum only, , (4) Charge, total energy and momentum, , Sol. Answer (4), Fact., , N, because, P, (1) Electrostatic repulsion dominate over nuclear attraction, , 17. A heavy nucleus is unstable for any value of, , (2) Nuclear repulsion dominate over nuclear attraction, (3) Nuclear forces are absent in heavy nucleus, (4) Nuclear force is long range force, Sol. Answer (1), In heavy nuclei repulsion between the lots of protons in the nucleus makes the nucleus unstable., 18. When 90Th228 gets converted into 83Bi212, then the number of - and -particles emitted will respectively be, (1) 4, 7, , (2) 4, 1, , (3) 8, 7, , (4) 4, 4, , Sol. Answer (2), Initial nucleus = 90Th228, Mass reduces by = 228 – 212, = 16 nucleons, 16, = 4 particles, 4, This results in atomic number reduction by 2 × 4 = 8, , Hence, alpha particles released are =, , 212, Now nucleus after alpha decays = x82, , After 1 decay Z increases by 1, Number of decays = 83 – 82 = 1, Hence answer is 4 and 1 decays., 19. In the radioactive decay of an element it is found that the count rate reduces from 1024 to 128 in, 3 minutes. Its half life will be, (1) 1 minute, , (2) 2 minute, , (3) 3 minute, , (4) 5 minute, , Sol. Answer (1), R, 128, =, R0 1024, R, 1, =, R0 8, , or, , R ⎛ 1⎞, =⎜ ⎟, R0 ⎝ 2 ⎠, , 3, , n=3, 3 half lives in 3 minutes, 1 half life in 1 minute, 20. If a radioactive material remains 25% after 16 days, then its half life will be, (1) 32 days, , (2) 8 days, , (3) 64 days, , (4) 28 days, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 47 :
Solution of Assignment, , Nuclei, , 151, , Sol. Answer (2), N, ⎛ N ⎞, =⎜, ⎟, N0 ⎝ 4 × N ⎠, , N, ⎛ 1⎞, =⎜ ⎟, N0 ⎝ 2 ⎠, , 2, , or n = 2, 2 half lives in 16 days, 1 half life is in 8 days, 21. The count rate of a radioactive source at t = 0 was 1600 count/s and at t = 8 s, it was 100 count/s. The count rate, (in counts) at t = 6 s was, (1) 150, , (2) 200, , (3) 300, , (4) 400, , Sol. Answer (2), R, ⎛ 1⎞, =⎜ ⎟, R0 ⎝ 2 ⎠, , 1 ⎛ 1⎞, =⎜ ⎟, 16 ⎝ 2 ⎠, , n, , n, , n=4, T1/2 = 2 s, 3, , at t = 6 s,, , R ⎛ 1⎞, , R R0 1600 200, R0 ⎜⎝ 2 ⎟⎠, 8, 8, , 22. The radioactivity of a sample is R1 at a time T1 and R2 at a time T2. If the half life of the specimen is T, the, number of atoms that have disintegrated in the time (T2 – T1) is proportional to, (1) (R1 T1 – R2T2), , (2) (R1 – R2), , (3) (R1– R2)/T, , (4) (R1 – R2)T, , Sol. Answer (4), R1 = N1, , R2 = N2, , N = N1 – N2 =, , ∆N =, , R1 − R2, λ, , (R1 − R2 )T, 0.693, , ( = 0.693), , 23. A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minute,, the rate is 1250 disintegrations per minute. The decay constant (per minute) is, (1) 0.8 ln 2, , (2) 0.4 ln 2, , (3) 0.2 ln 2, , (4) 0.1 ln 2, , Sol. Answer (2), R, ⎛ 1⎞, =⎜ ⎟, R0 ⎝ 2 ⎠, , n, , 1250 ⎛ 1 ⎞, =⎜ ⎟, 5000 ⎝ 2 ⎠, 1 ⎛ 1⎞, =⎜ ⎟, 4 ⎝2⎠, , n, , n, , n=2, , T1/2 =, , 5, = 2.5 minute, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 48 :
152, , Nuclei, , Solution of Assignment, , t, , T, , T, , t, , T, , t, , (4), , logeA, , (3), , logeA, , (2), , logeA, , (1), , logeA, , 24. At time t = 0 some radioactive gas is injected into a sealed vessel. At time T some more of the gas is injected, into the vessel. Which one of the following graphs best represents the logarithm of the activity A of the gas, with time t?, , T, , t, , Sol. Answer (3), A = A0e–t, ln A = ln A0 – t, Answer is (3) as it is a linear relation between ln A and t., 25., , 40, , K isotope of potassium has a half life of 1.37 × 109 years and decays to an isotope of argon which is stable. In a, particular sample of moon rock, the ratio of potassium atoms to argon atoms was found to be 1 : 7. The age of the, rock, assuming that originally there was no argon present, is, , (1) 4.11 × 109 year, , (2) 2.74 × 109 year, , (3) 5.48 × 109 year, , (4) 1.37 × 109 year, , Sol. Answer (1), T1/2 = 1.37 × 109 year, Ratio – Potassium : Argon = 1 : 7, Amount of Potassium left =, , 1, 8, , n, , 1, ⎛ 1⎞, ⎜⎝ ⎟⎠ =, 2, 8, n or number of half lives = 3, 26. Two radioactive isotopes P and Q have half lives 10 minutes and 15 minutes respectively. Freshly prepared, samples of each isotope initially contain the same number of atoms. After 30 minutes, the ratio, , number of atoms of P, will be, number of atoms of Q, (1) 0.5, , (2) 2.0, , (3) 1.0, , (4) 3.0, , Sol. Answer (1), TP = 10 minute, , TQ = 15 minute, , After 30 minute P has gone 3 half lives and Q = 2 half lives, , ⎛ 1⎞, NP = N0 ⎜ ⎟, ⎝2⎠, , 3, , ⎛ 1⎞, NQ = N0 ⎜ ⎟, ⎝2⎠, , 2, , NP : NQ = 1 : 2, or, , NP, = 0.5, NQ, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 49 :
Solution of Assignment, , Nuclei, , 153, , 27. In a radioactive decay let N be the number of residual active nuclei, D the number of daughter nuclei, R the rate, of decay and M the mass of active sample at any time t. Below are shown four curves., , N, , D, , M, , (ii), , (i), , (iii), , (iv), , t, , t, , t, , The correct ones are, (1) (i), (ii) and (iv), , (2) (ii), (iii) and (iv), , (3) (iii), (iv) and (i), , (4) All of these, , Sol. Answer (1), Daughter nuclei increase exponentially Mass of active sample and number of active nuclei decreases exponentially., Rate decreases exponentially but since it is not shown as such (iii) is wrong., 28. A freshly-prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible, safe level. The minimum time after which it would be possible to work safely with this source is, (1) 6 h, , (2) 12 h, , (3) 24 h, , (4) 128 h, , Sol. Answer (2), T1/2 = 2 hours, To work safely the number of reacting molecules must decrease by 64 times, , ⎛ 1⎞, N = N0 ⎜ ⎟, ⎝2⎠, N, ⎛ 1⎞, =⎜ ⎟, N0 ⎝ 2 ⎠, , 1 ⎛ 1⎞, =⎜ ⎟, 64 ⎝ 2 ⎠, , n, , n, , n, , n=8, Time it will take to T1/2 × 8 or 16 hours, 29. -decay occurs when, (1) Pair annihilation takes place, (2) Energy is released due to conversion of neutron into proton, (3) Energy is released due to de-excitation of nucleus, (4) None of these, Sol. Answer (3), Fact., 30. The sample of a radioactive substance has 106 nuclei. Its half life is 20 s. The number of nuclei that will be, left after 10 s is nearly, (1) 1 × 105, , (2) 2 × 105, , (3) 7 × 105, , (4) 11 × 105, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 50 :
154, , Nuclei, , Solution of Assignment, , Sol. Answer (3), , ⎛ 1⎞, N = N0 ⎜ ⎟, ⎝2⎠, n=, , n, , 1, 2, , ⎛ 1 ⎞, N = 106 ⎜, ⎝ 2 ⎠⎟, , N 0.732 × 106, N 7 × 105, 31. Half life of radioactive element depends upon, (1) Amount of element present, , (2) Temperature, , (3) Pressure, , (4) The nature of the element, , Sol. Answer (4), Fact., 32. Neutrino is a particle which, (1) Has no charge and no spin, (2) Has no charge but has spin, (3) Is charged like an electron and has spin, (4) Has no charge but has mass nearly equal to that of a proton, Sol. Answer (2), Fact., 33. Heavy water instead of ordinary water is used as a moderator in nuclear reactor because ordinary water, (1) Cannot slow down neutron, , (2) Absorbs neutrons, , (3) Is expensive, , (4) Accelerates neutron, , Sol. Answer (2), Fact., 34. Out of the following, which one is not emitted by a natural radioactive substance?, (1) Electrons, (2) Electromagnetic radiations, (3) Helium nuclei with charge equal to that of two protons, (4) Neutrons, Sol. Answer (4), Fact., 35. In each fission of 92U235 energy of 200 MeV is released. How many acts of fission must occur per second to, produce a power of 1kW?, (1) 3.1 × 1013, , (2) 1.3 × 1016, , (3) 1.3 × 1015, , (4) 3.1 × 1016, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 51 :
Solution of Assignment, , Nuclei, , 155, , Sol. Answer (1), Each fission 200 MeV is released, or energy released = 200 × 106 × 1.6 × 10–19 J, Power needed = 1000 W, Number of fission =, , 1000, 6, , 200 × 10 × 1.6 × 10−19, , = 3.1 × 1013, 36. If 1 g hydrogen is converted into 0.993 gm of helium in a thermonuclear reaction, the energy released in the, reaction is, (1) 63 × 107 J, , (2) 63 × 1010 J, , (3) 63 × 1014 J, , (4) 63 × 1020 J, , (3) 1 MeV, , (4) 0.03 MeV, , Sol. Answer (2), 1 g hydrogen converted to 0.993 g helium., m = 0.007g, m = 7 × 10–6 g, E = mC2, E = 7 × 10–6 × 9 × 1016 = 63 × 1010, 37. Thermal neutrons are those whose energy is about, (1) 1 J, , (2) 0.03 eV, , Sol. Answer (2), Fact., 38. A neutron strikes a 92U235 nucleus and as a result 36Kr93 and 56Ba140 are produced with, (1) -particle, , (2) 1-neutron, , (3) 3-neutron, , (4) 2--particle, , Sol. Answer (3), In the reaction sum of atomic number remains the same but mass reduces by 3., Hence 3 neutrons were produced., 39. Control rods used in nuclear reactors are made of, (1) Stainless steel, , (2) Graphite, , (3) Cadmium, , (4) Plutonium, , Sol. Answer (3), Fact., 40. In the equation, 27, 4, 30, 15, P, 13 Al 2He , , X,, , The correct symbol for X is, (1), , 0, 1e, , (2), , 1, 1H, , (3), , 4, 2 He, , (4), , 1, 0n, , Sol. Answer (4), Conserving mass and charge, net mass reduces by 1 by charge does not change. Hence, a neutron must, have been released., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 52 :
156, , Nuclei, , Solution of Assignment, , SECTION - B, Objective Type Questions, 1., , A certain stable nucleide, after absorbing a neutron, emits -particle and the new nucleide splits spontaneously, into two -particles. The nucleide is, (1), , 4, 2 He, , (2), , 7, 3 Li, , (3), , 6, 4 Be, , (4), , 6, 3 Li, , Sol. Answer (2), After absorbing neutron it undergoes decay. Also it decays into 2 alpha particles, Hence, net charge after decay must have been 4, Before decay it must have been 3, Since it was stable initially, it must be a common nucleus of Z = 3 which is 73 Li, 2., , After 3 hours, only 0.25 mg of a pure radioactive material is left. If initial mass was 2 mg then the half life of the, substance is, (1) 1.5 hr, , (2) 1 hr, , (3) 0.5 hr, , (4) 2 hr, , Sol. Answer (2), Initial mass = 2 mg, Final mass = 0.25 mg, , , N, ⎛ 1⎞, =⎜ ⎟, N0 ⎝ 2 ⎠, , n, , 0.25 ⎛ 1 ⎞, =⎜ ⎟, ⎝2⎠, 2, , 1 ⎛ 1⎞, =⎜ ⎟, 8 ⎝2⎠, , n, , n, , n = 3 and time is 3 hours, Half life is therefore 1 hour, 3., , Pauli suggested the emission of nutrino during + decay to explain, (1) Continuous energy distribution of positrons, , (2) Conservation of linear momentum, , (3) Conservation of mass-energy, , (4) All of these, , Sol. Answer (1), Fact., 4., , If a heavy nucleus has N/Z ratio higher than that required for stability, then, (1) It emits –, , (2) It emits +, , (3) It emits particle, , (4) It will undergo K electron capture, , Sol. Answer (1), If, , N, ratio is higher it will try to increase number of protons by decay., Z, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 53 :
Solution of Assignment, , 5., , Nuclei, , 157, , Half lives for and emission of a radioactive material are 16 years and 48 years respectively. When material, decays giving and emission simultaneously then time in which, (1) 29 years, , (2) 24 years, , 3, th of the material decays is, 4, , (3) 64 years, , (4) 12 years, , Sol. Answer (2), 1 material is giving two products, Let initial number be = N0, Let time when, , 3, N decay be t, 4 0, , T1T2, Effective half life = T + T = 12 years, 1, 2, N, ⎛ 1⎞, =⎜ ⎟, N0 ⎝ 2 ⎠, , 1 ⎛ 1⎞, =⎜ ⎟, 4 ⎝2⎠, , n, , n, , n=2, , Hence, time will be 24 years, 6., , Two radioactive samples A and B have half lives T1 and T2 (T1 > T2) respectively. At t = 0, the activity of B was twice, the activity of A. Their activity will become equal after a time, , T1T2, (1) T T, 1, 2, , (2), , T1 T2, 2, , (3), , T1 T2, 2, , T1T2, (4) T T, 1, 2, , Sol. Answer (1), 2RA = RB, 21N1 = 2N2, , ....(i), , Radio-activity is same after say time t, λ1N1e −λ1t = λ 2N2e −λ2t, , ....(ii), , Dividing (i) by (ii), 2e λ1t = e λ2t, 2 = e( λ2 −λ1 )t, , Taking ln on both sides, 0.693 = (2 – 1)t, ⎛ 1 1⎞, 1= ⎜ − ⎟t, ⎝ T2 T1 ⎠, T2T1, =t, T1 − T2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 54 :
158, 7., , Nuclei, , Solution of Assignment, , Choose the correct statement, (1) The nuclear force becomes strong if the nucleus contains too many protons compared to neutrons, (2) The nuclear force becomes strong if the nucleus contains too many neutrons compared to protons, (3) Nuclei with atomic number less than 82 shows a tendency to disintegrate, (4) The nuclear force becomes weak if the nucleus contains a large number of nucleons, , Sol. Answer (4), Nuclear force being a short range force becomes unstable with too many nucleons., 8., , N atoms of a radioactive element emit n number of -particles per second. Mean life of the element in seconds, is, (1), , n, N, , (2), , N, n, , (3) 0.693, , N, n, , (4) 0.693, , n, N, , Sol. Answer (2), n is the rate of decay n N, or n = N, , 1, N, or, λ, n, , Mean life is, 9., , Ten percent of a radioactive sample has decayed in 1 day. After 2 days, the decayed percentage of nuclei will be, (1) 81%, , (2) 19%, , (3) 20%, , (4) 100%, , Sol. Answer (2), N1 = N0 – N0e–t, , N0, 10, , Since N1 =, , N0 e −λt =, , , 9N0, 10, , e −λt =, , 9, 10, , Amount left =, , 9N0, 10, , N2 = N0 – N0e–2t, 81, N0, 100, or N2 = 19% of N0, , or, , N2 = N0 −, , 10. A sample of radioactive element has a mass of 10 gm at an instant t = 0. The approximate mass of this element in, the sample after two mean lives is, (1) 2.50 gm, , (2) 1.35 gm, , (3) 6.30 gm, , (4) 3.70 gm, , Sol. Answer (2), m = 10 g at t = 0, m = m0e–t, where t =, m=, , 2, λ, , m0, e2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 55 :
Solution of Assignment, , Nuclei, , 159, , 11. During mean life of a radioactive element, the fraction that disintegrates is, (1) e, , (2), , e 1, e, , (3), , 1, e, , (4), , e, e 1, , Sol. Answer (2), In mean life t =, , 1, λ, , N N0 e, Fraction that disintegrates is 0, N0, , , , 1, , , ⎛ 1 e ⎞, ⎛ 1 e ⎞, Or magnitude ⎜, ⎜, ⎟, ⎟, ⎝ e ⎠, ⎝ e ⎠, , 12. An element A decays by a two step process into element C, A B + 2He4, B C + 2e–, Then,, (1) A and B are isobars, , (2) A and C are isobars, , (3) A and B are isotopes(4) A and C are isotopes, , Sol. Answer (4), First A loses two protons. Then by minus decay it gains two protons., Hence, atomic number Z is same for A and C and they are isotopes., 13. After five half lives percentage of original radioactive atoms left is, (1) 1%, , (2) 0.3%, , (3) 3.125%, , (4) 0.2%, , Sol. Answer (3), n, , ⎛ 1⎞, N = N0 ⎜ ⎟ where n = 5, ⎝2⎠, N=, , N0, 32, , N = 3.125% of N0, 14. The radioactivity of a certain radioactive elements drops to, , 1, of its initial value in 30 seconds. Its half life, 64, , is, (1) 8 seconds, , (2) 15 seconds, , (3) 7.5 seconds, , (4) 5 seconds, , Sol. Answer (4), N, ⎛ 1⎞, =⎜ ⎟, N0 ⎝ 2 ⎠, , 1 ⎛ 1⎞, =⎜ ⎟, 64 ⎝ 2 ⎠, , n, , n, , n=6, Hence half life is 5 seconds, 15. Find the decay rate of the substance having 4 × 1015 atoms. Half life of a radioactive substance in -decay, is 1.2 × 107 s, (1) 2.3 × 108 atom/s, , (2) 3.2 × 108 atom/s, , (3) 2.3 × 1011 atom/s, , (4) 3.2 × 1011 atom/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 56 :
160, , Nuclei, , Solution of Assignment, , Sol. Answer (1), N0 = 4 × 1015 atoms, T1/2 = 1.2 × 107 s =, , , or, , λ=, , 0.693, λ, , 0.693, 1.2 × 107, , −, , dN, = λN0, dt, , −, , dN, 0.693, =, × 4 × 1015, dt, 1.2 × 107, , dN, = 2.3 × 108 atom/s, dt, , 16. The average binding energy per nucleon in the nucleus of atom is approximately, (1) 8 J, , (2) 8 KeV, , (3) 8 eV, , (4) 8 MeV, , Sol. Answer (4), Fact., 17. For the nuclear fusion reaction 12 H 13 H 24 He 10 n temperature to which gases must be heated is 3.7 × 109 K., Potential energy between two nuclei is closest to (Boltzmann’s constant k = 1.38 ×10–23 J/K), (1) –10–10 J, , (2) –10–12 J, , (3) –10–14 J, , (4) –10–16 J, , Sol. Answer (4), KE of nuclei =, , 3, kt = 7.659 × 10–14, 2, , (for fusion), , Potential energy between nuclei must be much less than the initiating KE so that the nuclei have enough, KE for reaction to take place., 18. A nucleus 220X at rest decays emitting an -particle. If energy of daughter nucleus is 0.2 MeV, Q value of the, reaction is, (1) 10.8 MeV, , (2) 10.9 MeV, , (3) 11 MeV, , (4) 11.1 MeV, , Sol. Answer (3), Energy of daughter nucleus = 0.2 MeV, 0.2 MeV =, , mα, Q, mα + mD, , 0.2 MeV =, , 4, Q, 220, , 0.2 × 220, MeV = Q, 4, 2 × 55, =Q, 10, , Q = 11 MeV, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 57 :
Solution of Assignment, , Nuclei, , 161, , 19. Radioactive nuclei P and Q disintegrate into R with half lives 1 month and 2 months respectively. At time t =, 0, number of nuclei of each P and Q is x. Time at which rate of disintegration of P and Q are equal, number, of nuclei of R is, (1) x, , (2) 1.25 x, , (3) 1.5 x, , (4) 1.75 x, , Sol. Answer (2), Let time be t, λ1 × e −λ1t = λ 2 × e −λ2t, λ1e −λ1t = λ 2e −λ2t, λ1, = e( λ1 −λ2 )t, λ2, ln, , λ1, = ( λ1 − λ 2 )t, λ2, , ln 1 – ln 2 = (1 – 2)t, 0.693 = (1 – 2)t, 20. A radioactive element X emits six -particles and four -particles leading to end product, (1), , 238, 92 U, , (2), , 230, 90 Th, , (3), , 232, 90 Th, , (4), , 208, 82 Pb ., , X is, , 239, 92 U, , Sol. Answer (3), Calculating Zx = 82 + 6 × 2 – 4, = 90, Ax = 208 + 6 × 4 = 232, 21. In nature, ratio of isotopes of Boron, 5B10 and 5B11, is (given that atomic weight of boron is 10.81), (1) 81 : 19, , (2) 21 : 44, , (3) 19 : 81, , (4) 44 : 21, , Sol. Answer (3), Atomic weight = 10.81, Weighted mean is hence = 10.81, , 10.81 =, , 10 × x + 11 × (100 − x ), 100, , 1081 = 10x + 1100 – 11x, x = 19, Hence, B10 : B11 = 19 : 81, 22. Q-value of the decay, , 22, 11 Na, , , , 22, , 10 Ne e, , is, , 22, 22, (1) [m (11, Na) m(10, Ne)] c 2, , 22, 22, (2) [m(11, Na) m(10, Ne) me ] c 2, , 22, 22, (3) [m(11, Na) m(10, Ne) 2me ] c 2, , 22, 22, (4) [m(11, Na) m(10, Ne) 3me ] c 2, , Sol. Answer (3), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 58 :
162, , Nuclei, , Solution of Assignment, , 23. Which of the alternatives gives correct match of Column-I with Column-II?, Column-I, , Column-II, , a. Binding energy per nucleon for 56Fe, , (i), , 5.5 M eV, , b. Energy of -particle in Geiger Marsden experiment, , (ii), , 200 M eV, , c. Energy of photon of visible light, , (iii), , 8.75 M eV, , d. Energy released in fission of a uranium nucleus, , (iv), , 2 eV, , (1) a(i), b(iii), c(iv), d(ii), , (2) a(iii), b(i), c(ii), d(iv), , (3) a(iii), b(i), c(iv), d(ii) (4) a(i), b(iv), c(ii), d(iii), , Sol. Answer (3), Fact., 24. Correct increasing order of penetrating powers of , particles and -rays, all moving with same kinetic energy, is, (1) , , , (2) , , , (3) , , , (4) All have same penetrating power as all have same kinetic energy, Sol. Answer (1), Fact., 25. In proton-proton cycle, four hydrogen atoms combine to release energy, (1) 2.67 MeV, , (2) 2.67 KeV, , (3) 26.7 MeV, , (4) 26.7 KeV, , Sol. Answer (3), Fact., In proton-proton cycle four hydrogen nuclei combine to form helium., 26. 37 Rutherford equals, (1) 1 milli bacquerel, , (2) 1 milli curie, , (3) 1 micro bacquerel, , (4) 1 micro curie, , Sol. Answer (1), Fact., 27. Which of these is incorrect about nuclear forces?, (1) They are independent of charge, (2) Nuclear forces are derived from quark-quark interaction, (3) Hadrons do not experience strong nuclear force, (4) Nuclear force is not a central force, Sol. Answer (3), Fact., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 60 :
164, 3., , Nuclei, , Solution of Assignment, , The Binding energy per nucleon of 73 Li and 42 He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear, reaction 73 Li + 11H 42 He + 42 He + Q, the value of energy Q released is, (1) 19.6 MeV, , (2) –2.4 MeV, , [AIPMT-2014], , (3) 8.4 MeV, , (4) 17.3 MeV, , Sol. Answer (4), Q = 2 (4 × 7.06) – (7 × 5.60), Q = 17.3 MeV, 4., , A radio isotope X with a half life 1.4 × 109 years decays to Y which is stable. A sample of the rock from a cave, was found to contain X and Y in the ratio 1 : 7. The age of the rock is, [AIPMT-2014], (1) 1.96 × 109 years, , (2) 3.92 × 109 years, , (3) 4.20 × 109 years, , (4) 8.40 × 109 years, , Sol. Answer (3), X : Y = 1 : 7 then X , , 1, 7, , Y , 8, 8, , n, , 1, ⎛ 1⎞, ⎜2⎟ 8 n = 3, ⎝ ⎠, t, t1/2, , 3 t 3 t1/2 = 3 × 1.4 × 109, , t = 4.2 × 109 years., 5., , A certain mass of Hydrogen is changed to Helium by the process of fusion. The Mass defect in fusion reaction, is 0.02866 u. The energy liberated per u is:(Given 1 u = 931 MeV), [NEET-2013], (1) 26.7 MeV, , (2) 6.675 MeV, , (3) 13.35 MeV, , (4) 2.67 MeV, , Sol. Answer (2), 6., , The half life of a radioactive isotope X is 20 years. It decays to another element Y which is stable. The two, elements X and Y were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated, to be, [NEET-2013], (1) 60 years, , (2) 80 years, , (3) 100 years, , (4) 40 years, , Sol. Answer (1), 7., , If the nuclear radius of, , 27Al, , is 3.6 Fermi, the approximate nuclear radius of, , 64Cu, , in Fermi is, [AIPMT (Prelims)-2012], , (1) 4.8, , (2) 3.6, , (3) 2.4, , (4) 1.2, , Sol. Answer (1), R A1/3, 8., , A mixture consists of two radioactive materials A1 and A2 with half lives of 20 s and 10 s respectively. Initially, the mixture has 40 g of A1 and 160 g of A2. The amount of the two in the mixture will become equal after, [AIPMT (Prelims)-2012], (1) 20 s, , (2) 40 s, , (3) 60 s, , (4) 80 s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 62 :
166, , Nuclei, , Solution of Assignment, , 11. The half life of a radioactive isotope X is 50 years. It decays to another element , Y which is stable. The two, elements X and Y were found to be in the ratio of 1:15 in a sample of a given rock. The age of the rock was, estimated to be, [AIPMT (Prelims)-2011], (1) 100 years, , (2) 150 years, , (3) 200 years, , (4) 250 years, , Sol. Answer (3), Tx = 50 years, Amount of x left =, , x0, ⎛ 1⎞, = x0 ⎜ ⎟, ⎝2⎠, 16, , 1, 16, , n, , n=4, Hence, age of the rock is 200 years, 12. Fusion reaction takes place at high temperature because, , [AIPMT (Prelims)-2011], , (1) Molecules break up at high temperature, (2) Nuclei break up at high temperature, (3) Atoms get ionised at high temperature, (4) Kinetic energy is high enough to overcome the coulomb repulsion between nuclei, Sol. Answer (4), Fact., 13. A nucleus, (1), , m, n, , X emits one -particle and two – particles. The resulting nucleus is, , m4, n 2 Y, , (2), , m 6, n 4 Z, , (3), , m 6, nZ, , (4), , [AIPMT (Prelims)-2011], m4, n X, , Sol. Answer (4), Final number of nucleons = m – 4, Final number of protons = n – 2 + 2 = n, Resulting nucleus =, , m−4, nZ, , 14. Two radioactive nuclei P and Q in a given sample decay into a stable nucleus R. At time t = 0, number of P, species are 4N0 and that of Q are N0. Half-life of P (for conversion to R) is 1 minute whereas that of Q is 2, minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be :, [AIPMT (Mains)-2011], (1), , 5N0, 2, , (2) 2N0, , (3) 3N, , (4), , 9N0, 2, , Sol. Answer (4), Number of nuclei of P = 4N0, Number of nuclei of Q = N0, 4N0e–2t = N0e–t, 4 = et, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 64 :
168, , Nuclei, , Solution of Assignment, , 18. The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When, two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is, [AIPMT (Mains)-2010], (1) 23.6 MeV, , (2) 2.2 MeV, , (3) 28.0 MeV, , (4) 30.2 MeV, , Sol. Answer (1), E = (28 – 4.4) MeV, E = 23.6 MeV, 19. In the nuclear decay given below:, A, ZX, , Z A1Y AZ –– 41B * AZ –– 41B, the particles emitted in the sequence are:, , (1) , , , , (2) , , , , (3) , , , , [AIPMT (Prelims)-2009], (4) , , , , Sol. Answer (4), First particle reduces atomic nucleus without changing mass., Then alpha decay occurs but is in an excited state., gets to lower energy state by emitting a photon., 20. The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted, by it. The resulting daughter is an:, [AIPMT (Prelims)-2009], (1) Isomer of parent, , (2) Isotone of parent, , (3) Isotope of parent, , (4) Isobar of parent, , Sol. Answer (3), Atomic number of initial and final nuclei will be same., 21. In a Rutherford scattering experiment when a projectile of charge z1 and mass M1 approaches a target nucleus, of charge z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is:, [AIPMT (Prelims)-2009], (1) Directly proportional to z1z2, , (2) Inversely proportional to z1, , (3) Directly proportional to mass M1, , (4) Directly proportional to M1 × M2, , Sol. Answer (1), 22. Two radioactive materials X1 and X2 have decay constants 5 and respectively. If initially they have the same, number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be, , 1, after a time, e, , [AIPMT (Prelims)-2008], (1), , e, , , (2) , , (3), , 1, , 2, , (4), , 1, 4, , Sol. Answer (4), 23. Two nuclei have their mass numbers in the ratio of 1:3. The ratio of their nuclear densities would be, [AIPMT (Prelims)-2008], (1) 1 : 1, , (2) 1 : 3, , (3) 3 : 1, , (4) (3)1/3 : 1, , Sol. Answer (1), Nuclear densities are always roughly same., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 65 :
Solution of Assignment, , Nuclei, , 24. If M(A; Z), Mp and Mn denote the masses of the nucleus, = 931.5 MeV /, , C2), , A, ZX, , 169, , , proton and neutron respectively in units of u (1u, , and BE represents its bonding energy in MeV, then, , [AIPMT (Prelims)-2008], , (1) M(A, Z) = ZMp + (A – Z)Mn + BE/C2, , (2) M(A, Z) = ZMp + (A – Z)Mn – BE/C2, , (3) M(A, Z) = ZMp + (A – Z)Mn + BE, , (4) M(A, Z) = ZMp + (A – Z)Mn – BE, , Sol. Answer (2), 25. If the nucleus, , 27, 13 Al, , has a nuclear radius of about 3.6 fm, then, , 125, 32Te, , would have its radius approximately as, [AIPMT (Prelims)-2007], , (1) 4.8 fm, , (2) 6.0 fm, , (3) 9.6 fm, , (4) 12.0 fm, , Sol. Answer (2), R A1/3, RAL A1/3, AL, = 1/3, RTe, ATe, , RAL ×, , 1/3, ATe, , 3.6fm ×, , 3.6 ×, , = RTe, , A1/3, Al, , 1251/3, 271/3, , = RTe, , 5, = RTe, 3, , RTe = 6 fm, 26. In a radioactive decay process, the negatively charged emitted – particles are, , [AIPMT (Prelims)-2007], , (1) The electrons produced as a result of the decay of neutrons inside the nucleus, (2) The electrons produced as a result of collisions between atoms, (3) The electrons orbiting around the nucleus, (4) The electrons present inside the nucleus, Sol. Answer (1), By definition., 27. A nucleus, , A, ZX, , has mass represented by M(A, Z). If Mp and Mn denote the mass of proton and neutron respectively, , and BE the binding energy in MeV, then, , [AIPMT (Prelims)-2007], , (1) BE = M(A, z) – ZMp – (A – Z)Mn, , (2) BE = [M(A, z) – ZMp – (A – Z)Mn]c2, , (3) BE = [zMp + (A, z)Mn – M(A, Z)]c2, , (4) BE = [zMp + AMn – M(A, Z)]c2, , Sol. Answer (3), Binding energy = Mass defect (m)C2, m = Mass of individual protons + Mass of individual neutrons – Mass of nucleus, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 66 :
170, , Nuclei, , Solution of Assignment, , 28. Two radioactive substances A and B have decay constants 5 and respectively. At t = 0 they have the same, number of nuclei. The ratio of number of nuclei of A to those of B will be (1/e)2 after a time interval, [AIPMT (Prelims)-2007], 1, 2, , (1), , (2), , 1, 4, , (3) 4 , , (4) 2 , , Sol. Answer (1), NA = N0e–5t, , NB = N0e–t, , NA : NB = 1 : e2, N0 e −5 λt, N0 e, , −λt, , 1, , e 4 λt =, , =, , 1, e2, , 1, , e2, 4t = 2, t=, , 1, 2λ, , 29. The binding energy of deuteron is 2.2 MeV and that of 42 He is 28 MeV. If two deuterons are fused to form one, 4, 2 He, , then the energy released is, , (1) 25.8 MeV, , (2) 23.6 MeV, , [AIPMT (Prelims)-2006], (3) 19.2 MeV, , (4) 30.2 MeV, , Sol. Answer (2), Binding energy = 2.2 MeV for 2H1 and = 28 MeV for 24 He, Energy released = Final BE – Initial BE, = 28 – 2 × 2.2 MeV = 23.6 MeV, 30. In a radioactive material the activity at time t1 is R1 and at a later time t2, it is R2. If the decay constant of the, material is , then, [AIPMT (Prelims)-2006], (1) R1 R2e – (t1 t2 ), , (2) R1 R2e(t1 t2 ), , ⎛t ⎞, (3) R1 R2 ⎜ 2 ⎟, ⎝ t1 ⎠, , (4) R1 = R2, , Sol. Answer (1), Simply by formula of radioactive disintegration, R1 = R2e −λ( t1 − t2 ), 31. The radius of germanium (Ge) nuclide is measured to be twice the radius of 94 Be . The number of nucleons in, Ge are, [AIPMT (Prelims)-2006], (1) 73, , (2) 74, , (3) 75, , (4) 72, , Sol. Answer (4), 32. In the reaction 21H 31H 42 He 01n , if the binding energies of 21H , 31H and 42 He are respectively a, b and c (in, MeV), then the energy (in MeV) released in this reaction is, [AIPMT (Prelims)-2005], (1) c + a – b, , (2) c – a – b, , (3) a + b + c, , (4) a + b – c, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 67 :
Solution of Assignment, , Nuclei, , 171, , Sol. Answer (2), Binding energy of reactants = Binding energy of product + Energy released, 33. The nuclei of which one of the following pairs of nuclei are isotones?, (1), , 74, 71, 34Se , 31Ga, , (2), , 92, 92, 42Mo , 40Zr, , (3), , 38Sr, , 84,, , 38Sr, , [AIPMT (Prelims)-2005], 86, , (4), , 40, 32, 20Ca , 16S, , Sol. Answer (1), Isotones have same nuclei numbers but different atomic number., 34. Fission of nuclei is possible because the binding energy per nucleon in them, , [AIPMT (Prelims)-2005], , (1) Increases with mass number at high mass numbers, (2) Decreases with mass number at high mass numbers, (3) Increases with mass number at low mass numbers, (4) Decreases with mass number at low mass numbers, Sol. Answer (2), The mpre the binding energy the more stable the nucleus. Energy is released when fission occurs by going, from less stable to more stable configuration., massof fission products, 35. In any fission process the ratio massof parent nucleus is:, , [AIPMT (Prelims)-2005], , (1) Less than 1, , (2) Greater than 1, , (3) Equal to 1, , (4) Depends on the mass of parent nucleus, , Sol. Answer (1), In fission process energy is released by converting same mass into energy., Hence, mass of fission product < mass of parent nucleus., 36. The volume occupied by an atom is greater than the volume of the nucleus by a factor of about, (1) 101, , (2) 105, , (3) 1010, , (4) 1015, , (2) Positron, , (3) Protons, , (4) Ionized helium atoms, , Sol. Answer (4), Fact., 37. Alpha particles are, (1) Neutrally charged, Sol. Answer (4), Fact., 38. The mass number of a nucleus is, (1) Always less than its atomic number, (2) Always more than its atomic number, (3) Sometimes equal to its atomic number, (4) Sometimes less than and sometimes more than its atomic number, Sol. Answer (3), It is sometimes equal to its atomic number, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 68 :
172, , Nuclei, , Solution of Assignment, , 39. The radius of germanium (Ge) nuclide is measured to be twice the radius of 94 Be . The number of nucleons, in Ge are, (1) 72, , (2) 73, , (3) 74, , (4) 75, , Sol. Answer (1), R A1/3, 1/3, RBe ABe, = 1/3, RGe AGe, , 1 91/3, = 1/3, 2 AGe, , or, , 1/3, AGe, = 2 × 91/3, , Taking cube on both sides, AGe = 8 × 9 = 72 nucleons, 40. What is the respective number of and -particles emitted in the following radioactive decay?, X90 168Y80, , 200, , (1) 8 and 8, , (2) 8 and 6, , (3) 6 and 8, , (4) 6 and 6, , Sol. Answer (2), Number of alpha particles :, , 200 − 168, =8, 4, , After alpha decay number of protons left = 90 – 8 × 2 = 74, Number of particles = 80 – 74 = 6, 41. A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to 2 : 1. What will be the ratio, of their nuclear size (nuclear radius)?, (1) 31/2 : 1, , (2) 1 : 31/2, , (3) 21/3 : 1, , (4) 1 : 21/3, , (3) X-rays, , (4) -rays, , Sol. Answer (4), m1v1 = m2v2, or, , r13, r23, , =, , 1, 2, , or, , m1 v 2, =, m2 v1, , or, , r1, 1, =, r2 21/3, , 42. The most penetrating radiation out of the following are, (1) -rays, , (2) -rays, , Sol. Answer (2), Gamma rays are most penetrating., 43. Complete the equation for the following fission process 92U235 + 0n1 38Sr90 + …, (1), , X142 + 30n1, , 57, , (2), , X145 + 30n1, , 54, , (3), , X143 + 30n1, , 54, , (4), , X142 + 0n1, , 54, , Sol. Answer (3), U235 + 0n1 38Sr90, , 92, , Conerving mass :, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 69 :
Solution of Assignment, , Nuclei, , 173, , Initial mass 236, Final mass 90 + mx + Nn01, 146 = mx + N01, Among option only (3) meets the criteria., 44. After 1 and 2–emissions, (1) Mass number reduces by 6, , (2) Mass number reduces by 4, , (3) Mass number reduces by 2, , (4) Atomic number reduces by 4, , Sol. Answer (2), In alpha emission mass number reduces by 4. In decay no subsequent reduction of mass takes place., 45. For the given reaction, the particle X is 6C11 5B11 + + + X, (1) Neutron, , (2) Anti-neutrino, , (3) Neutrino, , (4) Proton, , Sol. Answer (3), Mass since this is a positive decay x will be an neutrino., 46. X(n, ), (1), , 7, 3 Li ,, , then X will be, , 10, 5 B, , (2), , 9, 5B, , (3), , 11, 4 Be, , (4), , 4, 2 He, , Sol. Answer (1), , x + n01 ⎯⎯, → 73 Li + 24 He, Conserving charge Zx = 5, Conserving mass = Mx = 7 + 4 – 1 = 10, x must be, , 10, 5B, , 47. Mn and Mp represent the mass of neutron and proton respectively. An element having mass M has N-neutron, and Z-protons, then the correct relation will be, (1) M < {N . Mn+ Z . Mp}, , (2) M > {N . Mn + Z . Mp}, , (3) M = {N . Mm + Z . Mp}, , (4), , M = N {Mn + Mp}, , Sol. Answer (1), Mass M = N + Z, Due to mass defect M < NMn + ZMp, 48. Which rays contain (positive) charged particle?, (1) -rays, , (2) -rays, , (3) -rays, , (4) X-rays, , Sol. Answer (1), Fact., 49. A deutron is bombarded on 8O16 nucleus then -particle is emitted. The product nucleus is, (1) 7N13, , (2) 5B10, , (3) 4Be9, , (4) 7N14, , Sol. Answer (4), 2, O16, → 24He + X 816+1+−22− 4, 8 + 1 H ⎯⎯, , X = N714, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 70 :
174, , Nuclei, , Solution of Assignment, , 50. A nuclear reaction given by, (1) -decay, , X A , , z, , z 1Y, , A, , 1e 0 represents, , (2) -decay, , (3) Fussion, , (4) Fission, , Sol. Answer (1), It represents decay as atomic number changes with a change in mass., 51. If in a nuclear fusion process the masses of the fusing nuclei be m1 and m2 and the mass of the resultant, nucleus be m3, then, (1) m3 = m1 + m2, , (2) m3 = |m1 – m2|, , (3) m3 < (m1 + m2), , (4) m3 > (m1 + m2), , Sol. Answer (3), Due to mass defect m3 < (m1 + m2), 52. Mp denotes the mass of a proton and Mn that of a neutron. A given nucleus, of binding energy B, contains Z, protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is the velocity of light), (1) M(N, Z) = NMn + ZMp – Bc2, , (2) M(N, Z) = NMn + ZMp + Bc2, , (3) M(N, Z) = NMn + ZMp – B/c2, , (4) M(N, Z) = NMn + ZMp + B/c2, , Sol. Answer (3), Mp Denotes mass of proton, Mn Neutron, Binding energy B, M(N, Z) = NMn + ZmD –, , B, C, , 2, , as, , B, C2, , is the mass defect, , 53. The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy, of, , 4, 2 He, , is (Given helium nucleus mass = 4.0015 u), , (1) 0.0305 J, , (2) 0.0305 erg, , (3) 28.4 MeV, , (4) 0.061 n, , Sol. Answer (3), Mass of He 4.0015, Mass defect = 2 × 1.0073 + 2 × 1.0087 – 4.0015 = 2.0146 + 2.0174 – 4.0015 = 232 = 0.0305 amu, Binding energy = 0.0305 × 931.5 MeV = 28.4 MeV, 54. A nucleus represented by the symbol, , A, ZX, , has, , (1) Z neutrons and (A – Z) protons, , (2) Z protons and (A – Z) neutrons, , (3) Z protons and A neutrons, , (4) A protons and (Z – A) neutrons, , Sol. Answer (2), Number of neutrons = (A – Z), Atomic number = Z = Number of protons, 55. How many elementary particles are emitted when, (1) One, , (2) Two, , 14, 6C, , transforms to, (3) Three, , 14, 7N?, , (4) Four, , Sol. Answer (2), 14, 6C, , can be converted to nitrogen by a single decay in which an electron and a neutrino get emitted., , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 71 :
Solution of Assignment, , Nuclei, , 175, , 56. Q value of a nuclear reaction is positive. The reaction is, (1) Exothermic, , (2) Endothermic, , (3) Elastic, , (4) Both exothermic and endothermic, , Sol. Answer (1), Fact., 6 C14 ....., 57. Choose the correct product of nuclear reaction, 7 N14 0 n1 , (1) Proton, , (2) Neutron, , (3) Deutron, , (4) Electron, , Sol. Answer (1), Conserving mass and charge in the reactants and products, answer will be proton., 58. When two nuclei (with A = 8) join to form a heavier nucleus, the binding energy (B.E.) per nucleon of the heavier, nuclei is, (1) More than the B.E. per nucleon of the lighter nuclei, (2) Same as the B.E. per nucleon of the lighter nuclei, (3) Less than the B.E. per nucleon of the lighter nuclei, (4) Double the B.E. per nucleon of the lighter nuclei, Sol. Answer (1), The daughter nucleus is more stable hence binding energy per nucleon will be higher, 59. When helium nuclei bombard beryllium nuclei, then, (1) Electrons are emitted, , (2) Protons are emitted, , (3) Neutrons are emitted, , (4) Protons and neutrons are emitted, , Sol. Answer (3), Fact., 60. The binding energies per nucleon for a deuteron and an -particle are x1 and x2 respectively. The energy Q, released in the reaction 2H1 + 2H1 4He2 + Q, is, (1) 4(x1 + x2), , (2) 4(x2 – x1), , (3) 2(x2 – x1), , (4) 2(x1 + x2), , Sol. Answer (2), Final Binding energy – Initial binding energy, = 4 × x2 – 4 × x1, = 4(x2 – x1), 61. The count rate of a Geiger Muller counter for the radiation of the a radioactive material of half-life of 30 minutes, decreases to 5 second–1 after 2 hours. The initial count rate was, (1) 80 second–1, , (2) 625 second–1, , (3) 20 second–1, , (4) 25 second–1, , Sol. Answer (1), T1/2 = 30 minutes, Time = 2 hours, , ⎛ 1⎞, R = R0 ⎜ ⎟, ⎝2⎠, , n, , n=, , 2 × 60, =4, 30, , 5 × 24 = R0, R0 = 80 second–1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 72 :
176, , Nuclei, , Solution of Assignment, , 62. Half-lives of two radioactive substances A and B are respectively 20 minutes and 40 minutes. Initially the, samples of A and B have equal number of nuclei. After 80 minutes the ratio of remaining numbers of A and, B nuclei is, (1) 1 : 4, , (2) 4 : 1, , (3) 1 : 16, , (4) 1 : 1, , 1, (3) T1/ 2 , , , (4) ( + T1/2) = In 2, , Sol. Answer (1), N0 is same, , ⎛ 1⎞, N A = N0 ⎜ ⎟, ⎝2⎠, , 4, , ⎛ 1⎞, NB = N0 ⎜ ⎟, ⎝2⎠, , 2, , NA : NB = 1 : 4, 63. The relation between and T1/2 as (T1/2 half life), (1) T1/2 , , ln 2, , , (2) T1/2 ln 2 , , Sol. Answer (1), Fact., , T1/2 =, , ln2, λ, , 64. Nuclear-Fission is best explained by, (1) Liquid drop model, , (2) Yukawa -meson theory, , (3) Independent particle model of the nucleus, , (4) Proton-proton cycle, , Sol. Answer (1), Fact., 65. Half life of a radioactive element is 12.5 hour and its quantity is 256 gm. After how much time its quantity, will remain 1 g?, (1) 50 hrs, , (2) 100 hrs, , (3) 150 hrs, , (4) 200 hrs, , Sol. Answer (2), N, ⎛ 1⎞, =⎜ ⎟, N0 ⎝ 2 ⎠, , n, , 1, ⎛ 1⎞, =⎜ ⎟, 256 ⎝ 2 ⎠, , n, , n=8, Hence, time = T1/2 × 8 = 12.5 × 8 = 100 hours, 66. Energy released in nuclear fission is due to, (1) Some mass is converted into charge, (2) Total binding energy of fragments is more than the binding energy of parental element, (3) Total binding energy of fragments is less than the binding energy of parental element, (4) Total binding energy of fragments is equals to the binding energy of parental element, Sol. Answer (2), Fact., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 73 :
Solution of Assignment, , Nuclei, , 177, , 67. A sample of radioactive element contains 4 x 1016 active nuclei. Half life of element is 10 days, then number, of decayed nuclei after 30 days, (1) 0.5 × 1016, , (2) 2 × 1016, , (3) 3.5 × 1016, , (4) 1 x 1016, , Sol. Answer (3), N, ⎛ 1⎞, =⎜ ⎟, N0 ⎝ 2 ⎠, , 3, , N = 4 × 1016 ×, , 30, as n = T, =3, 1/2, , 1, 8, , N = 0.5 × 1016, 68. A sample of radioactive element has a mass of 10 g at an instant t = 0. The approximate mass of this element, in the sample after two mean lives is, (1) 1.35 g, , (2) 2.50 g, , (3) 3.70 g, , (4) 6.30 g, , Sol. Answer (1), , 2, λ, m = m0e–t, t=, , m=, , m0, e2, , 69. The half life of radium is about 1600 years. Of 100 g of radium existing now, 25 g will remain unchanged after, (1) 4800 years, , (2) 6400 years, , (3) 2400 years, , (4) 3200 years, , Sol. Answer (4), T1/2 = 1600, m ⎛ 1⎞, =⎜ ⎟, m0 ⎝ 2 ⎠, , 1 ⎛ 1⎞, =⎜ ⎟, 4 ⎝2⎠, , n, , n, , n=2, Hence, answer is 1600 × 2 = 3200, 70. When 90Th288 gets converted into 83Bi272, then the number of and -particle emitted will be respectively, (1) 4, 7, , (2) 4, 1, , (3) 8, 7, , (4) 4, 4, , Sol. Answer (2), (288 − 272), = 4 particles, 4, Atomic number will reduce by = 4 × 2 = 8, , Number of alpha particles emitted =, , Number of decays = 1, 71. A radioactive substance has 108 nuclei. Its half life is 30 s. The number of nuclei left after 15 s is nearly, (1) 2 × 105, , (2) 3 × 106, , (3) 7 × 107, , (4) 5 × 108, , Sol. Answer (2), N = N0e–t, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 74 :
178, , Nuclei, , Solution of Assignment, , N = 108 e, , −, , 0.693 × t, T, , N = 108e–0.693/2, N=, , 108, 2, , 72. A certain stable nucleide, after absorbing a neutron, emits -particle and the new nucleide splits spontaneously, into two -particles. The nucleide is, 4, 2 He, , (1), , (2), , 7, 3 Li, , (3), , 6, 4 Be, , (4), , 6, 3 Li, , Sol. Answer (2), Let nucleus be PXA, It adds a neutron and a proton also due to decay, XA + 1, , P+1, , Since it splits into 2 particles, A+1=8, P+1=4, , , P, , XA = 3Li7, , 73. Pauli suggested the emission of nutrino during + decay to explain, (1) Continuous energy distribution of positrons, , (2) Conservation of linear momentum, , (3) Conservation of mass-energy, , (4) All of these, , Sol. Answer (1), Fact., 74. In a nuclear reaction transforming a nucleus into another with the emission of a positron, the neutron proton, ratio, (1) Decreases, , (2) Increases, , (3) Remains same, , (4) May decrease or increase, , Sol. Answer (2), + decays occurs when proton reduces to neutron., Hence,, , N, ratio will increase., P, , 75. If a heavy nucleus has N/Z ratio higher than that required for stability, then, (1) It emits –, , (2) It emits +, , (3) It emits particle, , (4) It will undergo K electron capture, , Sol. Answer (1), – decay will reduce its, , N, ratio, Z, , 76. The half-life of I131 is 8 days. Given a sample of I131 at time t = 0, we can assert that, (1) No nucleus will decay before t = 4 days, , (2) No nucleus will decay before t = 8 days, , (3) All nuclei will decay before t = 16 days, , (4) A given nucleus may decay at t = 0, , Sol. Answer (4), Nuclear reactions are completely spontaneous and unpredictable., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 75 :
Solution of Assignment, , Nuclei, , 179, , 77. Which of the following is used as a moderator in nuclear reactor?, (1) Cadmium, , (2) Plutonium, , (3) Uranium, , (4) Heavy water, , Sol. Answer (4), Fact., 78. Which of the following are suitable for the fusion process?, (1) Light nuclei, (2) Heavy nuclei, (3) Element must be lying in the middle of the periodic table, (4) Middle elements, which are lying on binding energy curve, Sol. Answer (1), Fact., 79. Solar energy is mainly caused due to, (1) Burning of hydrogen in the oxygen, (2) Fission of uranium present in the Sun, (3) Fusion of protons during synthesis of heavier elements, (4) Gravitational contraction, Sol. Answer (3), Fusion reaction in the sun is the fusion of hydrogen nuclei., , SECTION - D, Assertion-Reason Type Questions, 1., , A : Uncertainty principle demands that an electron confined to a nucleus must have very high energy so that, the electron cannot reside in a nucleus., R : The electrostatic attraction between electron and proton is large at such a small distance but is not enough, to bind such a high-energy electron., , Sol. Answer (1), 2., , A : A free proton is stable but inside a nucleus, a proton gets converted into a neutron, positron and neutrino, (p n + e+ + )., R : Inside a nucleus, neutron decay (n p + e– + ) as well as proton decay are possible, since other, nucleons can share energy and momentum to conserve energy as well as momentum and both the decays, are in dynamic equilibrium., , Sol. Answer (1), 3., , A : Exothermic reactions are possible when two light nuclei fuse or when a heavy nucleus undergoes fission, into intermediate mass nuclei., R : The nature of nuclear binding energy curve is such that it rises for lighter nuclei and slightly decreasing, for heavier nuclei., , Sol. Answer (1), 4., , A : For fusion, the light nuclei must have sufficient initial energy to cross the Coulomb barrier. Hence, fusion, requires high temperature, however, the actual temperature required is somewhat less than expected, classically., R : It is due to quantum mechanical tunneling of the potential barrier., , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 76 :
180, 5., , Nuclei, , Solution of Assignment, , A : Only in low or medium energy nuclear reactions, the number of protons and number of neutrons are, separately conserved., R : In high energy reactions, protons and neutrons can be converted into other particles and a new quantum, number, the Baryon number is however, always conserved., , Sol. Answer (2), 6., , A : Nuclear density is almost same for all nuclei., R : The radius (r) of a nucleus depends only on the mass number (A) as r A1/3., , Sol. Answer (1), Mass = A, , v=, , 4 3, πr, 3, , v=, , 4, KA, 3, , Density =, , 7., , r ∝ A1/3, , A, 1, =, = constant, 4, V, πK, 3, , A : During radioactive disintegration an -particle and a-particle do not emit simultaneously from any nucleus., R : An -particle emits from a nucleus when the N/Z ratio is less than the stability range (where N = number, of neutrons and Z = number of protons in a nucleus)., , Sol. Answer (2), The two common modes of radioactive decay are alpha and beta types. In both these decays the other, particle is not ejected., 8., , A : In -decay an electron is emitted by the nucleus., R : Electrons are not present inside the nucleus., , Sol. Answer (2), 9., , A : A radioactive substance has half life of 1 hour. Therefore, if two nuclei of the substance are present initially,, after 1 hour only one will remain undissociated., R : When a nucleus makes a transition from excited state to ground state, it emits a -particle., , Sol. Answer (4), Both the statements are wrong. Nuclear reaction being spontaneous may occur at any time., 10. A : Fast moving neutrons do not cause fission of a uranium nucleus., R : A fast moving neutron spends very little time inside the nucleus., Sol. Answer (1), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 77 :
Chapter, , 27, , Atoms, Solutions, SECTION - A, Objective Type Questions, 1., , An alpha particle colliding with one of the electrons in a gold atom loses, 1, rd of its momentum, 3, , (1) Most of its momentum, , (2), , About, , (3) Little of its energy, , (4), , Most of its energy, , Sol. Answer (3), The mass of an electron is hundred of times lesser than the mass of an alpha particle. Hence the alpha, particles does not transfer much of its energy on collision with the electron., 2., , According to classical theory, Rutherford atom was, (1) Electrostatically stable, , (2), , Electrodynamically unstable, , (3) Semi stable, , (4), , Stable, , Sol. Answer (1), Rutherford designed his theory to be electrostatically stable., 3., , The angular momentum of an electron in a hydrogen atom is proportional to (where r is radius of orbit), (1), , 1, r, , (2), , 1, r, , (3), , r, , (4), , r2, , Sol. Answer (3), The angular momentum of an electron is quantised as, nh, 2π, While radius of the nth orbit is r = n2r0 where, r0 = 0.53 Å, , mvr =, , 4., , When a hydrogen atom is raised from the ground state to third state, (1) Both kinetic energy and potential energy increase, (2) Both kinetic energy and potential energy decrease, (3) Potential energy increases and kinetic energy decreases, (4) Potential energy decreases and kinetic energy increases, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 78 :
116, , Atoms, , Solution of Assignment, , Sol. Answer (3), When hydrogen atom is raised from the ground state to third state, E0, , E=, , 3, , or, , 2, , or PE is, , E=, , E0, 9, , −2E0, E0, and KE is, while initially, PE was –2E0 and KE was E0., 9, 9, , Hence, potential energy increases and kinetic energy decreases., 5., , What is the angular momentum of an electron in Bohr’s hydrogen atom whose energy is –3.4 eV?, (1), , h, , , (2), , 2h, , , (3), , h, 2, , (4), , 1, 4, , Sol. Answer (1), E=−, , E0, n2, , 13.6eV, , 3.4 =, , n2, , n2 =, , 13.6, 3.4, , n2 =, , 68, =4, 17, , or n = 2, Hence mvr =, , 6., , nh, h, or, 2π, π, , E, and 2E respectively., 3, A photon of wavelength is emitted for a transition 3 1. What will be the wavelength of emission for transition, 2 1?, , The energy levels of a certain atom for first, second and third levels are E, 4, , (1), , , 3, , (2), , 3, , (3), , 3, 4, , (4), , 4, 3, , Sol. Answer (2), , hc, E = 2E – E = , 1, hc, E= , 1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 79 :
Solution of Assignment, , 117, , 4E, hc, E , 3, 2, , Similarly, , 7., , Atoms, , , , E hc, , 3 2, , , , = 3, , The ground state energy of H - atom is –13.6 eV. The energy needed to ionise H - atom from its second excited, state is, (1) 1.51 eV, , (2), , 3.4 eV, , (3), , 13.6 eV, , (4), , 12.1 eV, , Sol. Answer (1), Energy for nth excited state =, , E0, , (n + 1)2, , To ionise H-atom from its second excited state =, 8., , 13.6, = 1.51 eV, 9, , If element with principal quantum number n > 4 were not allowed in nature, then the number of possible, elements would be, (1) 60, , (2), , 32, , (3), , 4, , (4), , 64, , Sol. Answer (1), Number of electron possible in a shell is given by, N = 2n2, where n is number of the shell, N1 = 2, , N2 = 8, , N3 = 18, , N4 = 32, , Total N = 2 + 8 + 18 + 32 = 60 atoms, 9., , The angular speed of electron in the nth orbit of hydrogen atom is, (1) Directly proportional to n2, , (2), , Directly proportional to n, , (3) Inversely proportional to n3, , (4), , Inversely proportional to n, , Sol. Answer (3), mvr =, , nh, 2π, , 2, m ωr =, , ω∝, , nh, 2π, , n, r2, , As, r n2, , , ω∝, , 1, n3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 80 :
118, , Atoms, , Solution of Assignment, , 10. Of the various series of the hydrogen spectrum, the one which lies completely in the ultraviolet region is, (1) Lyman series, , (2), , Balmer series, , (3), , Paschen series, , (4), , Brackett series, , Sol. Answer (1), The series will the highest frequencies is the Lyman series. Thus it is almost always in the electromagnetic., 11. As the n (number of orbit) increases, the difference of energy between the consecutive energy levels, (1) Remains the same, , (2), , Increases, , (3) Decreases, , (4), , Sometimes increases and sometimes decreases, , Sol. Answer (3), The difference in energy, between consecutive energy levels keeps reducing as easily predicted by the, equation of the line spectra, for hydrogen atom, , ν=, , R⎛ 1, 1⎞, −, ⎜, c ⎝ nf2 ni2 ⎟⎠, , 12. The magnetic field induction produced at the centre of orbit due to an electron revolving in nth orbit of hydrogen, atom is proportional to, (1) n–3, , (2), , n–5, , (3), , n5, , (4), , n3, , (4), , c/137, , Sol. Answer (2), mvr =, , I = qe, , nh, nh, 1, → m ωr 2 =, →ω∝ 3, 2π, 2π, n, , I, 2π, , B = µ0, , as r n2, , I, 2r, , B=, , μ0 ω, again as r n2, 2π 2r, , , , B∝, , 1, n5, , 13. The speed of an electron in the orbit of hydrogen atom in the ground state is, (1) c, , (2), , c/10, , (3), , c/2, , Sol. Answer (4), The speed of electron in ground state of an hydrogen atom is about 2.2 × 106 m/s. Which is approximately, c, , when c is the speed of light., 137, 14. If the radius of the first orbit of hydrogen atom is 5.29 × 10–11 m, the radius of the second orbit will be, (1) 21.16 × 10–11 metre, , (2), , 15.87 × 10–11 metre, , (3), , 10.58 × 10–11 metre (4), , 2.64 × 10–11 metre, , Sol. Answer (1), Radius of nth orbit = n2r0, Radius of 2nd orbit = 4r0, [r0 5.3 × 10–11 m], Hence radius needed 21.16 × 10–11 m, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 81 :
Solution of Assignment, , Atoms, , 119, , 15. The ratio of minimum to maximum wavelength of radiation emitted by transition of an electron to ground state, of Bohr’s hydrogen atom is, (1), , 3, 4, , (2), , 1, 4, , (3), , 1, 8, , (4), , 3, 8, , Sol. Answer (1), Minimum wavelength (1), 1, ⎛ 1 1⎞, = R⎜ 2 − ⎟, ⎝1, λ1, ∞⎠, , ....(i), , Maximum wavelength (2), 1, 1⎞, ⎛1, = R⎜ 2 − 2 ⎟, ⎝1, λ2, 2 ⎠, , ....(ii), , λ1 3, =, λ2 4, , 16. In Bohr’s model of the hydrogen atom, the ratio between the period of revolution of an electron in the orbit of, n = 1 to the period of revolution of the electron in the orbit n = 2 is, (1) 1 : 2, , (2), , 2:1, , (3), , 1:4, , (4), , 1:8, , (4), , 6.57 × 1015, , Sol. Answer (4), Period of electron in nth orbit, , Tn =, , 2πrn, vn, , Tn =, , 2πr0 n 3, v0, , T1 : T2 = 1 : 8, 17. How many times does the electron go round the first Bohr orbit in a second?, (1) 6.57 × 105, , (2), , 6.57 × 1010, , (3), , 6.57 × 1013, , Sol. Answer (4), T0 in ground state of hydrogen is 1.51 × 10–16 s, Number of revolution = (f) =, , 1, 1, =, 6.57 × 1015, T0 1.5 × 10 −16, , 18. The ratio of the energies of the hydrogen atom in its first excited state to second excited state is, (1) 1/4, , (2), , 4/9, , (3), , 9/4, , (4), , 4, , Sol. Answer (3), Energy in its 1st excited =, , E0, 4, , Energy in its 2nd excited state =, , E0, 9, , E1 : E2 = 9 : 4, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 82 :
120, , Atoms, , Solution of Assignment, , 19. Which of the following transitions in a hydrogen atom emit photons of lowest frequency?, (1) n = 2 to n = 1, , (2), , n = 4 to n = 2, , (3), , n = 4 to n = 3, , (4), , n = 3 to n = 1, , Sol. Answer (3), The pair for which the value, , ⎛ 1, 1⎞, f = cR ⎜ 2 − 2 ⎟, ⎝ nf ni ⎠, is lowest is 4 and 3, 20. The energy of hydrogen-atom in its ground state is –13.6 eV. The energy of the level corresponding to n = 5, is, (1) –0.544 eV, , (2), , –5.40 eV, , (3), , –0.85 eV, , (4), , –2.72 eV, , Sol. Answer (1), Energy in a hydrogen atom varies as E =, E0 = –13.6 eV, E=, , E0, n2, , n=5, , −13.6, 52, , or E = –0.544 eV, 21. If the electron in hydrogen atom jumps from third orbit to second orbit, the wavelength of the emitted radiation, is given by, (1) , , 36, 5R, , (2), , , , 5R, 36, , (3), , , , 5, R, , (4), , , , R, 6, , Sol. Answer (1), , ⎛ 1, 1, 1⎞, = R⎜ 2 − 2 ⎟, λ, ⎝ nf ni ⎠, 1, ⎛ 1 1⎞, = R⎜ − ⎟, ⎝4 9⎠, λ, , 1, ⎛9 − 4⎞, = R⎜, ⎝ 36 ⎟⎠, λ, λ=, , 36, 5R, , 22. When an electron is excited to nth energy state in hydrogen, the possible number of spectral lines emitted, are, (1) n, , (2), , 2n, , (3), , n2 n, 2, , (4), , n2 n, 2, , Sol. Answer (3), n, , C2 , , n(n 1), 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 83 :
Solution of Assignment, , Atoms, , 121, , 23. Which series of hydrogen atom lie in infra red region?, (1) Lyman, , (2), , Balmer, , (3) Brackett, Paschen and Pfund, , (4), , All of these, , Sol. Answer (3), The Brackett, Paschen and Pfund series are series which lie in the infrared region., 24. Using Bohr’s formula for energy quantization, the ionisation potential of the ground state of Li++ atoms is, (1) 122 V, , (2), , 13.6 V, , (3), , 3.4 V, , (4), , 10.2 V, , Sol. Answer (1), Energy in a shell of an atom is given by, , E = E0, , Z2, n2, , ELi = 13.6 × 9, , or, , 122 V, , 25. The total energy of an electron in the hydrogen atom in the ground state is –13.6 eV. The kinetic energy of, this electron is, (1) 13.6 eV, , (2), , 0, , (3), , –13.6 eV, , (4), , 6.8 eV, , Sol. Answer (1), Total energy = –PE + KE, where PE = –2KE, KE = 13.6 eV, 26. The wavelength of first member of Balmer series in hydrogen spectrum is . Calculate the wavelength of first, member of Lyman series in the same spectrum, (1) (5/27), , (2), , (4/27), , (3), , (27/5), , (4), , (27/4), , Sol. Answer (1), 1, 1⎞, ⎛ 1, = R⎜ 2 − 2 ⎟, ⎝2, λ1, 3 ⎠, , [For 1st members of Balmer series], , 1, 1⎞, ⎛1, = R⎜ 2 − 2 ⎟, ⎝1, λ2, 2 ⎠, , [For 1st members of Lyman series], , λ1 36 3, =, ×, λ2, 5 4, λ 2 = λ1 ×, , 5, 27, , 27. Which state of triply ionised beryllium (Be3+) has the same orbital radius as that of the ground state of, hydrogen?, (1) 1, , (2), , 2, , (3), , 3, , (4), , 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 84 :
122, , Atoms, , Solution of Assignment, , Sol. Answer (2), Radius = 0.53, , n2, Å, Z, , For radius of Beryllium to be same as hydrogen's ground state =, , n2, =1, Z, , n2 = 4, n=2, 28. Energy levels A, B, C of a certain atom correspond to increasing values of energy i.e. EA < EB < EC. If 1, 2, and 3 be the wavelength corresponding to the transitions C to B, B to A and C to A respectively, then which, of the following is correct?, (1) 3 = 1 + 2, , (2), , 3 , , 1 2, ( 1 2 ), , (3), , 1 = 23/2 + 3, , (4), , 32/12 + 22, , Sol. Answer (2), EC > EB > EA, , ⎛ 1, 1, 1 ⎞, = R⎜ 2 − 2 ⎟, λ1, ⎝ nB nC ⎠, , ....(i), , ⎛ 1, 1, 1 ⎞, = R⎜ 2 − 2 ⎟, λ2, ⎝ nA nB ⎠, , ....(ii), , ⎛ 1, 1, 1 ⎞, = R⎜ 2 − 2 ⎟, λ3, ⎝ nA nC ⎠, , ⎛ 1, 1, 1, 1 ⎞, (i) + (ii) λ + λ = R ⎜ 2 − 2 ⎟, ⎝ nA nC ⎠, 1, 2, λ3 =, , λ1λ 2, λ1 + λ 2, , 29. The minimum wavelength of the X-rays produced at accelerating potential V is . If the accelerating potential, is changed to 2V, then the minimum wavelength would become, (1) 4, , (2), , 2, , (3), , /2, , (4), , /4, , Sol. Answer (3), , λmin =, , 12400, V, , If potential is changed to 2V, New min =, , λ, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 85 :
Solution of Assignment, , Atoms, , 123, , 30. X-rays of a particular wavelength are used to irradiate sodium and copper surfaces in two separate experiments, and stopping potentials are determined. The stopping potentials are, (1) Equal in both cases, , (2), , Greater for sodium, , (3), , Greater for copper, , (4), , Infinite in both cases, , Sol. Answer (2), The stopping potential is greater for sodium as it is easier to ionise sodium, than copper., 31. An X-ray tube has a short wavelength end at 0.45Å. The voltage of tube is, (1) 450000 V, , (2), , 9600 V, , (3), , 27500 V, , (4), , 60600 V, , Sol. Answer (3), λmin =, , 12400, V, , 32. The frequencies of X-rays, -rays and U.V. rays are respectively a, b and c. Then, (1) a < b, b < c, , (2), , a < b, b > c, , (3), , a > b, b > c, , (4), , a > b, b < c, , Sol. Answer (2), Frequency of the gamma rays as highest followed by X-rays and then UV rays., 33. The wavelength of the K line for an element of atomic number 43 is . The wavelength of the K line for an element, of atomic number 29 is, ⎛ 43 ⎞, ⎟, (1) ⎜, ⎝ 29 ⎠, , (2), , ⎛ 42 ⎞, ⎜, ⎟, ⎝ 28 ⎠, , (3), , ⎛9⎞, ⎜ ⎟, ⎝4⎠, , (4), , ⎛4⎞, ⎜ ⎟, ⎝9⎠, , Sol. Answer (3), , f ∝ (Z − 1), or f (Z – 1)2, , , , , hc, , Z 12, , λ1 (Z2 − 1)2, =, λ 2 (Z1 − 1)2, λ1 ⎛ 28 ⎞, =⎜ ⎟, λ 2 ⎝ 42 ⎠, λ1 ⎛ 2 ⎞, =⎜ ⎟, λ2 ⎝ 3 ⎠, , 2, , 2, , λ1 4, =, λ2 9, λ2 =, , 9, λ1, 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 86 :
124, , Atoms, , Solution of Assignment, , 34. X-rays incident on a material, (1) Will exert a force on it, , (2), , Will transfer energy to it, , (3) May cause emission of electrons, , (4), , All of these, , Sol. Answer (4), All the option are true phenomenon observed in the case of X-rays., 35. Penetrating power of X-rays increases with increase in, (1) Accelerating potential, , (2), , Wavelength, , (3) Mass number of the target material, , (4), , Filament current, , Sol. Answer (1), Penetrating power depends on energy of the x-wave which depends on the accelerating potential of the X-ray tube., 36. If the potential difference V applied to the coolidge tube is doubled, then the cut off wavelength, (1) Is doubled, , (2), , Is halved, , (3) Remains unchanged, , (4), , Is quadrupled, , (1) Highly coherent, , (2), , Highly monochromatic, , (3) Highly directional, , (4), , All of these, , (3), , X-ray, , Sol. Answer (2), , λmin =, , 12400, if V is doubled min is halved., V, , 37. Laser is/are, , Sol. Answer (4), All the above facts are true for a laser., 38. A situation of population inversion is related to, (1) Matter wave, , (2), , -ray, , (4), , LASER, , Sol. Answer (4), A situation of population inversion means when more atoms in a group are in the excited state than lower energy states. This is used in LASER., 39. In He-Ne laser, metastable state exists in, (1) He, , (2), , Ne, , (3), , Both (1) & (2), , (4), , Neither He nor Ne, , Sol. Answer (2), Fact., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 87 :
Solution of Assignment, , Atoms, , 125, , 40. If the emitted radiation falls in the microwave region, the device is termed as, (1) LASER, , (2), , MASER, , (3), , Both (1) & (2), , (4), , None of these, , Sol. Answer (2), Stimulated emitted radiation in the microwave form is called MASER., , SECTION - B, Objective Type Questions, 1., , The wavelength of radiation emitted is 0 when an electron jumps from third to second orbit of hydrogen atom. For, the electron to jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted, will be, (1), , 16, 0, 25, , (2), , 20, 0, 27, , (3), , 27, 0, 20, , (4), , 25, 0, 16, , Sol. Answer (2), 1, 1⎞, ⎛ 1, = R⎜ 2 − 2 ⎟, ⎝2, λ0, 3 ⎠, , 1, 5, =R, λ0, 36, 1, 1⎞, ⎛ 1, = R⎜ 2 − 2 ⎟, ⎝2, λ1, 4 ⎠, , 1, ⎛1 1 ⎞, = R⎜ −, ⎝ 4 16 ⎟⎠, λ1, , 1, 3, =R, λ1, 16, λ1, 5 16, =, ×, λ0 36 3, λ1 =, , 2., , 20, λ0, 27, , An electron in a hydrogen atom makes a transition from n1 to n2. If the time period of electron in the initial, state is eight times that in the final state then, (1) n1 = 3n2, , (2), , n1 = 4n2, , (3), , n1 = 2n2, , (4), , n1 = 5n2, , Sol. Answer (3), Let T1 = n13T0, T2 = n23T0, T1 n13, =, T2 n23, , n1 = 2n2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 88 :
126, 3., , Atoms, , Solution of Assignment, , When a hydrogen atom emits a photon of energy 12.09 eV, its orbital angular momentum changes by (where, h is Planck’s constant), 3h, , , (1), , (2), , 2h, , , (3), , h, , , (4), , 4h, , , Sol. Answer (3), This is the case of electron jumping from 3rd orbit to 1st orbit, n1h nh, −, 2π 2π, , ∆L =, , =, , 4., , 3h h, h, −, =, 2π 2π π, , If the frequency of K X-rays emitted from the element with atomic number 31 is then the frequency of K X-rays, emitted from the element with atomic number 51 would be, 5, , 3, , (1), , (2), , 51, , 31, , (3), , 25, , 9, , (4), , 9, , 25, , Sol. Answer (3), , ν = K α (31 − 1), , ....(i), , ν x = K α (51 − 1), , ....(ii), , Dividing (ii) by (i), , νx, , =, , ν, νx =, , 5., , 50, 30, , 25, ν, 9, , A hydrogen atom is in ground state. In order to get six lines in its emission spectrum, wavelength of incident, radiation should be, (1) 800 Å, , (2), , 825 Å, , (3), , 970 Å, , (4), , 1025 Å, , Sol. Answer (3), To get six possible emission lines the electron must be excite to the third level as, , 6=, , n, (n + 1), 2, , n=3, , 1⎞, ⎛, Frequency of incident radiation is such that hf = 13.6 ⎜1 − 2 ⎟, ⎝ 3 ⎠, or, , hc, 8, = 13.6 ×, λ, 9, , or, , λ=, , hc × 9, 13.6 × 8, , or = 970 Å, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 89 :
Solution of Assignment, , 6., , Atoms, , 127, , The X-ray beam coming from the X-rays tube will be, (1) Monochromatic, (2) Dichromatic, (3) Having all wavelengths greater than a certain minimum wavelength, (4) Having all wavelengths between a minimum and maximum wavelengths, , Sol. Answer (3), Only the minimum wavelength is fixed in the case of an X-ray from X-ray tube. In this case X-rays are formed, from emission by a substance and all possible transitions need to be accounted for. The only limit is the, maximum energy supplied to material by incident light which depends on incident light., 7., , If the energy in the first excited state in hydrogen atom is 23.8 eV then the potential energy of a hydrogen atom in the, ground state can be assumed to be, (1) 10 eV, , (2), , 23.3 eV, , (3), , – 13.6 eV, , (4), , Zero, , Sol. Answer (4), 8., , If energy required to remove one of the two electrons from He atom is 29.5 eV, then what is the value of energy, required to convert a helium atom into -particle?, (1) 54.4 eV, , (2), , 83.9 eV, , (3), , 29.5 eV, , (4), , 24.9 eV, , (4), , Infinite, , (4), , Zero to infinity, , Sol. Answer (2), Energy needed to convert He+ ion to He2+ is = E0Z2, = 13.6 × 4, Hence total energy to create alpha particle of out of helium atom is, Etotal = 13.6 × 4 + 29.5, = 83.9 eV, 9., , The maximum wavelength that a sample of hydrogen atoms can absorb is, (1) 912 Å, , (2), , 1216 Å, , (3), , 1028 Å, , Sol. Answer (2), 1, 1⎞, ⎛1, = R⎜ 2 − 2 ⎟, ⎝1, λmax, 2 ⎠, , Solve for max., 10. The lines in Balmer series have their wavelengths lying between, (1) 1266 Å to 3647 Å, , (2), , 642 Å to 3000 Å, , (3), , 3647 Å to 6563 Å, , Sol. Answer (3), 1, ⎛ 1 1⎞, = R⎜ 2 − ⎟, ⎝2, λmax, ∞⎠, 1, 1⎞, ⎛ 1, = R⎜ 2 − 2 ⎟, ⎝2, λmax, 3 ⎠, , solve for the range, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 90 :
128, , Atoms, , Solution of Assignment, , 11. If an electron in hydrogen atom jumps from third orbit to second orbit, the frequency of the emitted radiation, is given by (c is speed of light), (1), , 3Rc, 29, , (2), , 5Rc, 36, , (3), , 7Rc, 36, , (4), , 8Rc, 31, , Sol. Answer (2), , 1, 1⎞, ⎛ 1, = R⎜ 2 − 2 ⎟, ⎝2, λ, 3 ⎠, ⎛ 1 1⎞, f = Rec ⎝⎜ − ⎠⎟, 4 9, f =, , Rc 5 5Rc, =, 36, 36, , 12. Let F1 be the frequency of second line of Lyman series and F2 be the frequency of first line of Balmer series, then frequency of first line of Lyman series is given by, (1) F1 – F2, , (2), , F1 + F2, , (3), , F2 – F1, , (4), , F1F2, F1 F2, , Sol. Answer (1), , ⎡ 1, 1⎤, F1 = Rc ⎢ 2 − 2 ⎥, ⎣⎢ n1 n3 ⎥⎦, , ....(i), , ⎡ 1, 1⎤, F2 = Rc ⎢ 2 − 2 ⎥, ⎢⎣ n2 n3 ⎥⎦, , ....(ii), , Subtracting (ii) from (i), , ⎡ 1, 1⎤, F = Rc ⎢ 2 − 2 ⎥, ⎣⎢ n1 n2 ⎥⎦, F = F1 – F2, 13. Identify the incorrect relationship, (1) Number of waves in an orbit, n =, , 2r, , , vn, (2) Number of revolutions of an electron per second in nth orbit = 2r, n, , (3) Wavelength of an electron =, , h, p, , (4) Speed of a (de-Broglie wavelength) particle accelerated by a potential difference V is v , , 2eV, m, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 91 :
Solution of Assignment, , Atoms, , 129, , Sol. Answer (4), Kinetic energy gained by an electron accelerated by potential V is, KE = eV, 1, mv 2 = eV, 2, , or, , v=, , 2eV, m, , Hence (4) is wrong., 14. If the difference between (n + 1)th Bohr radius and nth Bohr radius is equal to the (n – 1)th Bohr radius then, find the value of n, (1) 4, , (2), , 3, , (3), , 2, , (4), , 1, , Sol. Answer (1), Bohr radius is given by r = r0n2, [(n + 1)2 – n2]r0 = (n – 1)2r0, , Now, , 2n + 1 = (n – 1)2, n2 – 2n + 1 = 2n + 1, n2 – 4n = 0, n(n – 4) = 0, or n = 4, 15. If radius of first orbit of hydrogen atom is 5.29 × 10–11 m, the radius of fourth orbit will be, (1) 8.46 Å, , (2), , 10.23 Å, , (3), , 9.22 Å, , (4), , 9.48 Å, , Sol. Answer (1), 5.3 × 10–11, or 0.53 Å, Radius of nth orbit = 0.53 × n2, Radius of 4th orbit = 0.53 × 16 Å, = 8.46 Å, 16. Ratio of magnetic dipole moment to the angular momentum for hydrogen like atoms is (e and m are electronic, charge and mass respectively), (1), , e, m, , (2), , 2e, m, , (3), , e, 2m, , (4), , e, 4m, , Sol. Answer (3), Angular momentum of electron = mvr, Magnetic moment of electron., 17. The ratio of energies of hydrogen atom in its first excited state to third excited state is, (1), , 1, 4, , (2), , 4, 1, , (3), , 3, 4, , (4), , 4, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 92 :
130, , Atoms, , Solution of Assignment, , Sol. Answer (2), Energy in nth excited state =, , −13.6, (n + 1)2, , Energy in 1st excited state =, , −13.6, 4, , Energy in 3rd excited state =, , −13.6, 16, , Ratio E1 : E3 = 4 : 1, 18. What should be the ratio of minimum to maximum wavelength of radiation emitted by transition of an electron, to ground state of Bohr’s hydrogen atom?, (1), , 3, 4, , (2), , 1, 4, , (3), , 1, 8, , (4), , 3, 8, , Sol. Answer (1), In maximum wavelength electron falls from 1st excited state., 1, λmax, , 1⎤, ⎡1, = R⎢ 2 − 2⎥, 2 ⎦, ⎣1, , In minimum wavelength the electron comes from infinity, 1, λmin, , 1 ⎤, ⎡1, = R⎢ 2 − 2⎥, ∞ ⎦, ⎣1, , λmax 4, =, Hence λ, 3, min, , 19. The product of angular speed and tangential speed of electron in nth orbit of hydrogen atom is, (1) Directly proportional to n2, , (2), , Directly proportional to n3, , (3) Inversely proportional to n4, , (4), , Independent of n, , Sol. Answer (3), vn, Angular velocity = r, n, , vn, Product of angular velocity and tangential velocity = r × vn, n, or, , or, , v n2, rn, v0, r0 n, , v0, ⎛, ⎞, and rn n 2 r0 ⎟, ⎜∵ v n , n, ⎝, ⎠, , 4, , Hence product , , 1, n4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 93 :
Solution of Assignment, , Atoms, , 131, , 20. Ground state energy of H-atom is –13.6 eV. The energy needed to ionise H-atom from its second excited state, is, (1) 1.51 eV, , (2), , 3.4 eV, , (3), , 13.6 eV, , (4), , 12.1 eV, , Sol. Answer (1), Total energy of nth excited state is =, Energy needed = +, , −13.6, (n + 1)2, , 13.6, ≈ 1.51 eV, 9, , 21. The energy of hydrogen atom in its ground state is –13.6 eV, the energy of the level corresponding to n = 7, is, (1) –0.544 eV, , (2), , –5.40 eV, , (3), , –0.85 eV, , (1) Lyman, , (2), , Balmer, , (3) Brackett, Paschen and Pfund, , (4), , All of these, , (4), , –0.28 eV, , Sol. Answer (4), Energy of nth shell is given by E =, , E=−, , E0, n2, , 13.6, 49, , E –0.28 eV, 22. Which series of hydrogen atom lie in infrared region?, , Sol. Answer (3), In hydrogen atom. The Brackett, Paschen and Pfund series all lie in the infrared region., 23. Total energy of an electron in the hydrogen atom in the ground state is –13.6 eV. The potential energy of this, electron is, (1) 13.6 eV, , (2), , 0, , (3), , –27.2 eV, , (4), , –13.6 eV, , Sol. Answer (3), Total energy =, , Potential energy, = −KE, 2, , Hence potential energy = 2 × –13.6, = –27.2 eV, 24. If potential energy of an electron in a hydrogen atom in first excited state is taken to be zero, kinetic energy, (in eV) of an electron in ground state will be, (1) 13.6 eV, , (2), , 10.2 eV, , (3), , 3.4 eV, , (4), , 5.1 eV, , Sol. Answer (1), If zero of potential energy is changed, KE does not change and continues to be + 13.6 eV., When 2nd shell is taken as reference, ground state PE = –13.6 eV × 2 + 6.8 eV = –20.4 eV, KE =, , –PE, = 10.2 eV, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 94 :
132, , Atoms, , Solution of Assignment, , a, 25. Time period of revolution of an electron in nth orbit in a hydrogen like atom is given by T T0 n . Z = atomic, Zb, number, , (1) T0 = 1.5 × 10–16 s, a = 3, , (2), , T0 = 6.6 × 1015 s, a = 3, , (3) T0 = 1.51 × 10–16 s, b = 3, , (4), , T0 = 6.6 × 1015 s, b = 3, , Sol. Answer (1), , T =, , T0 n a, , Tn =, , Zb, , T0 = 1.51 × 10–16 s, , T0 n 3, Z, a=3, , Which is a fact., 26. Maximum wavelength in balmer series of hydrogen spectrum is, (1) 912 Å, , (2), , 3645 Å, , (3), , 6561 Å, , (4), , 8201 Å, , Sol. Answer (3), 27. In Rutherford’s experiment, number of particles scattered at 90º angle are x per second. Number particles, scattered per second at angle 60º is, (1) x, , (2), , 4x, , (3), , 8x, , (4), , 16 x, , Sol. Answer (2), N() , , Z2, ⎛θ⎞, sin4 ⎜ ⎟ E 2, ⎝2⎠, , or N() , , 1, ⎛θ⎞, sin4 ⎜ ⎟, ⎝2⎠, , = 60º and 90º, , , N (90) sin4 30, , N (60) sin4 45, , N(90º) = x, N(60º) = 4x, 28. Compton effect supports that, (1) X-rays are transverse waves, (2) X-rays have high frequency compared to visible light, (3) X-rays can easily penetrate matter, (4) Photons have momentum, Sol. Answer (4), Compton effect is the scattering of particles due to radiation and prove that photons have momenta., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 95 :
Solution of Assignment, , Atoms, , 133, , 29. Which is the correct relation between de-Broglie wavelength of an electron in the nth Bohr orbit and radius of, the orbit R?, (1) = n2R, , (2), , , , 2R, n, , (3), , , , 4R, n, , (4), , , , 2R, nh, , Sol. Answer (2), λ=, , nh, h, in 2rn =, mv n, mv, , 30. Atomic number of anticathode material in an X-ray tube is 41. Wavelength of K X-ray produced in the tube, is, (1) 0.66 Å, , (2), , 0.76 Å, , (3), , 0.82 Å, , (4), , 0.88 Å, , Sol. Answer (2), , f = K α (Z − 1), , f = (40), f = 2.48 × 1015 × 40, f = 2.48 × 1015 × 1600, , c, = 2.48 × 1015 × 1600, λ, = 0.76 × 10–10, , = 0.76 Å, , or, , 31. Hydrogen atoms are excited from ground state to the principal quantum number 5. Number of spectral lines, observed will be, (1) 5, , (2), , 4, , (3), , 10, , (4), , 8, , Sol. Answer (3), Number of spectral lines are given by (N) =, , n(n − 1), 2, , 5×4, 2, , or N =, , or N = 10 lines, 32. If in Bohr’s atomic model, it is assumed that force between electron and proton varies inversely as r4, energy, of the system will be proportional to, (1) n2, , (2), , n4, , (3), , n6, , (4), , n8, , Sol. Answer (3), dE = Fdr, k, , ∫ dE = ∫ r 4 dr, E=, E=, , k1, , r3, , k2, , n6, , as r n2, , (Where, k, k1 and k2 are constants), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 96 :
134, , Atoms, , Solution of Assignment, , 33. Which of the following may be representing graph between number of scattered particles detected (N) and, scattering angle () in Rutherford’s experiment?, , N, , N, , (1), , N, , (2), , N, , (3), , , , (4), , , , , , , , Sol. Answer (2), , N() =, , k, 4, , sin (θ / 2), , Hence N() decreases with increase in angle ., , SECTION - C, Previous Years Questions, 1., , 3rd, , He+, , Consider, orbit of, (Helium) using non-relativistic approach the speed of electron in this orbit will be (given, K = 9 × 109 constant Z = 2 and h (Planck’s constant) = 6.6 × 10–34 Js)., [AIPMT-2015], (1) 3.0 × 108 m/s, , (2), , 2.92 × 106 m/s, , (3), , 1.46 × 106 m/s, , (4), , 0.73 × 106 m/s, , Sol. Answer (3), For hydrogen like atom, , 2., , Vn , , 2KZe 2, nh, , Vn , , C Z 2.2 106 2, , 1.46 106 m/s, 137 n, 3, , Hydrogen atom in ground state is excited by a monochromatic radiation of = 975 Å. Number of spectral lines, in the resulting spectrum emitted will be, [AIPMT-2014], (1) 3, , (2), , 2, , (3), , 6, , (4), , 10, , Sol. Answer (3), ⎛ 1, ⎛ 1, 1, 1 ⎞, 1, 1 ⎞, ⎟ , R⎜, , ⎟, 1.097 107 ⎜ , 2, 2, , 10, 2, ⎜, ⎟, ⎜, , 975 10, n22 ⎟⎠, ⎝ n1 n2 ⎠, ⎝1, , n2 = 4, Number of spectral line =, 3., , n(n 1), 6, 2, , Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is:, [NEET-2013], (1), , 3, 23, , (2), , 7, 29, , (3), , 9, 31, , (4), , 5, 27, , Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 97 :
Solution of Assignment, , 4., , Atoms, , 135, , Electron in hydrogen atom first jumps from third excited state to second excited state and then from second, excited to the first excited state. The ratio of the wavelengths 1 : 2 emitted in the two cases is, [AIPMT (Prelims)-2012], (1), , 27, 5, , (2), , 20, 7, , 7, 5, , (3), , (4), , 27, 20, , Sol. Answer (2), , 1 1, , 2 9 16, , 1 1, 1, , 4 9, , , 5., , 1 20, , 2, 7, , An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that, the atom acquired as a result of photon emission will be, [AIPMT (Prelims)-2012], , 25 m, (1) 24 hR, , (2), , 24 m, 25 hR, , 24 hR, 25 m, , (3), , (4), , 25 hR, 24 m, , Sol. Answer (3), As per conservation of momentum, Momentum of photon = Momentum of atom, , 6., , , , h, mv, , , , , v, , h, h, ⎡1 1⎤, RZ 2 ⎢ 2 2 ⎥, m m, ⎣1 5 ⎦, , , , v, , hR 24, 25 m, , The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation.Infrared, radiation will be obtained in the transition from, [AIPMT (Mains)-2012], (1) 2 1, , (2), , 32, , (3), , 52, , (4), , 42, , Sol. Answer (4), Energy released in this range is of infrared frequency range., 7., , The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer, series for a hydrogen like ion. The atomic number Z of hydrogen like ion is, [AIPMT (Prelims)-2011], (1) 2, , (2), , 3, , (3), , 4, , (4), , 1, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 98 :
136, , Atoms, , Solution of Assignment, , Sol. Answer (1), There are 2 equation and 2 unknown, 1, 1⎤, ⎡1, = R⎢ 2 − 2⎥, λ, 2 ⎦, ⎣1, , ...(i), , 1, 1⎤, ⎡1, = RZ 2 ⎢ 2 − 2 ⎥, λ, 4 ⎦, ⎣2, , also,, , ....(ii), , on comparing (i) and (ii), we get Z = 2., 8., , An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted, illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron, is 10 V, then the value of n is, [AIPMT (Mains)-2011], (1) 5, , (2), , 2, , (3), , 3, , (4), , 4, , Sol. Answer (4), –13.6 + 10 + 2.75 = –0.85 =, , 13.6, n2, , n=4, 9., , Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according, to Bohr's atomic model?, [AIPMT (Mains)-2011], (1) 13.6 eV, , (2), , 0.65 eV, , (3), , 1.9 eV, , (4), , 11.1 eV, , Sol. Answer (4), Only the fourth is not available for any ionisation., 10. The energy of a hydrogen atom in the ground state is –13.6 eV. The energy of a He+ ion in the first excited, state will be, [AIPMT (Prelims)-2010], (1) –6.8 eV, , (2), , –13.6 eV, , (3), , –27.2 eV, , (4), , –54.4 eV, , Sol. Answer (2), E, , He+, , =, , E0, n, , 2, , Z2, , = −13.6 ×, , 4, 4, , = −13.6 eV, , 1, mv 2 bombards a heavy nuclear target of charge Ze. Then the distance of closest, 2, approach for the alpha nucleus will be proportional to, [AIPMT (Prelims)-2010], , 11. An alpha nucleus of energy, , (1), , 1, Ze, , (2), , v2, , (3), , 1, m, , (4), , 1, v4, , Sol. Answer (3), , r0 , , 1, ze 2, 40 1, mv 2, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 99 :
Solution of Assignment, , Atoms, , 137, , 12. The electron in the hydrogen atom jumps from excited state (n = 3) to its ground state (n = 1) and the photons, thus emitted irradiate a photosensitive material. If the work function of the material is 5.1 eV, the stopping, potential is estimated to be (the energy of the electron in nth state En , (1) 5.1 V, , (2), , 12.1 V, , (3), , 17.2 V, , 13.6, n2, , eV ), , (4), , [AIPMT (Mains)-2010], 7V, , Sol. Answer (4), ⎡1 1 ⎤, Energy of radiation = 13.6 ⎢ 1 − 2 ⎥, ⎣1 3 ⎦, , ⎡8⎤, = 13.6 ⎢ ⎥, ⎣9⎦, = 12.08 eV, KEmax = E – w0, = 12.08 – 5.1, = 6.98 eV, 13. The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited, to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds, to the transition between:, [AIPMT (Prelims)-2009], (1) n = 3 to n = 1 states, , (2), , n = 2 to n = 1 states, , (3) n = 4 to n = 3 states, , (4), , n = 3 to n = 2 states, , Sol. Answer (3), The shell as state of an atom is, , n ⋅ (n − 1), =6, 2, , n=4, The lowest energy change will correspond the maximum wavelength., 14. In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears, as a result of, [AIPMT (Prelims)-2008], (1) Collision between different electrons of the atoms of the gas, (2) Excitation of electrons in the atoms, (3) Collision between the atoms of the gas, (4) Collisions between the charged particles emitted from the cathode and the atoms of the gas, Sol. Answer (2), 15. The ground state energy of hydrogen atom is –13.6 eV. When its electron is in the first excited state, its, excitation energy is, [AIPMT (Prelims)-2008], (1) Zero, , (2), , 3.4 eV, , (3), , 6.8 eV, , (4), , 10.2 eV, , Sol. Answer (4), 1⎞, ⎛, Energy absorbed for electron to go to 2nd shell is E = 13.6 ⎜1 − 2 ⎟, ⎝ 2 ⎠, , E = 10.2 eV, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 100 :
138, , Atoms, , Solution of Assignment, , 16. The total energy of electron in the ground state of hydrogen atom is –13.6 eV. The kinetic energy of an electron, in the first excited state is, [AIPMT (Prelims)-2007], (1) 1.7 eV, , (2), , 3.4 eV, , (3), , 6.8 eV, , (4), , 13.6 eV, , Sol. Answer (2), KE =, , KE =, , −PE, 2, 13.6, 2 × n2, , n=2, , = 1.7 eV, 17. Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic, radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be:, [AIPMT (Prelims)-2006], (1) Two, , (2), , Three, , (3), , Four, , (4), , One, , Sol. Answer (2), Electrons are excited by 12.1 eV radiation. New PE of electron = –(13.6 – 12.1) = –1.5 eV, This is the energy corresponding to n = 3, Hence number of spectral lines, , n(n − 1), =3, 2, , 18. In a discharge tube ionization of enclosed gas is produced due to collisions between :, [AIPMT (Prelims)-2006], (1) Positive ions and neutral atoms/molecules, , (2), , Negative electrons and neutral atoms/molecules, , (3) Photons and neutral atoms/molecules, , (4), , Neutral gas atoms/molecules, , Sol. Answer (2), 19. Energy levels A, B and C of a certain atom correspond to increasing values of energy i.e.,, EA < EB < EC. If 1, 2 and 3 are wavelengths of radiations corresponding to transitions C to B, B to A and C, to A respectively, which of the following relations is correct ?, [AIPMT (Prelims)-2005], (1) 3 = 1 + 2, , (2), , 1 2, 3 = , 1, 2, , (3), , 1 + 2 + 3 = 0, , (4), , 32 12 22, , Sol. Answer (2), , ⎡ 1, 1, 1⎤, = R⎢ 2 − 2 ⎥, λ3, ⎣⎢ nC nA ⎥⎦, , ⎡ 1, 1, 1⎤, = R⎢ 2 − 2 ⎥, λ1, ⎣⎢ nC nB ⎥⎦, , ⎡ 1, 1, 1⎤, = R⎢ 2 − 2 ⎥, λ2, ⎣⎢ nB nA ⎥⎦, 1, 1, 1, =, +, λ3 λ 2 λ1, , λλ, 1, = 1 2, λ3 λ1 + λ 2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 101 :
Solution of Assignment, , Atoms, , 139, , 20. The total energy of an electron in the first excited state of hyrogen is about – 3.4 eV. Its kinetic energy in this, state is :, [AIPMT (Prelims)-2005], (1) –3.4 eV, , (2), , –6.8 eV, , (3), , 6.8 eV, , (4), , 3.4 eV, , Sol. Answer (4), 21. In a Rutherford scattering experiment when a projectile of charge z1 and mass M1 approaches a target nucleus, of charge z2 and mass M2, the distance of closest approach is r0. The energy of the projectile is, (1) Directly proportional of mass M1, , (2), , Directly proportional of M1 × M2, , (3) Directly proportional of z1z2, , (4), , Inversely proportional to z1, , Sol. Answer (3), 22. An electron makes a transition from orbit n = 4 to the orbit n = 2 of a hydrogen atom. What is the wavelength, of the emitted radiations? (R = Rydberg’s constant), (1), , 16, 4R, , 16, 5R, , (2), , (3), , 16, 2R, , (4), , 16, 3R, , Sol. Answer (4), Transition is from n = 4 to n = 2 of hydrogen atom, 1, 1⎞, ⎛ 1, = R⎜ 2 − 2 ⎟, ⎝, λ, 2, 4 ⎠, , 1 R, =, λ 16, λ=, , 16, 3R, , 23. When a hydrogen atom is raised from the ground state to an excited state,, (1) Both K.E. and P.E. increase, , (2), , Both K.E. and P.E, decrease, , (3) The P.E. decreases and K.E. increases, , (4), , The P.E. increases and K.E. decreases, , Sol. Answer (4), The potential energy increases when the electron is taken to a higher shell., The kinetic energy or velocity of the electron decreases., 24. The figure indicates the energy level diagram of an atom and the origin of six spectral lines in emission (e.g., line number 5 arises from the transition from level B to A). Which of the following spectral lines will also occur, in the absorption spectrum?, C, B, A, , 1 2 3 4 5 6, (1) 4, 5, 6, , (2), , 1, 2, 3, 4, 5, 6, , (3), , X, 1, 2, 3, , (4), , 1, 4, 6, , Sol. Answer (3), Only 1, 2, 3rd lines will be there in the absorption spectrum as those are the only line which include the ground, state. In absorption a non-excited or ground-state atom absorbs energy and the electron goes into an excited, state., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 102 :
140, , Atoms, , Solution of Assignment, , 25. The energy of a hydrogen atom in its ground state is – 13.6 eV. The energy of the level corresponding to the, quantum number n = 2 in the hydrogen atom is, (1) 0.54 eV, , (2), , – 3.4 eV, , (3), , – 2.72 eV, , (4), , – 0.85 eV, , Sol. Answer (2), Energy in ground state hydrogen atom = –13.6 eV, , In n = 2, E = –, , 13.6, n2, , = –3.4 eV, , 26. According to Bohr’s principle, the relation between principal quantum number (n) and radius of orbit is, (1) r , , 1, n, , (2), , r, , 1, n2, , rn, , (3), , (4), , r n2, , Sol. Answer (4), r = 0.53n2, Here r n2, 27. When hydrogen atom is in its first excited level, its radius is ……………. of the Bohr radius., (1) Twice, , (2), , 4 times, , (3), , Same, , (4), , Half, , Sol. Answer (2), r = 0.53n2, At first excited level, n = 2, r = 0.53 × 4, r = 2.1 Å, 28. Atomic weight of Boron is 10.81 and it has two isotopes, nature would be, (1) 15 : 16, , (2), , 10 : 11, , (3), , B10 and, , 5, , 19 : 81, , B11. Then the ratio of 5B10 : 5B11 in, , 5, , (4), , 81 : 19, , Sol. Answer (3), Taking weighed mean, 10.81 = 10x + 11(1 – x), 10.81 = 10x +11 – 11x, 10.81 = 11 – x, x = 11 – 10.81, x = 0.19, or 19%, (x) : (1 – x) = 19 : 81, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 103 :
Solution of Assignment, , Atoms, , 141, , 29. In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between, the proton and the electron. If a0 is the radius of the ground state orbit, m is the mass and e is the charge, on the electron and 0 is the vacuum permittivity, the speed of the electron is, e, , (1), , 4 0 a0 m, , (2), , e, 0 a0 m, , (3), , Zero, , (4), , 4 0 a0 m, e, , Sol. Answer (1), mv 2, 1 e2, =, 4 πε0 a02, a0, , e, , v=, , 4 πε0 a0 ⋅ m, , 30. Maximum frequency of emission is obtained for the transition, (1) n = 2 to n = 1, , (2), , n = 6 to n = 2, , (3), , n = 1 to n = 2, , (4), , n = 2 to n = 6, , Sol. Answer (1), Maximum frequency is obtained for maximum energy difference in levels, , ⎛ 1, 1⎞, E = 13.6 ⎜ 2 − 2 ⎟, ⎝ nf ni ⎠, The values of nf and ni for which this is maximum is nf = 1 and ni = 2., 31. When an electron do transition from n = 4 to n = 2, then emitted line in spectrum will be, (1) First line of Lyman series, , (2), , Second line of Balmer series, , (3) First line of Paschen series, , (4), , Second line of Paschen series, , Sol. Answer (2), Fact., 32. The energy of hydrogen atom in nth orbit is En then the energy in nth orbit of singly ionised helium atom will, be, (1) 4En, , (2), , En/4, , (3), , 2En, , (4), , En/2, , Sol. Answer (1), Energy in hydrogen like atom is given by En × Z2, Since Z = 2 in helium, E = En × 4, 33. In which of the following systems will the radius of the first orbit (n = 1) be minimum?, (1) Doubly ionized lithium, , (2), , Singly Ionized helium, , (3) Deuterium atom, , (4), , Hydrogen atom, , Sol. Answer (1), , r ∝, , 1, Z, , Since in lithium Z = 3, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 104 :
142, , Atoms, , 34., , The Bohr model of atoms, , Solution of Assignment, , (1) Assumes that the angular momentum of electrons is quantized, (2) Uses Einstein’s photo-electric equation, (3) Predicts continuous emission spectra for atoms, (4) Predicts the same emission spectra for all types of atom, Sol. Answer (1), Fact., 13.6, eV . The energy of a, n2, photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately, , 35. Energy E of a hydrogen atom with principal quantum number n is given by E , , (1) 1.5 eV, , (2), , 0.85 eV, , (3), , 3.4 eV, , (4), , 1.9 eV, , 1: 8, , (4), , 1:3, , 13.6, eV, (11)2, , (4), , 13.6 × (11)2 eV, , Sol. Answer (4), 1⎤, ⎡1, E = −13.6 ⎢ 2 − 2 ⎥ = 1.9 eV, 3 ⎦, ⎣2, , 36. The ratio of radii of first shell of H atom and that of fourth shell of He+ ion is, (1) 1 : 8, , (2), , 1:4, , (3), , Sol. Answer (1), r, , n2, Z, , If rH is 1 rHe =, , 42, =8, 2, , rH : rHe = 1 : 8, 37. The ionisation energy of 10 times ionised sodium atom is, , (1) 13. 6 eV, , (2), , 13.6 × 11 eV, , (3), , Sol. Answer (4), E = E0 × Z2 for hydrogen like atom., 38. The wavelength of radiation emitted is 0 when an electron jumps from third to second orbit of hydrogen atom. For, the electron to jump from the fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted, will be, , (1), , 16, 0, 25, , (2), , 20, 0, 27, , (3), , 27, 0, 20, , (4), , 25, 0, 16, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 105 :
Solution of Assignment, , Atoms, , 143, , Sol. Answer (2), 1, 1 ⎤ 5R, ⎡1, = R⎢ 2 − 2⎥ =, λ0, 3 ⎦ 36, ⎣2, , 1, 1⎤, ⎡1, ⎡ 4 − 1⎤, = R⎢ 2 − 2⎥ = R⎢, ⎥, λ, 4 ⎦, ⎣2, ⎣ 16 ⎦, , λ=, , or, , 1 R3, =, λ 16, , 20, λ0, 27, , 39. The ratio of wavelengths of the 1st line of Balmer series and that of the 1st line of Paschen series is, (1) 20 : 7, , (2), , 7 : 20, , (3), , 7:4, , (4), , 4:7, , (3), , 1, R, , (4), , 3, 4R, , Sol. Answer (2), 1, 1⎤, ⎡1, = R⎢ 2 − 2⎥, λB, 2 ⎦, ⎣1, 1, 1⎤, ⎡1, = R⎢ 2 − 2⎥, λP, 4 ⎦, ⎣3, , Solve B : P, 40. The shortest wavelength of Balmer series of H-atom is, , (1), , 4, R, , (2), , 36, 5R, , Sol. Answer (1), 1, λmin, , 1, λmin, , ⎡ 1 1⎤, = R⎢ 2 − ⎥, ∞⎦, ⎣1, , =, , 1, R, , 41. An electron in a hydrogen atom makes a transition from n = n1 to n = n2. The time period of the electron in the initial, n1, state is eight times that in the final state. Find the ratio n, 2, , (1) 2, , (2), , 3, , (3), , 4, , (4), , 8, , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 106 :
144, , Atoms, , Solution of Assignment, , 42. Assuming Bohr’s model for Li++ atom, the first excitation energy of ground state of Li++ atom is, (1) 10.2 eV, , (2), , 91.8 eV, , (3), , 13.6 eV, , (4), , 3.4 eV, , Sol. Answer (2), 1⎤, ⎡1, E = 13.6 × Z2 ⎢ 2 − 2 ⎥, 2 ⎦, ⎣1, , Put Z = 3 and solve for energy to get the answer., 43. The absorption transition between the first and the fourth energy states of hydrogen atom are 3. The emission, transition between these states will be, (1) 3, , (2), , 4, , (3), , 5, , (4), , 6, , Sol. Answer (4), The emission transition lines between any two shells is given by, , n(n + 1), 2, , Since there are 3 absorption lines the shell 3 is mentioned here., 44. When a hydrogen atom emits a photon of energy 12.1 eV, its orbital angular momentum changes by (where, h is Planck’s constant), (1), , 3h, , , (2), , 2h, , , (3), , h, , , (4), , 4h, , , Sol. Answer (3), This is the transition from 3rd shell to the first, Change in angular momentum =, , 3h h, h, −, =, 2π 2π π, , SECTION - D, Assertion-Reason Type Questions, 1., , A : Both the Thomson's as well as the Rutherford's models constitute an unstable system., R : Thomson's model is unstable electrostatically while Rutherford's model is unstable because of, electromagnetic radiation of orbiting electrons., , Sol. Answer (1), 2., , A : Bohr's orbits are regions where the electron may be found with large probability., R : The orbital picture in Bohr's model of the hydrogen atom was inconsistent with the uncertainty principle., , Sol. Answer (1), 3., , A : Bohr's model with its planet-like electron is not applicable to many electron atoms., R : Unlike the situation in the solar system, where planet-planet gravitational forces are very small as compared, to the gravitational force of the sun on each planet, the electron-electron electric force interaction is, comparable in magnitude to the electron nucleus electric force., , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 107 :
Solution of Assignment, , 4., , Atoms, , 145, , A : In Bohr model, the frequency of revolution of an electron in its orbit is not connected to the frequency of, spectral line for smaller principal quantum number n., R : For transitions between large quantum number the frequency of revolution of an electron in its orbit is, connected to the frequency of spectral line, as per Bohr's Correspondence principle., , Sol. Answer (2), , 5., , A : Spectral analysis can differentiate between isotopes as per the equation, , 1 ⎤, ⎡ 1, 1, RZ 2 ⎢ 2 2 ⎥ ., , ⎣ n1 n2 ⎦, , R : Rydberg's constant R is not a universal constant and is dependent on the mass of nuclei as well., Sol. Answer (1), 6., , A : If the accelerating potential in an X-ray machine is decreased, the minimum value of the wavelength of the, emitted X-rays gets increased., R : The minimum value of the wavelength of the emitted X-rays is inversely proportional to the accelerating, potential., , Sol. Answer (1), 7., , A : According to Bohr’s atomic model the ratio of angular momenta of an electron in first excited state and, in ground state is 2 : 1., R : In a Bohr’s atom the angular momentum of the electron is directly proportional to the principal quantum, number., , Sol. Answer (1), 8., , A : If a beam of photons of energy 10.0 eV each, is incident on a sample of hydrogen gas containing all atoms, in the ground state, then the beam of the photons is completely transmitted through the gas without, absorption., R : The minimum energy required by an electron to make a transition to an excited state is, 10.2 eV., , Sol. Answer (1), 9., , A : The nature of the characteristic X-rays does not depend on accelerating potential., R : X-rays are electromagnetic radiation., , Sol. Answer (2), The nature of the characteristic X-rays does not depend on characteristic accelerates potential but on the, material. Both statements are true, independent facts., 10. A : If vacuum is not created inside an X-ray tube, X-rays will not be produced., R : Without vacuum inside the X-ray tube the electrons are not emitted by the filament., Sol. Answer (3), If vacuum is created inside the tube the emitted electrons will not hit the molecules of air to produce X-rays., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 108 :
146, , Atoms, , Solution of Assignment, , 11. A : Gases are insulators at ordinary pressure but they start conducting at very low pressure., R : At low pressures, ions have a chance to reach their respective electrodes and constitute a current but at, ordinary pressures, ions undergo collision with gas molecules and recombination., Sol. Answer (1), 12. A : The oil-drops of Millikan’s experiment should be of microscopic size., R : For larger drops the electric fields needed in the experiment will be impractically high., Sol. Answer (1), 13. A : Stoke’s formula for viscous drag is not really valid for oil-drops of extremely minute sizes., R : Stoke’s formula is valid for motion through a homogeneous continuous medium and the size of the drop, should be much larger than the intermolecular separation in the medium for this assumption to be valid., Sol. Answer (1), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 109 :
Chapter, , 26, , Dual Nature of Radiation and Matter, Solutions, SECTION - A, Objective Type Questions, 1., , When photon of energy 3.8 eV falls on metallic surface of work function 2.8 eV, then the kinetic energy of, emitted electrons are, (1) 1 eV, , (2), , 6.6 eV, , (3), , 0 to 1 eV, , (4), , 2.8 eV, , (1) Is different for different materials, , (2), , Is same for all metals, , (3) Depends upon frequency of the incident light, , (4), , Depends upon intensity of the incident light, , Sol. Answer (3), h = w0 + KEmax, 3.8 – 2.8 = KEmax, KEmax = 1 eV, 2., , The photoelectric work function, , Sol. Answer (1), Work function is the property of material., 3., , The phenomenon of photoelectric effect was first explained by, (1) Hallwach, , (2), , Einstein, , (3), , Planck, , (4), , Bohr, , Sol. Answer (2), Einstein won the Nobel Prize for predicting photoelectric effect by proposing the dual nature of light., 4., , The energy of the most energetic photoelectrons emitted from a metal target depends upon, (1) Threshold frequency of the metal, , (2), , Photoelectric work function of the metal, , (3) Wavelength of the incident radiation, , (4), , All of these, , Sol. Answer (4), h – w0 = KEmax, Hence, all options are correct., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 110 :
84, 5., , Dual Nature of Radiation and Matter, , Solution of Assignment, , Threshold wavelength for sodium is 6 × 10–7m. Then photoemission occurs for light of wavelength if, (1) 6 × 10–7 metre, , (2), , 6 × 10–7 metre, , (3) = 5 × 1014 metre, , (4), , Frequency 5 × 1014 hertz, , Sol. Answer (2), threshold, 6., , In photoelectric effect, the slope of stopping potential versus frequency of incident light for a given surface will, be, (1) h e–1, , (2), , eh, , (3), , e, , (4), , h, , Sol. Answer (1), V0 , , 7., , w, h, f 0, e, e, , When a metallic surface is illuminated with light of wavelength , the stopping potential is x volt. When the, same surface is illuminated by light of wavelength 2, the stopping potential is, , x, . Threshold wavelength for, 3, , the metallic surface is, (1), , 4, 3, , (2), , 4, , (3), , 6, , (4), , 8, 3, , Sol. Answer (2), From Einstein photoelectric equation., h.v = + Kmax, hc, = + ev0, , , (v0 = stopping potential), , hc, = + ex, , x, hc, = e, 2, 3, , On solving, (work function) =, , hc, 4, , hc, hc, 0 = 4, 0 = 4., 8., , A monochromatic point source of light is placed at a distance d from a metal surface. Photo electrons are, ejected at a rate n per second, and with maximum kinetic energy E. If the source is brought nearer to, distance d/2, the rate and the maximum kinetic energy per photoelectron become nearly, (1) 2n and 2E, , (2), , 4n and 4E, , (3), , 4n and E, , (4), , n and 4E, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 111 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 85, , Sol. Answer (3), Intensity , , 1, d2, , So when the source of light is brought closer, Intensity becomes 4 times., Hence, rate Intensity the rate will become 4n., KEmax depends on incident frequency which does not change and will remain E., 9., , Given that a photon of light of wavelength 10000 Å has energy 1.23 eV. Now when light of intensity I0 and, wavelength 5000 Å falls on a photo cell the saturation current and stopping potential are 0.40 A and 1.36 V, respectively. The work function is, (1) 0.43 eV, , (2), , 1.10 eV, , (3), , 1.36 eV, , (4), , 2.47 eV, , Sol. Answer (2), = 10000 Å has E = 1.23 eV, ∵ Energy of 5000 Å photon = 2.46 eV, 2.46 = + 1.36, = 1.1 ev, 10. If the frequency of light incident on a metallic plate be doubled, how will the maximum kinetic energy of the, photoelectrons change?, (1) It becomes more than double, , (2), , It becomes less than double, , (3) It becomes exactly double, , (4), , It does not change, , Sol. Answer (1), hf = w0 + KEmax, KEmax = hf – w0, Since w0 is constant., KEmax will become more than double, 11. Wave nature of light cannot explain photoelectric effect because in photoelectric effect, it is seen that, (1) For the frequency of light below a certain value, the photoelectric effect does not take place, irrespective, of intensity, (2) Maximum kinetic energy of ejected electrons is independent of intensity of radiation, (3) There is no time lag between the incidence of radiation and emission of electrons, (4) All of these, Sol. Answer (4), (Laws of photo electric emission), 12. The photoelectric threshold for a certain metal surface is 330 Å. What is the maximum kinetic energy of the, photoelectrons emitted, if radiations of wavelength 1100 Å are used ?, (1) 1 eV, , (2), , 2 eV, , (3), , 7.5 eV, , (4), , No electron is emitted, , Sol. Answer (4), Since incident is greater than threshold wavelength. No electron will be ejected., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 112 :
86, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 13. When a point source of light is at a distance of 50 cm from a photoelectric cell, the cut-off voltage is found to be V0., If the same source is placed at a distance of 1 m from the cell, then the cut-off voltage will be, (1) V0/4, , (2), , V0/2, , (3), , V0, , (4), , 2 V0, , Sol. Answer (3), Stopping potential depends on incident light and will not change., 14. In photoelectric effect when photons of energy h fall on a photosensitive surface (work function h0) electrons, are emitted from the metallic surface. It is possible to say that, (1) All ejected electrons have same kinetic energy equal to h – h0, (2) The ejected electrons have a distribution of kinetic energy from zero to (h – h0), (3) The most energetic electrons have kinetic energy equal to h, (4) All ejected electrons have kinetic energy h0, Sol. Answer (2), KEmax = h – h0 and hence the energies of electrons can range anywhere between these values., 15. Light of frequency 1.5 times the threshold frequency is incident on photo-sensitive material. If the frequency, is halved and intensity is doubled, the photo current becomes, (1) Quadrupled, , (2), , Doubled, , (3), , Halved, , (4), , Zero, , Sol. Answer (4), If frequency is halved, the initial frequency will become less than incident frequency, 16. The work function of a substance is 4eV. The longest wavelength of light that can cause the emission of, photoelectrons from this substance is approximately, (1) 540 nm, , (2), , 400 nm, , (3), , 310 nm, , (4), , 220 nm, , Sol. Answer (3), hc, = , 0, , 17. In photoelectric effect, the curve between photoelectric current and anode potential V (for different frequencies), is shown in figure, then, , Photoelectric current, v3, , v2 v, 1, , –V03 –V02 –V01, (1) v1 > v2 > v3, , (2), , v1 < v2 < v3, , Anode potential V, (3), , v1 = v2 = v3, , (4), , v1 > v2 < v3, , Sol. Answer (2), As is evident from the graph v1 < v2 < v3, V1, V2, V3 in this case are the stopping potential., 18. X-rays of wavelength 22 pm are scattered from a carbon target at an angle of 85° to the incident beam. The, Compton shift for X-rays is, (1) 2.2 pm, , (2), , 1.1 pm, , (3), , 0.55 pm, , (4), , 4.4 pm, , Sol. Answer (1), λ' − λ =, , h, (1 − cos φ ), m0 c, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 113 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 87, , 19. The cathode of a photocell is changed such that the work function changes from w1 to w2 (w2 > w1). If the saturation, currents before and after the change are I 1 and I 2 and all other conditions are unchanged, then, (assuming h > w2), (1) I1 = I2, , (2), , I1 < I2, , (3), , I1 > I2, , (4), , I1 < I2 < 2I2, , (4), , It has no momentum, , Sol. Answer (1), Saturation current depends on intensity which does not change. Hence, I1 = I2, 20. If the energy of a photon is E, then its momentum is (c is velocity of light), (1) E/c, , (2), , E/2c, , (3), , 2E/c, , Sol. Answer (1), , h, hc, for photon, = E and momentum is, λ, λ, P of a photon =, , E, C, , 21. The de Broglie wavelength of an electron in the nth Bohr orbit is related to the radius R of the orbit as, (1) n = R, , (2), , n , , 3, R, 2, , (3), , n = 2R, , (4), , n = 4R, , Sol. Answer (3), n = 2r (According to Bohr's theory), 22. A photon and an electron both have wavelength 1Å. The ratio of energy of photon to that of electron is, (1) 1, , (2), , 0.012, , (3), , 82.7, , (4), , 10–10, , (4), , 1 fermi, , Sol. Answer (3), E(Photon) =, , hc, , , E(electron) =, , P2, 1, ⎛h⎞, ⎜ ⎟, 2 me ⎝ ⎠ 2 me, , 2, , 23. Wavelength of an electron accelerated through a potential difference of 1 volt is, (1) 12.27 Å, , (2), , 1.234 Å, , (3), , 1 micron, , Sol. Answer (1), Energy gained by one electron = 1 eV, P = 2mE and λ =, , Hence, λ =, , , , λ=, , h, P, , h, 2mE, , h, 2me, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 114 :
88, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 24. For a proton accelerated through V volts, de Broglie wavelength is given as =, (1), , 12.27, V, , Å, , 0.101, , (2), , V, , Å, , 0.286, , (3), , V, , Å, , (4), , 12400, Å, V, , (4), , Charge, , Sol. Answer (3), Energy gained by proton = 1 eV, , h, , λ=, , 2mE, , or, , λ=, , h, 2m e, , 1, , ×, , V, , 25. The wavelength of matter waves is independent of, (1) Mass, , (2), , Velocity, , (3), , Momentum, , Sol. Answer (4), The wavelength of matter waves is given by =, , h, mv, , Hence, it is independent of charge., 26. Which one of the following statements is not true about de-Broglie waves ?, (1) All atomic particles in motion have waves of a definite wavelength associated with them, (2) The higher the momentum, the longer is the wave-length, (3) The faster the particle, the shorter is the wave-length, (4) For the same velocity, a heavier particle has a shorter wavelength, Sol. Answer (2), =, , h, . Hence, higher the momentum shorter is the wavelength., P, , 27. A proton and an alpha particle are accelerated through the same potential difference. The ratio of de Broglie, wavelength of the proton to that of the alpha particle will be, (1) 2 : 1, , (2), , 1:1, , (3), , 1:2, , (4), , 22 : 1, , (4), , 2:1, , Sol. Answer (4), Energy gained by protons = 1 eV, Energy gained by alpha particle = 2 eV, , λP =, , h, 2mE, , λα =, , h, 2 × 4m × 2 eV, , λP 2 2, =, λα, 1, 28. The ratio of specific charge e/m of a proton to that of an -particle is, (1) 1 : 4, , (2), , 1:2, , (3), , 4:1, , Sol. Answer (4), Charge by mass ratio of proton =, , e, mP, , Charge by mass ratio of particle =, , 2e, 4mP, , Ratio of specific charge = 2 : 1, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 115 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 89, , 29. Choose the only correct statement out of the following, (1) Only a charged particle in motion is accompanied by matter waves, (2) Only subatomic particles in motion are accompanied by matter waves, (3) Any particle in motion, whether charged or uncharged, is accompanied by matter waves, (4) No particle, whether at rest or in motion, is ever accompanied by matter waves, Sol. Answer (3), 30. Neglecting variation of mass with velocity, the wavelength associated with an electron having a kinetic energy, E is proportional to, 1, , (1) E 2, , (2), , E, , (3), , E, , , , 1, 2, , (4), , E–2, , Sol. Answer (3), , λ=, , h, 2mE, , E–1/2, 31. Of the following having the same kinetic energy, the one which has the largest wavelength is, (1) An alpha particle, , (2), , A neutron, , (3), , A proton, , (4), , An electron, , Sol. Answer (4), , λ=, , λ∝, , h, 2mE, 1, 2m, , Hence, answer is electron., 32. The de Broglie wavelength associated with a proton increases by 25%, if its momentum is decreased by, p0. The initial momentum was, (1) 4 p0, , (2), , p0, 4, , (3), , 5 p0, , (4), , p0, 5, , Sol. Answer (3), 2 1, 25, , 1, 100, , 2, P1, 5, 5, , , , 1, P, 4, 4, 2, P2 – P1 = –P0, 4P1, P1 = –P0, 5, , P1 = 5P0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 116 :
90, , Dual Nature of Radiation and Matter, , Solution of Assignment, , SECTION - B, Objective Type Questions, 1., , If in a photoelectric cell, the wavelength of incident light is changed from 4000 Å to 3000 Å then change in stopping, potential will be, (1) 0.66 V, , (2), , 1.03 V, , (3), , 0.33 V, , (4), , 0.49 V, , Sol. Answer (2), Wavelength is changed from 4000 Å to 3000 Å, hc, = KEmax + w 0, λ1, hc, = KEmax(2) + w 0, λ2, hc hc, −, = KEmax − KEmax(2), λ1 λ 2, hc hc, −, = eV, λ1 λ 2, , ⎛ hc hc ⎞ 1, ⎜⎝ λ − λ ⎟⎠ × e = V, 1, 2, , V = 1.03 V, 2., , Find the number of electrons emitted per second by a 24 W source of monochromatic light of wavelength 6600 Å,, assuming 3% efficiency for photoelectric effect (take h = 6.6 × 10–34 Js), (1) 48 × 1019, , (2), , 48 × 1017, , (3), , 8 × 1019, , (4), , 24 × 1017, , Sol. Answer (4), Energy per photon = hf or, , hc, λ, , Number of photon's per second =, , Number of electrons emitted =, , 3., , 24, hf, , 3, 24, ×, ×λ, 100 hc, , In photoelectric effect, if a weak intensity radiation instead of strong intensity of suitable frequency is used then, (1) Photoelectric effect will get delayed, , (2), , Photoelectric effect will not take place, , (3) Maximum kinetic energy will decrease, , (4), , Saturation current will decrease, , Sol. Answer (4), Saturation current which depends on intensities will decrease., 4., , The work function of tungsten is 4.50 eV. The wavelength of fastest electron emitted when light whose photon, energy is 5.50 eV falls on tungsten surface, is, (1) 12.27 Å, , (2), , 0.286 Å, , (3), , 12400 Å, , (4), , 1.227 Å, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 117 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 91, , Sol. Answer (3), KEmax =, , hc, − w 0 = 5.5 – 4.5, λ, , KEmax = 1 eV, , h, , λ=, 5., , 2mE, , A proton is accelerated through 225 V. Its de Broglie wavelength is, (1) 0.1 nm, , (2), , 0.2 nm, , (3), , 0.3 nm, , (4), , 0.4 nm, , (4), , 1 2m, c E2, , Sol. Answer (2), Energy it gains = 225 eV, , λ=, 6., , h, =, mv, , λh, 2mE, , For same energy, find the ratio of photon and electron (Here m is mass of electron), (1) c, , 2m, E, , 1 2m, c E, , (2), , (3), , 1, c2, , 2m, E, , Sol. Answer (1), Energy of photon =, , λP =, , hc, λP, , hc, E, , λe =, , h, 2mE, , λP, 2m, =c, λe, E, 7., , What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength, 5200 Å?, (1) 700 m/s, , (2), , 1000 m/s, , v=, , λ, λme, , (3), , 1400 m/s, , (4), , 2800 m/s, , Sol. Answer (3), Momentum of photon =, h, = mev, λ, , 8., , or, , h, λ, , Figure shows four situations in which an electron is moving in electric / magnetic field. In which case the de Broglie, wavelength of electron is increasing?, , –, (2), , (1), , E, Sol. Answer (2), , , –, , –, , –, (3), , E, , (4), , B, , B, , 1, v, , de Broglie wavelength of an electron depends on its velocities. Velocity of the electron is only decreasing in, the IInd case., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 119 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 93, , 13. Wavelength of a 1 keV photon is 1.24 × 10–9 m. What is the frequency of 1 MeV photon?, (1) 1.24 × 1015 Hz, , (2), , 2.4 × 1020 Hz, , (3), , 1.24 × 1018 Hz, , (4), , 2.4 × 1023 Hz, , Sol. Answer (2), Wavelength of 1 keV photon is 1.24 × 10–9 m, , E∝, , 1, λ, , E1 λ 2, =, E2, λ1, f2 =, , c, λ2, , 14. Which of the following statements is not correct?, (1) Photographic plates are sensitive to infrared rays, (2) Photographic plates are sensitive to ultraviolet rays, (3) Infrared rays are invisible but can cast shadows like visible light, (4) Infrared photons have more energy than photons of visible light, Sol. Answer (4), Infrared photons have less frequency than visible light. Hence, they have less energy., 15. Ultraviolet radiations of 6.2 eV fall on an aluminium surface (work function 4.2 eV). The kinetic energy (in joule), of the fastest electron emitted is approximately, (1) 3.2 × 10–21, , (2), , 3.2 × 10–19, , (3), , 3.2 × 10–17, , (4), , 3.2 × 10–15, , Sol. Answer (2), Energy incident = 6.2 eV, Work function 4.2 eV, KEmax = Energy incident – Work function, = 2 eV, = 3.2 × 10–19 J, 16. The work function of a metallic surface is 5.01 eV. The photo electrons are emitted when light of wavelength, 2000 Å falls on it. The potential difference applied to stop the fastest photo electrons is [h = 4.14 × 10–15 eV s], (1) 1.2 volt, , (2), , 2.24 volt, , (3), , 3.6 volt, , (4), , 4.8 volt, , Sol. Answer (1), w0 = 5.01 eV, KEmax =, , KEmax =, , hc, – 5.01, λ, , 4.14 × 10−15 C, 2000 × 10−10, , − 5.01 eV, , = 2.07 × 10–8 C – 5.01 eV, = 6.21 – 5.01, = 1.2 eV, Stopping potential =, , KEmax, = 1.2 V, e, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 120 :
94, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 17. A radio transmitter operates at a frequency of 880 kHz and a power of 10 kW. The number of photons emitted, per second is, (1) 1.72 × 1031, , (2), , 1.327 × 1034, , (3), , 13.27 × 1034, , (4), , 0.075 × 10–34, , Sol. Answer (1), Power = 10 × 103 watt, Frequency = 880 × 103 Hz, Energy per photon = hf, Number of photons =, , 10 × 103, hf, , 18. Assuming photoemission to take place, the factor by which the maximum velocity of the emitted photo electrons, changes when the wavelength of the incident radiation is increased four times, is (assuming work function to, be negligible in comparison to hc/), (1) 4, , (2), , 1, 4, , (3), , 2, , (4), , 1, 2, , Sol. Answer (4), hc, – w0, λ, Work function is negligible, , KEmax =, , 1, hc, mv 2 =, 2, λ, 1, λ, If is increased to 4 times velocities become half., , , v2 ∝, , 19. The stopping potential V for photoelectric emission from a metal surface is plotted along Y-axis and frequency, of incident light along X-axis. A straight line is obtained as shown. Planck’s constant is given by, , Y, V, , , , X, , (1) Slope of the line, (2) Product of slope on the line and charge on the electron, (3) Product of intercept along Y-axis and mass of the electron, (4) Product of slope and mass of electron, Sol. Answer (2), h = KEmax + w0, KEmax = eV, h = eV + w0, , v=, , hν, − w0, e, , Slope(m ) =, , h, e, , h = me, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 121 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 95, , 20. If the work function of a metal is ‘’ and the frequency of the incident light is ‘’, there is no emission of photo, electron for, (1) , , , h, , (2), , , , , h, , (3), , , , , h, , (4), , >=<, , , h, , Sol. Answer (1), Frequency = , work function = , h + KEmax, KEmax = 0, , ν=, , φ, φ, , hence, no emission occurs when ν =, h, h, , 21. Specific heat of water is 4.2 J/g°C. If light of frequency 3 × 109 Hz is used to heat 400 gm of water from 20°C, to 40°C, the number of photons needed will be, (1) 1.69 × 1029, , (2), , 1.69 × 1028, , (3), , 2.80 × 104, , (4), , 2.80 × 105, , Sol. Answer (2), S = 4.231 gºC, = 3 × 109 Hz, m = 400 g, Energy of one photon = h, Heat needed for 20ºC temp. change = msT = 400 × 4.2 × 20 J, Number of photons =, , 400 × 4.2 × 20, 400 × 4.2 × 20, =, = 1689 × 1025 = 1.69 × 1028, 9, −34, hν, × 3 × 10, 6.63 × 10, , 22. The work function of a certain metal is 2.3 eV. If light of wave number 2 × 106 m–1 falls on it, the kinetic energies, of fastest and slowest ejected electron will be respectively, (1) 2.48 eV, 0.18 eV, , (2), , 0.18 eV, zero, , (3), , 2.30 eV, 0.18 eV, , (4), , 0.18 eV, 0.18 eV, , Sol. Answer (2), w0 = 2.3 eV, n=, , 1, = 2 × 106 m–1 (wave number), λ, , hcn = KEmax + w0, hcn = KEmax + 2.3 eV, 23. When the electromagnetic radiations of frequencies 4 × 1015 Hz and 6 × 1015 Hz fall on the same metal in, different experiments, the ratio of maximum kinetic energy of electrons liberated is 1 : 3. The threshold frequency, for the metal is, (1) 2 × 1015 Hz, , (2), , 1 × 1015 Hz, , (3), , 3 × 1015 Hz, , (4), , 1.67 × 1015 Hz, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 122 :
96, , Dual Nature of Radiation and Matter, , Solution of Assignment, , Sol. Answer (2), Frequencies given are 4 × 1015 and 6 × 1015 Hz, hf1 = E1 + w0, hf2 = E2 + w2, E1 : E 2 = 1 : 3, E2 = 3E1, E1 + w0 = hf1, , ....(i), , 3E1 + w0 = hf2, , ....(ii), , Solve for w0., 24. How many photons are emitted by a laser source of 5 × 10 –3 W operating at 632.2 nm in 2 s?, (h = 6.63 × 10–34 Js), (1) 3.2 × 1016, , (2), , 1.6 × 1016, , (3), , 4 × 1016, , (4), , 0.4 × 1016, , Sol. Answer (1), Number of photons per second =, , Energy of one photon =, , Total energy emitted/sec, Energy of one photon, , hc, hc, =, λ, 632.2 × 10−9, , Energy per second emitted = 5 × 10–3 w, Total number of photons = Photons emitted per second × Time, When time = 2 s, , 25. V (stopping potential) is plotted against, , 1, , where is wavelength of incident radiations, for two metals, , , Metal 1, , V, , 1, , Metal 2, , 2, , 1, , (1) Metal 1 may be gold and metal 2 may be cesium, (2) 1 > 2, if metal -1 is gold and metal-2 is cesium, (3) 1 = 2, for any two metals, (4) 1 > 2, if metal -1 and metal-2 are gold and copper respectively, Sol. Answer (3), hc, hc, =, λ, 632.2 × 10−9, , hc, = eV + w 0, λ, eV =, , hc, − w0, λ, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 123 :
Solution of Assignment, , V =, , Dual Nature of Radiation and Matter, , 97, , hc 1 w 0, × −, e λ e, , tan θ =, , hc, = constant, e, , [Slope of V vs, , 1, graph], λ, , is same for all metals., 26. Variation of momentum of particle (p) with associated de-Broglie wavelength () is shown correctly by, , p, , p, , (1), , p, , (2), , p, , (3), , , , (4), , , , , , , , Sol. Answer (1), , λ=, , h, which is a hyperbolic graph like(1)., P, , photon, 27. A photon and a proton have equal energy E. Ratio of wavelengths , is proportional to, proton, , (1) E, , (2), , (3), , E, , 1, E, , (4), , Eº, , Sol. Answer (3), λPhoton =, , hc, E, , λProton =, , h, 2mE, , λPhoton, 2m, 1, =c, ∝, λProton, E, E, 28. Violet light is falling on a photosensitive material causing ejection of photoelectrons with maximum kinetic, energy of 1 eV. Red light falling on metal will cause emission of photoelectrons with maximum kinetic energy, (approximately) equal to, (1) 1.2 eV, , (2), , 0.9 eV, , (3) 0.5 eV, , (4), , Zero, that is no photoemission, , Sol. Answer (4), When violet light falls on photosensitive material, hc, = Ev + w 0, λv, , , , hc, − 1 = w0, λv, , hc, − w 0 < 0 as difference in energies of the violet and red light is greater than 1 eV., λv, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 124 :
98, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 29. For a proton accelerated through potential difference of one volt, kinetic energy gained in eV is, (1) 1, , (2), , 1840, , (3), , 1, 1840, , (4), , 931.5 × 10–6, , Sol. Answer (1), Kinetic energy = qV, For one electron energy gained is therefore = eV, 30. If frequency of light falling on a photosensitive material doubles, (1) Saturation photocurrent doubled, (2) Saturation photocurrent becomes more than double, (3) Cut-off voltage becomes more than double, (4) Stopping potential doubles, Sol. Answer (3), h = E + w0, h – w0 = E1, When frequency is doubled, 2h – w0 = E2, or E2 = 2E1 + w0, E2 is more than double of E1, 31. Photoelectrons from metal do not come out with same energy. Most appropriate explanation is, (1) Some electrons loose energy in form of heat, (2) Work function of a metal is average energy required to pull out electrons, (3) Electrons in metal occupy different energy levels and work function is the minimum energy required for, electron in highest level of conduction band to get out of metal, (4) For some electrons, some part of energy gained during inelastic collision with photon is spent in over coming, attractive force by nucleus, Sol. Answer (3), 32. For an alpha particle, accelerated through a potential difference V, wavelength (in Å) of the associated matter, wave is, (1), , 12.27, , (2), , V, , 0.101, V, , (3), , 0.202, V, , (4), , 0.286, V, , Sol. Answer (2), Wavelength () =, or, , λ=, , h, P, , h, 2mE, , E = 2e × V for V potential difference, , λ=, λ=, , h, 4meV, 0.101, V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 125 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 99, , SECTION - C, Previous Years Questions, 1., , In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength, in the Balmer series is, [Re-AIPMT-2015], (1), , 5, 27, , (2), , 4, 9, , (3), , 9, 4, , (4), , 27, 5, , Sol. Answer (1), For longest wavelength of Lyman series, ⎛ 1, 1, 1 ⎞, R⎜ 2 2 ⎟, l, ⎝ n1 n2 ⎠, , n1 = 1, n2 = 2, 1, 3, ⎛1 1⎞, R ⎜ ⎟ Rz 2, l, 4, ⎝1 4 ⎠, , For longest wave length of Balmer series, n1 = 2, n2 = 3, 1, 1⎞, ⎛ 1, ⎛ 1 1⎞, R⎜ 2 2 ⎟ R⎜ ⎟ R 5, b, 2, 3, 36, ⎝, ⎠, ⎝4 9⎠, , 5, l, 5, 36, , , 3, b, 27, 4, , 2., , Light of wavelength 500 nm is incident on a metal with work function 2.28 eV. The de Broglie wavelength of, the emitted electron is, [Re-AIPMT-2015], (1) 2.8 1012 m, , (2), , 2.8 1010 m, , (3), , 2.8 109 m, , (4), , 2.8 109 m, , Sol. Answer (4), From Einstien's photoelectric equation, E = w0 + eV, , hc, = w0 + eV, λ, , 6.6 × 10−34 × 3 × 108, 500 × 10 −9 × 1.6 × 10 −19, , = 2.28 + eV, , V ≤ 0.2, , de-Broglie wavelength of electron, =, , 12.27, v, , Å, , λ ≥ 2.8 × 10−9 m, So, λ ≥ 2.8 × 10−9 m, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 126 :
100, 3., , Dual Nature of Radiation and Matter, , Solution of Assignment, , , . If the, 2, maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case,, the work function of the surface of the material is (h = Planck's constant, c = speed of light), , A photoelectric surface is illuminated successively by monochromatic light of wavelength and, , [Re-AIPMT-2015], (1), , hc, 3, , (2), , hc, 2, , (3), , hc, , , (4), , 2hc, , , Sol. Answer (2), 4., , Which of the following figures represent the variation of particle momentum and the associated de-Broglie, wavelength?, [AIPMT-2015], , p, , p, (2), , (1), , p, , p, (3), , , , (4), , , , , , , , Sol. Answer (3), Linear momentum = p, wavelength = , , , 5., , h, 1, ⇒ , p, p, , A certain metallic surface is illuminated with monochromatic light of wavelength . The stopping potential for, photo-electric current for this light is 3V0. If the same surface is illuminated with light of wavelength 2, the, [AIPMT-2015], stopping potential is V0. The threshold wavelength for this surface for photo-electric effect is, (1), , , 6, , (2), , 6, , (3), , 4, , (4), , , 4, , Sol. Answer (3), hc, W0 e3V0, , , ...(i), , hc, W0 eV0, 2, , (iii), , From (i) & (ii), eV0 , , hc, and W0 hc, 4, 4, , Threshold wavelength = 4, 6., , When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted, from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is, [AIPMT-2014], (1) 0.65 eV, , (2), , 1.0 eV, , (3), , 1.3 eV, , (4), , 1.5 eV, , Sol. Answer (2), E = W0 + 0.5, , ...(i), , 1.2E = W0 + 0.8, , ...(ii), , From (ii) – (i), E = 1.5 eV, W0 = 1 eV, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 127 :
Solution of Assignment, , 7., , Dual Nature of Radiation and Matter, , 101, , If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the deBroglie wavelength of the particle is:, [AIPMT-2014], (1) 25, , (2), , 75, , (3), , 60, , (4), , 50, , Sol. Answer (2), , K, , h2, , K1 22, 1 22, , 1, , , , , 2 , K 2 12, 16 12, 1 4, 2m 2, , 2 , , 1, 4, , , 1 1, 4 100 75%, % , 1, , 8., , For photoelectric emission from certain metal the cut off frequency is . If radiation of frequency 2 impinges on, the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass):, [NEET-2013], (1), , h, m, , 2h, m, , (2), , (3), , 2, , h, m, , (4), , h, 2m, , Sol. Answer (2), 9., , The wavelength e of an electron and p of a photon of same energy E are related by, (1) p e, , (2), , p e, , (3), , 1, p , e, , (4), , [NEET-2013], p e2, , Sol. Answer (4), 10. An -particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. The, de-Broglie wavelength associated with the particle will be, [AIPMT (Prelims)-2012], (1) 10 Å, , (2), , 0.01 Å, , (3), , 1Å, , (4), , 0.1 Å, , Sol. Answer (2), =, , qBr, m, , =, , h, h, , mv qBr, , 11. Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state, irradiates a photosensitive material. The stopping potential is measured to be 3.57 V. The threshold frequency, of the material is, [AIPMT (Prelims)-2012], (1) 1.6 × 1015 Hz, , (2), , 2.5 × 1015 Hz, , (3), , 4 × 1015 Hz, , (4), , 5 × 1015 Hz, , Sol. Answer (1), Stopping potential = 3.57 V, KEmax = 3.57 eV, 1⎤, ⎡1, Energy incident = 13.6 ⎢ 2 − 2 ⎥ = 3.4 × 3 = 10.2 eV, 2 ⎦, ⎣1, , or w0 = 10.2 eV – 3.57 eV, or, , ν=, , 6.63 × 1.6 × 10−19, h, , or, , w0 = 6.63 eV, , or, , = 1.6 × 1015, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 128 :
102, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 12. A 200 W sodium street lamp emits yellow light of wavelength 0.6 m. Assuming it to be 25% efficient in converting, electrical energy to light, the number of photons of yellow light it emits per second is, [AIPMT (Prelims)-2012], (1) 62 × 1020, , (2), , 3 × 1019, , (3), , 1.5 × 1020, , (4), , 6 × 1018, , Sol. Answer (3), Wavelength = 0.6 m, Power = 200 W, Light energy emitted =, , Number of photons =, , 25, × 200 W = 50 W, 100, , 50, 50 × 0.6 × 10−6, × 0.6 × 10−6 =, = 1.5 × 1020, hc, 6.63 × 10−34 × 3 × 108, , 13. If the momentum of an electron is changed by P, then the de-Broglie wavelength associated with it changes by, 0.5%. The initial momentum of electron will be, [AIPMT (Mains)-2012], (1) 200 P, , (2), , 400 P, , (3), , P, 200, , (4), , 100 P, , Sol. Answer (1), h, = P, 0, dP0, d, = P, , 0, , , , | d | | dP0 |, , P0, , , 0.5 P, 100 P0, P0= 200P, 14. Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface, of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is [AIPMT (Mains)-2012], (1) 1 : 4, , (2), , 1:2, , (3), , 1:1, , (4), , 1:5, , Sol. Answer (2), EPhoton = KEmax + w0, KE1 0.5 1, =, =, KE2, 2, 4, , 15. In photoelectric emission process from a metal of work function 1.8 eV, the kinetic energy of most energetic, electric electrons is 0.5 eV. The corresponding stopping potential is, [AIPMT (Prelims)-2011], (1) 2.3 V, , (2), , 1.8 V, , (3), , 1.3 V, , (4), , 0.5 V, , Sol. Answer (4), KEmax = 0.5 eV, Stopping potential only depends on maximum KE of electron ejected., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 129 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 103, , 16. Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to, 100 kV then the de-Broglie wavelength associated with the electrons would, [AIPMT (Prelims)-2011], (1) Increase by 4 times, , (2), , Increase by 2 times, , (3) Decrease by 2 times, , (4), , Decrease by 4 times, , Sol. Answer (3), Electrons voltage = 25 kV, Voltage is increased to 100 kV, λ=, , h, =, mv, , h, 2m KE, , Since KE becomes 4 KE, λ, 2, , λ′ =, , 17. Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively illuminate a metallic, surface whose work function is 0.5 eV successively. Ratio of maximum speeds of emitted electrons will be, [AIPMT (Prelims)-2011], (1) 1 : 5, , (2), , 1:4, , (3), , 1:2, , (4), , 1:1, , Sol. Answer (3), , v1, , v2, , 1– 0.5, 1, =, 2.5 – 0.5, 2, , 18. In the Davisson and Germer experiment, the velocity of electron of electrons emitted from the electron gun can, be increased by, [AIPMT (Prelims)-2011], (1) Decreasing the potential difference between the anode and filament, (2) Increasing the potential difference between the anode and filament, (3) Increasing the filament current, (4) Decreasing the filament current, Sol. Answer (2), Fact, 19. Photoelectric emission occurs only when the incident light has more than a certain minimum, [AIPMT (Prelims)-2011], (1) Frequency, , (2), , Power, , (3), , Wavelength, , (4), , Intensity, , Sol. Answer (1), Concept of threshold frequency, 20. The threshold frequency for a photosensitive metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident, on this metal, the cut-off voltage for the photoelectric emission is nearly, [AIPMT (Mains)-2011], (1) 5 V, , (2), , 1V, , (3), , 2V, , (4), , 3V, , Sol. Answer (3), h = h0 + KEmax, KEmax = h – h0, , V =, , h, ( ν − ν0 ), e, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 130 :
104, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 21. The potential difference that must be applied to stop the fastest photoelectrons emitted by a nickel surface,having, work function 5.01 eV, when ultraviolet light of 200 nm falls on it, must be, [AIPMT (Prelims)-2010], (1) 1.2 V, , (2), , 2.4 V, , (3), , –1.2 V, , (4), , –2.4 V, , Sol. Answer (1), Energy of incident light =, , hc, 200 × 10 −9, , w0 = 5.01 eV, Incident energy = KEmax + w0, hc, (eV) λ − 5.01 = eV, e, , 22. When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectron and their, maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted, electrons and their maximum kinetic energy are respectively, [AIPMT (Mains)-2010], (1) N and 2T, , (2), , 2N and T, , (3), , 2N and 2T, , (4), , N and T, , Sol. Answer (2), Number of photoelectrons = 2N, max. KE = T, Increase intensity of radiation increases number of photoelectrons emitted. Energy of photoelectron depends, on frequency of radiation., 23. The number of photo electrons emitted for light of a frequency v (higher than the threshold frequency v0) is, proportional to, [AIPMT (Prelims)-2009], (1) Threshold frequency (v0), , (2), , Intensity of light, , (3) Frequency of light (v), , (4), , v – v0, , Sol. Answer (2), 24. The Figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different, radiations. Which one of the following is a correct statement?, [AIPMT (Prelims)-2009], , Photo current, b, c, a, Retarding potential, , Anode potential, , (1) Curves (a) and (b) represent incident radiations of same frequency but of different intensity., (2) Curves (b) and (c) represent incident radiations of different frequencies but of different intensity., (3) Curves (b) and (c) represent incident radiations of same frequency having same intensity., (4) Curves (a) and (b) represent incident radiations of different frequencies and different intensities., Sol. Answer (1), Same stopping potential means incident light is of same frequency., 25. Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW., The number of photons arriving per sec. On the average at a target irradiated by this beam is, [AIPMT (Prelims)-2009], (1) 3 ×, , 1016, , (2), , 9×, , 1015, , (3), , 3×, , 1019, , (4), , 9 × 1017, , Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 131 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 105, , 26. The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for, which the stopping potential is 5 V lies in the, [AIPMT (Prelims)-2008], (1) X-ray region, , (2), , Ultraviolet region, , (3), , Visible region, , (4), , Infrared region, , Sol. Answer (2), w0 = 6.2 eV, , KEmax = 5 eV, , Net incident energy/photon = 5 + 6.2 = 11.2 eV, corrrspond to 11.2 eV energy belongs to ultraviolet region., 27. A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 × 106 ms–1. The, velocity of the particle is (mass of electron = 9.1 × 10–31 kg), [AIPMT (Prelims)-2008], (1) 2.7 × 10–21 ms–1, , (2), , 2.7 × 10–18 ms–1, , (3), , 9 × 10–2 ms–1, , (4), , 3 × 10–31 ms–1, , Sol. Answer (2), λe =, , λm =, , h, me × 3 × 106, h, 1 × 10 −6 × v, , λe, 10−6 v, =, λ m 3 × 106 × me, , or, , λe 3 × 106 × 9.1 × 10−31, =, =v, λm, 10−6, , or 3 × 1012 × 9.1 ×10–31 = v, or v = 2.7 × 10–18 m/s, 28. A 5 watt source emits monochromatic light of wavelength 5000 Å. When placed 0.5 m away, it liberates, photoelectrons from a photosensitive metallic surface. When the source is moved to a distance of 1 m, the, number of photoelectrons liberated will, [AIPMT (Prelims)-2007], (1) Be reduced by a factor of 2, , (2), , Be reduced by a factor of 4, , (3) Be reduced by a factor of 8, , (4), , Be reduced by a factor of 16, , Sol. Answer (2), Intensity , , 1, d2, , When distance is doubled intensity reduces by 4 times and does photocurrent., 29. Monochromatic light of frequency is 6.0 × 1014 Hz produced by a laser. The power emitted is 2 × 10–3 W. The, number of photons emitted, on the average, by the source per second is, [AIPMT (Prelims)-2007], (1) 5 × 1014, , (2), , 5 × 1015, , (3), , 5 × 1016, , (4), , 5 × 1017, , Sol. Answer (2), = 6 × 1014 Hz, Power = 2 × 10–3 W, Number of photons =, , 2 × 10−3, h × 6 × 1014, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 132 :
106, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 30. A photo-cell employs photoelectric effect to convert, , [AIPMT (Prelims)-2006], , (1) Change in the frequency of light into a change in electric voltage, (2) Change in the intensity of illumination into a change in photoelectric current, (3) Change in the intensity of illumination into a change in the work function of the photocathode, (4) Change in the frequency of light into a change in the electric current, Sol. Answer (2), 31. When photons of energy h fall on an aluminium plate (of work function E0),photoelectrons of maximum kinetic, energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected, photoelectrons will be, [AIPMT (Prelims)-2006], (2), , (1) K + E0, , 2K, , (3), , K, , (4), , K + h, , Sol. Answer (4), h = K + E0, 2h = K2 + E0, K2 = K + E 0, 32. The momentum of a photon of energy 1 MeV in kg m/s, will be, (1) 0.33 ×, , 106, , (2), , 7×, , 10–24, , (3), , 10–22, , [AIPMT (Prelims)-2006], (4), , 5 × 10–22, , Sol. Answer (4), 33. The work functions for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein’s, equation, the metals which will emit photoelectrons for a radiation of wavelength 4100 Å is/are, [AIPMT (Prelims)-2005], (1) C only, , (2), , A only, , (3), , A and B only, , (4), , All the three metals, , Sol. Answer (3), , hc, hc, = 3.01 eV, where = 4100 Å is such that, 4100 Å, λ, Hence, only A and B will show emission., 34. A photosensitive metallic surface has work function, h0. If photons of energy 2h0 fall on this surface, the electrons, come out with a maximum velocity of 4 × 106 m/s. When the photon energy is increased to 5h0, then maximum, velocity of photoelectrons will be :, [AIPMT (Prelims)-2005], (1) 2 × 106 m/s, , (2), , 2 × 107 m/s, , (3), , 8 × 105 m/s, , (4), , 8 × 106 m/s, , Sol. Answer (4), When incident energy is 2 h0, 2h = KEmax + w0, , 2h ν0 =, h ν0 =, , 1, × m × (4 × 106 )2 + h ν0, 2, , 1, m × (2 × 106 )2, 2, , When incident energy is 5h0, 5h ν0 − h ν0 =, 4h ν0 =, , 1, mv 22, 2, , 1, mv 22, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 133 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 107, , 35. Which one among the following shows particle nature of light?, (1) Photoelectric effect, , (2), , Interference, , (3), , Refraction, , (4), , Polarization, , Sol. Answer (3), 36. The slope of V versus graph where and V are the frequency of incident light and stopping potential for a, given surface will be, (1) h, , (2), , h, e, , (3), , eh, , (4), , e, , Sol. Answer (2), h = KEmax + w0, hν w 0, −, =V, e, e, , Slope =, , h, e, , 37. An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted, illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron, is 10 V, then the value of n is, (1) 5, , (2), , 2, , (3), , 3, , (4), , 4, , Sol. Answer (4), w0 = 2.75 eV, KEmax = 10 eV, Energy incident = KEmax + w0, hc, = 10 + 2.75 eV, λ, , ....(i), , 1, 1⎞, ⎛1, = R⎜ 2 − 2 ⎟, ⎝1, λ, n ⎠, , ....(ii), , Also,, , Solving (i) and (ii), 38. The maximum kinetic energy of the photoelectrons varies, (1) Inversely with the intensity and is independent of the frequency of the incident radiation, (2) Inversely with the frequency and is independent of the intensity of the incident radiation, (3) Linearly with the frequency and the intensity of the incident radiation, (4) Linearly with the frequency and is independent of the intensity of the incident radiation, Sol. Answer (4), h = KEmax + w0, Hence, KEmax varies linearly with frequency, 39. The work function for Al, K and Pt is 4.28 eV, 2.30 eV and 5.65 eV respectively. Their respective threshold, frequencies would be, (1) Pt > Al > K, , (2), , Al > Pt > K, , (3), , K > Al > Pt, , (4), , Al > K > Pt, , Sol. Answer (1), Higher work function means higher frequency as, w0 = h0, Pt > Al > K, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 134 :
108, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 40. If a photon has velocity c and frequency , which of the following represents its wavelength?, (1), , h, c, , (2), , 2, , h, , (3), , hc, , , (4), , c, , , Sol. Answer (4), 41. The velocity of photons is proportional to (where = frequency), 1, , (1), , (2), , , , 2, , (3), , 0, , (4), , , , Sol. Answer (3), 1, mv 2 + w 0, 2, Hence, there is no direct relationship between v and ., hν =, , 42. Which of the following statement is correct?, (1) The photocurrent increases with intensity of light, (2) The stopping potential increases with increase in intensity of incident light, (3) The current in photocell increases with increasing frequency, (4) The photocurrent is proportional to the applied voltage, Sol. Answer (1), Increase of intensity increases the number of photoelectron emiting from plate., 43. Light of wavelength 5000 Å falls on a sensitive plate with photoelectric work function of 1.9 eV. The kinetic, energy of the photoelectron emitted will be, (1) 1.24 eV, , (2), , 2.48 eV, , (3), , 0.58 eV, , (4), , 1.16 eV, , Sol. Answer (3), , hc, = KEmax + w 0, λ, KEmax =, , [w0 = 1.9 eV], , hc, − w0, λ, , 44. In a photo-emissive cell, with exciting wavelength , the fastest electron has speed v. If the exciting wavelength, 3, , the speed of the fastest emitted electron will be, is changed to, 4, (2) v(4/3)1/2, (3) v(3/4)1/2, (4) Greater than v(4/3)1/2, (1) Less than v(4/3)1/2, Sol. Answer (4), , hc 1, = mv 2 + w 0, λ, 2, 3λ, 4, , If becomes, , 4hc, 1, − w 0 = mv '2, 3λ, 2, 1/2, , , , ⎛4⎞, v' >v⎜ ⎟, ⎝3⎠, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 135 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 109, , 45. The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is about, (1) 3000 Å, , (2), , 2062.5 Å, , (3), , 4125 Å, , (4), , 6000 Å, , Sol. Answer (1), w0 = 4.125 eV, hc, = 4.125, λ0, λ0 =, , hc, 4.125 × 1.6 × 10−19, , 46. As the intensity of incident light increases, (1) Kinetic energy of emitted photoelectrons increases, (2) Photoelectric current decreases, (3) Photoelectric current increases, (4) Kinetic energy of emitted photoelectrons decreases, Sol. Answer (3), Photoelectric current Intensity, 47. A photo-cell is illuminated by a source of light, which is placed at a distance d from the cell. If the distance, becomes d/2. Then number of electrons emitted per second will be, (1) Remain same, , (2), , Four times, , (3), , Two times, , (4), , One-fourth, , Sol. Answer (2), Intensity , , 1, (Distance)2, , If distance is halved intensity becomes 4 times. Hence, photocurrent becomes 4 times., 48. The value of Planck’s constant is, (1) 6.63 × 10–34 J/s, , (2), , 6.63 × I0–34 kg m2/s2, , (3), , 6.63 × 10–34 kg m2, , (4), , 6.63 × 10–34 J s, , Sol. Answer (4), 49. A source S1 is producing 1015 photons per second of wavelength 5000 Å. Another source S2 is producing, 1.02 ×1015 photons per second of wavelength 5100 Å. Then (power of S2)/(power of S1) is equal to, (1) 0.98, , (2), , 1.00, , (3), , 1.02, , (4), , 1.04, , Sol. Answer (2), Power of source – 1 = 1015 ×, , hc, λ1, , Power of source – 2 = 1.02 × 1015 ×, , [1 = 5000 Å], hc, λ2, , [2 = 5100 Å], , 50. If in a photoelectric cell, the wavelength of incident light is changed from 4000 Å to 3000 Å then change in, stopping potential will be, (1) 0.66 V, , (2), , 1.03 V, , (3), , 0.33 V, , (4), , 0.49 V, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 136 :
110, , Dual Nature of Radiation and Matter, , Solution of Assignment, , Sol. Answer (2), hc, − w 0 = eV1, λ1, , ....(i), , hc, − w 0 = eV2, λ2, , ....(ii), , hc hc, −, = e(V1 − V2 ), λ1 λ 2, , 51. Find the number of electrons emitted per second by a 24 W source of monochromatic light of wavelength, 6600Å, assuming 3% efficiency for photoelectric effect (take h = 6.6 × 10–34 Js), (1) 48 × 1019, , (2), , 48 × 1017, , (3), , 8 × 1019, , (4), , 24 × 1017, , Sol. Answer (4), Power = 24 W, hc, [ = 6600 Å], λ, Energy incident on photographic plate = 0.03 × 24 watt = 0.072 watt, , Energy of one photon =, , ....(i), ....(ii), , Dividing (ii) by (i), n = 24 × 1017 photon, 52. The photo-electrons emitted from a surface of sodium metal are such that they, (1) All are of the same frequency, (2) Have the same kinetic energy, (3) Have the same de-Broglie wavelength, (4) Have their speeds varying from zero to a certain maximum value, Sol. Answer (4), Photo electrons they emit from different locations in metal plates., 53. In photoelectric effect, if a weak intensity radiation instead of strong intensity of suitable frequency is used, then, (1) Photoelectric effect will get delayed, , (2), , Photoelectric effect will not take place, , (3) Maximum kinetic energy will decrease, , (4), , Saturation current will decrease, , Sol. Answer (4), 54. The work function of tungsten is 4.50 eV. The wavelength of fastest electron emitted when light whose photon, energy is 5.50 eV falls on tungsten surface, is, (1) 12.27 Å, , (2), , 0.286 Å, , (3), , 12400 Å, , (4), , 1.227 Å, , Sol. Answer (1), w0 = 4.5 eV, Photon energy = 5.5 eV, KEmax = Photon energy – w0, or E = 1 eV, , λ=, , h, 2mE, , Solving = 12.27 Å, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 137 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 111, , 55. An electron of mass m, when accelerated through a potential difference V, has de-Broglie wavelength X. The, de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference,, will be, (1) , , M, m, , (2), , , , m, M, , (3), , , , M, m, , (4), , , , m, M, , Sol. Answer (4), Path will gain some energy E, , h, , x=, , 2mE, , λP =, , h, 2ME, m, M, , λP = x ×, , 56. If we consider electrons and photons of same wavelength, then they will have same, (1) Momentum, , (2), , Angular momentum, , (3), , Energy, , (4), , Velocity, , Sol. Answer (1), λ=, , h, always, P, , So, if they have same wavelengths they have same momentum, 57. An electron beam has a kinetic energy equal to 100 eV. Find its wavelength associated with a beam, if mass, of electron = 9.1 × 10–31 kg and 1 eV = 1.6 × 10–19 J. (Planck’s constant = 6.6 × 10–34 J-s), (1) 24.6 Å, , (2), , 0.12 Å, , (3), , 1.2 Å, , (4), , 6.3 Å, , Sol. Answer (3), KE = 100 eV, me = 9.1 × 10–31 kg, 1 eV = 1.6 × 10–19 J, , λ=, , 6.6 × 10 −34, 2mE, , =, , 6.6 × 10 −34, 2 × 9.1 × 10 −31 × 100 × e, , where e = 1.6 × 10–19, = 1.2 Å, 58. If particles are moving with same velocity, then which has maximum de-Broglie wavelength?, (1) Proton, , (2), , -particle, , (3), , Neutron, , (4), , -particle, , Sol. Answer (4), , λ=, , h, mv, , , , λ∝, , 1, m, , particle which has least mass will have maximum wavelength., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 138 :
112, , Dual Nature of Radiation and Matter, , Solution of Assignment, , 59. When ultraviolet rays fall on a metal plate then photoelectric effect does not occur, it occurs by, (1) Infrared rays, , (2), , X-rays, , (3), , Radiowave, , (4), , Light wave, , Sol. Answer (2), (x-rays) < (ultraviolet rays), 60. A metal surface is illuminated by the photons of energy 5 eV and 2.5 eV respectively. The ratio of their, wavelengths is, (1) 1 : 3, , (2), , 1:4, , (3), , 2:5, , (4), , 1:2, , Sol. Answer (4), Energy of photon =, , hc, λ, , λ=, , hc, E, , , , λ1 E2 2.5 1, =, =, =, λ 2 E1, 5, 2 =4, , 61. If the energy of a photon is 10 eV, then its momentum is, (1) 3.33 × 10–23 kg-m/s, , (2), , 3.33 × 10–25 kg-m/s, , (3), , 3.33 × 10–29 kg-m/s (4), , 3.33 × 10–27 kg-m/s, , Sol. Answer (4), Momentum =, , Momentum =, , E, c, 10 × 1.6 × 10−19, 3 × 10, , 8, , =, , 16, × 10 −27 = 3.33 × 10–27, 3, , 62. An electron and a proton have same kinetic energy. Ratio of their respective de-Broglie wavelength is about, (1), , 12.27, 0.286, , (2), , 0.101, 0.286, , (3), , 0.286, 12.27, , (4), , 0.101, 12.27, , h, c, , (4), , h, cv, , Sol. Answer (1), , λ=, , h, , , , 2ME, λe =, , λ∝, , 1, m, , M, × λP, m, , 63. Dynamic mass of the photon in usual notations is given by, (1), , hv, c, , (2), , h, c, , (3), , Sol. Answer (3), , h, cλ, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 139 :
Solution of Assignment, , Dual Nature of Radiation and Matter, , 113, , 64. de-Broglie wavelength associated with an electron revolving in the nth state of hydrogen atom is directly, proportional to, (1) n, , (2), , 1, n, , (3), , n2, , (4), , 1, n2, , (4), , 0.4 nm, , Sol. Answer (1), n = 2r, , , r, , r n2., n, , 65. A proton is accelerated through 225 V. Its de-Broglie wavelength is, (1) 0.1 nm, , (2), , 0.2 nm, , (3), , 0.3 nm, , Sol. Answer (2), Energy = 225 eV, , λ=, , h, 2mE, , = 0.2 nm, , SECTION - D, Assertion-Reason Type Questions, 1., , A : Every metal has a definite work function, still all photoelectrons do not come out with the same energy if, incident radiation is monochromatic., R : Work function is the minimum energy required for the electron in the highest level of the conduction band to, get out of the metal. Not all electrons in the metal belong to this level rather they occupy a continuous band, of levels., , Sol. Answer (1), 2., , A : Work function of aluminium is 4.2 eV. Emission of electrons will be possible by two photons, each of 2.5 eV, energy, striking the electron of aluminium., R : Energy of a photon can be less than the work function of the metal, for photoelectron emission., , Sol. Answer (4), Both the assertion and reason are incorrect. The quantised nature of light allows emission only when the photon, striking the electron has enough energy to eject the electron. There will be no emission due to photons with, energy less than work function., 3., , A : On increasing the intensity of light, the number of photoelectrons emitted is more. Also the kinetic energy of, each photon increases but the photoelectric current is constant., R : Photoelectric current is independent of intensity but increases with increasing frequency of incident radiation., , Sol. Answer (4), On increasing the intensity of light only number of photoelectrons increase and not the KE of electrons or, photons., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 140 :
114, 4., , Dual Nature of Radiation and Matter, , Solution of Assignment, , A : The process of photoelectron emission and thermionic emission of electrons is different., R : Photoelectric emission does not depend upon temperature, whereas thermionic emission is temperature, dependent., , Sol. Answer (1), The process of photoelectric emission depends of frequency of wave and thermionic emission occurs when, metal is heated., 5., , A : Wave nature of particles is not visible in daily life., R : In daily life, mass of particles is very high so their de Broglie wavelength is very small., , Sol. Answer (1), , h, The assertion is correct and reason is correct explanation =, and it is very low for masses in every, mv, day life as h is very small., 6., , A : If a stationary nucleus emits an -particle, the de Broglie wavelengths of the daughter nucleus and the, -particle are equal., R : The magnitudes of the linear momenta of the daughter nucleus and the -particle are the same., , Sol. Answer (1), 7., , A : When a photon of energy h is incident on an electron in a metal of work function (< h), the electron, will not necessarily come out of the metal., R : Work function is the minimum energy required to liberate an electron out of a metal. So some electrons may, require more energy for their liberation., , Sol. Answer (1), 8., , A : The photoelectric effect is a proof of the quantized nature of the light., R : Each photon in a light beam has same amount of energy., , Sol. Answer (3), The assertion is correct but reason is completely false. It can only be true for completely monochromatic laser., 9., , A : A photon cannot transfer all of its energy to an isolated electron., R : When energy of a photon is more than 1.02 MeV, it can materialize into two particles called electron and, positron., , Sol. Answer (2), 10. A : There is almost no time-lag between the incidence of light and the emission of photoelectrons., R : A photon transfers almost all its energy to a single electron in a metal., Sol. Answer (1), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 141 :
https://t.me/NEET_StudyMaterial, Chapter, , 25, , Wave Optics, Solutions, SECTION - A, Objective Type Questions, 1., , Huygens' concepts of secondary wavelets, (1) Allow us to find the focal length of a thin lens, , (2) Give the magnifying power of a microscope, , (3) Are a geometrical method to find a wavefront, , (4) Are used to determine the velocity of light, , Sol. Answer (3), Huygens' concepts of secondary wavelets are a geometrical method to find a wavefront., 2., , The intensity at a point at a distance r from a source which produces cylindrical wavefronts varies as, , (1) I 1, r2, Sol. Answer (2), , (2), , I, , 1, r, , (3), , I r0, , (4), , I, , 1, r3, , The intensity at a point at a distance r from a source varies as I 1, r, , This is because intensity is, 3., , Power, and the surface area of wavefronts changes by 2rl., Area, , Four waves are expressed as, (i) y1 = a1 sin t,, , (ii), , y2 = a2 sin 2t,, , (iii) y3 = a3 cos t, , (iv), , y4 = a4 sin (t + )., , (3), , (4), , Not possible at all, , The interference is possible between, (1) (i) and (iii), , (2), , (i) and (ii), , (ii) and (iv), , Sol. Answer (1), Four waves are expressed as y1 and y3 can interfere with each other among the options as they have same, frequency., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 142 :
54, 4., , Wave Optics, , Solution of Assignment, , Two waves having intensities in the ratio of 9 : 1 produce interference. The ratio of maximum to minimum, intensity is equal to, (1) 10 : 8, , (2), , 9:1, , (3), , 4:1, , (4), , 2:1, , Sol. Answer (3), I1 1 A12, = =, I2 9 A22, , I1 A12, , A1 1, A =3, 2, , I2 A22, Imax (A1 + A2)2, Imin (A2 – A1)2, , Imax (3 + 1)2 16 4, =, =, =, 4, 1, Imin, 22, Imax : Imin = 4 : 1, 5., , If the light is polarised by reflection, then the angle between reflected and refracted light is, (1) , , (2), , /2, , (3), , 2, , (4), , /4, , Sol. Answer (2), According to Brewster's law, 6., , In an interference pattern produced by two identical slits, the intensity at the site of the central maximum is, I. The intensity at the same spot when either of the two slits is closed is Io, then, (1) I = Io, , (2), , I = 2 Io, , (3) I = 4 Io, , (4), , I and Io are not related to each other, , Sol. Answer (3), Let amplitude of light from a slit be A., I0 A2, or I (A + A)2, , , I0 1, =, I, 4, , or I = 4I0, 7., , The fringe width in Young’s double-slit experiment can be increased if we decrease, (1) Separation of the slits, , (2), , Distance between the source and the screen, , (3) Wavelength of the source, , (4), , All of these, , Sol. Answer (1), Fringe width () =, , λD, d, , 1⎞, ⎛, Hence if we decrease d ⎜ β ∝ ⎟, ⎝, d⎠, , is increased., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 143 :
Solution of Assignment, , 8., , Wave Optics, , 55, , In case of Young’s experiment, (1) There are two virtual sources of light from same monochromatic source of light, (2) Both the slits get light from a single monochromatic source of light, (3) Two separate monochromatic sources of light of same wavelength are used, (4) None of these, , Sol. Answer (2), In case of Young's double slit. Experiment the waves from the slit should be coherent. Since true coherence, is from the same source always answer will be (2)., 9., , If light from a galaxy observed on the earth's surface has a red shift, then, (1) Galaxy is stationary w.r.t. the earth, , (2), , Galaxy is approaching the earth, , (3) Galaxy is receding from the earth, , (4), , Temperature of galaxy in increasing, , Sol. Answer (3), When there is Red shift. The wavelength is increased and frequency is decreased., 10. In Young’s double slit experiment, the separation between the slits is halved and the distance between the, slits and screen is doubled. The new fringe width is, (1) Unchanged, , (2), , Halved, , (3), , Doubled, , (4), , Quadrupled, , Sol. Answer (4), , β=, , Dλ, d, , If d is halved and D is doubled, β2 =, , λ × 2D, d /2, , 2 = 4, 11. If one of the two slits of a Young’s double slit experiment is painted over so that it transmits half the light, intensity of the other then, (1) The fringe system would disappear, (2) The bright fringes will be more bright and dark fringes will be more dark, (3) The dark fringes would be less dark and bright fringes would be less bright, (4) Bright as well as dark fringes would be more dark, Sol. Answer (3), The net intensity or energy per unit area interfering to form fringer will decrease. Hence observed brightness, will decrease., 12. Monochromatic light from a narrow slit illuminates two parallel slits producing an interference pattern on a screen., The separation between the screen and the slits is reduced to half. The fringe width, (1) Is doubled, , (2), , Becomes four times, , (3) Becomes one fourth, , (4), , Becomes half, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 144 :
56, , Wave Optics, , Solution of Assignment, , Sol. Answer (4), Monochromatic light, , β=, , λD, d, , If becomes, , β2 =, , D, 2, , λD β, =, 2d 2, , 13. Two slits separated by a distance of 1mm are illuminated with red light of wavelength, 6.5 × 10–7 m. The interference fringes are observed on a screen placed 1m from the slits. The distance between, the third dark fringe and the fifth bright fringe on the same side of central maxima is, (1) 0.65 mm, , (2), , 1.62 mm, , (3), , 3.25 mm, , (4), , 4.88 mm, , Sol. Answer (2), = 6.5 × 10–7 mm, d = 1 mm, D=1m, ⎛ nD λ ⎞, 5D λ, ⎜⎝, ⎟ 5th bright fringe =, d ⎠, d, , Dλ ⎞, ⎛, 5D λ, ⎜⎝ (2n − 1), ⎟ 3rd dark fringe, 2d ⎠, 2d, Distance (d) =, , =, , 5D λ 5 D λ, −, d, 2 d, 5 Dλ, 2 d, , 14. A double slit interference experiment is carried out in air and the entire arrangement is dipped in water. As a, result, (1) The fringe width decreases, , (2), , The fringe width increases, , (3) The fringe width remains unchanged, , (4), , Fringe pattern disappears, , Sol. Answer (1), In water or any other medium becomes, , β=, , λ, hence, μ, , λD, becomes, d, , β2 =, , λD, β, or β2 >, μd, μ, , So, fringe width decreases., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 145 :
Solution of Assignment, , Wave Optics, , 57, , 15. Double slit interference experiment is carried out with monochromatic light and interference fringes are observed., If now monochromatic light is replaced by white light; what change is expected in interference pattern?, (1) No change, (2) Pattern disappears, (3) White and dark fringes are observed throughout, (4) A few coloured fringes are observed on either side of central white fringe, Sol. Answer (4), Central fringe is white because all the wave lengths are super-imposing at central position with maxima. On, either sides all wavelengths make their maximas and minimas so they appeared coloured., 16. In a Young’s double slit experiment the wavelength of red light is 7.8 × 10–5 cm and that of blue light is, 5.2 × 10–5cm. The value of n for which (n + 1)th blue bright band coincides with nth bright red band is, (1) 1, , (2), , 2, , (3), , 3, , (4), , 4, , Sol. Answer (2), Distance of nth red band from central fringe =, , n λR D, d, , Distance of (n + 1)th band (blue) from fringe =, , (n + 1) λB D, d, , n λR D (n + 1)λB D, =, d, d, 17. In Young’s double-slit experiment, a glass plate of refractive index 1.5 and thickness 5 × 10–4 cm is kept in, the path of one of the light rays. Then, (1) There will be no shift in the interference pattern, (2) The fringe width will increase, (3) The fringe width will decrease, (4) The optical path of the ray will increase by 2.5 × 10–4 cm, Sol. Answer (4), Optical path of the ray will become = × , Hence new path = 7.5 × 10–4, Increase in optical path length = (7.5 – 5) × 10–4, = 2.5 × 10–4, 18. Which of the following is correct regarding microscope and telescope?, (1) Telescope provides magnification, whereas microscope provides resolution, (2) Telescope provides resolution whereas microscope provides magnification, (3) Both provide resolution, (4) Both provide magnification, Sol. Answer (2), A telescope produces images of far object nearer to our eye. Therefore objects which are not resolved at far, distance, can be resolved by looking at them through a telescope. A microscope on the other hand magnifies object and produced their large image., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 146 :
58, , Wave Optics, , Solution of Assignment, , 19. If Young’s experiment is performed using two separate identical sources of light instead of using two slits and, one light source then the, (1) Interference fringes will be darker, (2) Interference fringes will be brighter, (3) Fringes will not be obtained, (4) Contrast between bright and dark fringes increases, Sol. Answer (3), Two independent sources of light identical in nature are not coherent., 20. In a Young's double slit experiment, the intensity at the central maximum is I0. The intensity at a distance, /4 from the central maximum is ( is fringe width), (2), , (1) I0, , I0, 2, , (3), , I0, 2, , (4), , I0, 4, , Sol. Answer (2), Let the intensity of each source is I, I0 = 4I, , y=, , β, 4, , (Path difference at y distance from central maxima), yd λ, =, D, 4, , x =, , (phase difference) =, , λ ⎛ 2π ⎞ π, ⎜ ⎟=, 4⎝ λ ⎠ 2, , (Intensity at y distance), I = 2I (1 + cos ), π⎞, ⎛I ⎞ ⎛, I = 2 ⎜ 0 ⎟ ⎜1 + cos ⎟, ⎝ 4 ⎠⎝, 2⎠, , I =, , I0, 2, , 21. White light is used to illuminate the two slits in Young’s experiment. The separation between the slits is b, and the screen is at a distance d (>> b) from the slits. At a point directly in front of one of the slits, certain, wavelengths are missing. The missing wavelength(s) is/are, (1) = b2/d, , (2), , = b2/5d, , (3), , = b2/3d, , (4), , All of these, , Sol. Answer (4), Those wavelength will be missing for which the destructive interference occurs at the position, y =, , b, 2, , For destructive interference, d = b, D=d, , yd ′ (2m − 1) λ, =, 2, D, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 147 :
Solution of Assignment, , Wave Optics, , 59, , b ⎛ b ⎞ (2m − 1) λ, ⎜ ⎟=, 2 ⎝d ⎠, 2, , =, , b2, (2m − 1) d, , The possible wavelength are for m = 1, 2, 3, 22. Oil floating on water looks coloured due to interference of light. What should be the order of magnitude of the, thickness of oil layer in order that this effect may be observed?, (1) 10–6 m, , (2), , 10–2 m, , (3), , 10–10 m, , (4), , 10–8 m, , Sol. Answer (1), The order of thickness must be in the order of the wavelength of visible light., 23. When white light is incident normally on an oil film of thickness 10–4 cm and refractive index 1.4 then the, wavelength which will not be seen in the reflected system of light is, (1) 7000 Å, , (2), , 5600 Å, , (3), , 4000 Å, , (4), , All of these, , Sol. Answer (4), For the wavelength which will not be seen, 2t cos r = n, r = 0 (Normal incidence), 2(1.4) 10–4 × 10–2 cos0º = n, n = 2.8 × 10–6, n = 28000 × 10–10, =, , 28000, Å, n, , For possible wavelength n = 1, 2, 3, ....., 24. Imperfections in optical lenses can be observed with the help of, (1) Newton's Rings, , (2), , Fresnel's Biprism, , (3) Lloyd's single miror experiment, , (4), , Young's double slit experiment, , (3), , Dispersion of light, , Sol. Answer (1), 25. Dark and colour patterns on thin soap films are due to, (1) Interference of light, , (2), , Diffraction of light, , (4), , Polarization of light, , Sol. Answer (1), The colours are seen due to this film interference., 26. Choose the correct statement, (1) While watching television by means of an antenna, a passing nearby aeroplane can produce wavering ghost, images in the television picture, (2) Solar cells are often coated with a transparent thin film, such as silicon monoxide (SiO) to minimize, reflective losses., (3) Glass lenses used in cameras and other optical instruments are usually coated with a transparent thin film,, such as magnesium fluoride (MgF2) to reduce or eliminate unwanted reflection, (4) All of these, Sol. Answer (4), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 148 :
60, , Wave Optics, , Solution of Assignment, , 27. The beautiful iridescent (like a rainbow) colors on the wings of a tropical or morpho butterfly are due to, (1) Thin film interference of light, , (2), , Diffraction of light, , (3) Polarization of light, , (4), , Dispersion of light, , Sol. Answer (1), 28. Some currency notes (to avoid counterfeits) change their colour as you tilt them. This is due to, (1) Diffraction, , (2), , Polarization, , (3), , Interference, , (4), , Refraction, , (3), , Polarization, , (4), , Absorption, , (3), , X-rays, , (4), , All of these, , Sol. Answer (3), Thin film interferance, 29. Rainbows are classic example of the phenomenon of, (1) Interference, , (2), , Diffraction, , Sol. Answer (1), 30. The phenomenon of diffraction can be exhibited by, (1) Infrared waves, , (2), , Microwaves, , Sol. Answer (4), All electromagnetic waves can exhibit diffraction provided the size of aperture/obstacle is comparable to the, wavelength of light., 31. Though quantum theory of light can explain a number of phenomenon observed with light, it is necessary to, retain the wave nature of light to explain the phenomenon of, (1) Photoelectric effect, , (2), , Diffraction, , (3), , Compton effect, , (4), , Black body radiation, , Sol. Answer (2), 32. A diffraction pattern is obtained using a beam of red light. What happens, if the red light is replaced by blue, light?, (1) No change, (2) Diffraction bands become narrower and get crowded together, (3) Bands become broader and farther apart, (4) Bands disappear, Sol. Answer (2), Position of minima in diffraction pattern, , nλ, a, Decrease of makes the diffraction bands narrower., θ=, , 33. The main difference in the phenomenon of interference and diffraction in light waves is that, (1) Diffraction is due to interaction of light from the same wave-front whereas interference is the interaction of, waves from two isolated sources, (2) Diffraction is due to interaction of light from same wavefront, whereas the interference is the interaction of, two waves derived from the same source, (3) Diffraction is due to interaction of waves derived from the same source, whereas the interference is the, bending of light from the same wavefront, (4) Diffraction is caused by reflected waves from a source whereas interference is caused due to refraction, of waves from a surface, Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 149 :
Solution of Assignment, , Wave Optics, , 61, , 34. The condition for observing Fraunhofer diffraction from a single slit is that the wave front incident on the slit, should be, (1) Spherical, , (2), , Cylindrical, , (3), , Plane, , (4), , Elliptical, , Sol. Answer (3), Source is at far position from slit., 35. A parallel beam of monochromatic light of wavelength 5000 Å is incident normally on a single narrow slit of, width 0.001 mm. The light is focussed by a convex lens on a screen placed on focal plane. The first minimum, will be formed for the angle of diffraction equal to, (1) 0°, , (2), , 15°, , (3), , 30°, , (4), , 50°, , Sol. Answer (3), dsin = , sin =, sin =, , 5000 × 10−10, .001 × 10−3, , 1, 2, , = 30º, 36. Monochromatic light of wavelength 580 nm is incident on a slit of width 0.30 mm. The screen is 2m from the, slit. The width of the central maximum is, (1) 3.35 × 10–3 m, , (2), , 2.25 × 10–3 m, , (3), , 6.20 × 10–3 m, , (4), , 7.7 × 10–3 m, , Sol. Answer (4), W=, , 2 Dλ, a, , D = 2m, a = 0.3 × 10–3 m, = 580 × 10–9 m, 37. The resolving power of a compound microscope will be maximum when, (1) Red light is used to illuminate the object, (2) Violet light is used to illuminate the object instead of red light, (3) Infra red light is used to illuminate the object instead of visible light, (4) The microscope is in normal adjustment, Sol. Answer (2), Resolving power ∝, , 1, , > violet, λ red, , 38. Why a DVD stores almost 30 times more information than a CD?, (1) DVD uses shorter - wavelength lasers of 6350Å but CD uses an infrared laser of 7800 Å, (2) CD uses shorter wavelength laser compared to a DVD, (3) CD works on the principle of diffraction, (4) DVD works on diffraction of light, Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 150 :
62, , Wave Optics, , Solution of Assignment, , 39. If a classroom door is open just a small amount, we can hear sounds coming from the room but we can't see, what is going on inside the room because, (1) Diffraction of sound is easier as its wavelength is large, (2) Diffraction of light is easier as its wavelength is small, (3) Sound waves can be polarized, (4) Light waves can be polarized, Sol. Answer (1), Audible frequency of sound is less and wavelength is large., 40. When you look at a clear blue sky you see tiny specks and hair like structures floating in your view, called, "floaters". This is basically, (1) Interference pattern, , (2), , Diffraction pattern, , (3), , Emission spectra, , (4), , Absorption spectra, , Sol. Answer (2), 41. Unpolarised beam of light of intensity I is incident on two polarisers in contact. The angle between the axes, of the two polarisers is . Intensity of the light finally emerging from the combination is, (1) I cos2 , , (2), , ⎛I ⎞, ⎜ ⎟ cos 2 , ⎝2⎠, , (3), , I cos4 , , (4), , ⎛I ⎞, ⎜ ⎟ cos , ⎝2⎠, , Sol. Answer (2), After first polarisation intensity becomes, After second polarisation it is, , I, 2, , I, cos2 by law of malus., 2, , 42. A beam of light AO is incident on a glass slab (m = 1.54) in a direction as shown in the diagram. The reflected, ray OB is passed through a polariod. On viewing through the polariod, we find that on rotating the polariod, (Given tan 57° = 1.54), , B, , A, 33°, , 33°, O, , (1) The intensity is reduced down to zero and remains zero, (2) The intensity reduces down some what and rises again, (3) There is no change in intensity, (4) The intensity gradually reduces to zero and then again increases, Sol. Answer (4), The ray is incident at Brewster's angle so it reflected ray will be plane polarised., When passed through polarises the ray will display intensity according to the law of malus (I0 cos2 )., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 151 :
Solution of Assignment, , Wave Optics, , 63, , SECTION - B, Objective Type Questions, 1., , Two light rays initially in same phase travel through two media of equal length L having refractive index m1 and, m2 (m1 > m2) as shown in figure. If the wave length of light rays in air is l, the phase difference of the emerging, rays is given by, , Air, , µ1, , Air, , µ2, L, L1, (1) , 2, , (2), , (1 2 )L, 2, , (3), , 2(1 2 )L, , , (4), , Zero, , Sol. Answer (3), Path difference between the emerging rays = 1L – 2L, , (μ − μ2 )L, φ, = 1, 2π, λ, , , 2., , φ=, , (μ1 − μ 2 )L, λ, , Light wave travel in vacuum along the x-axis, which of the following may represent the wave front, (1) x = a, , (2), , y=a, , (3), , z=a, , (4), , x+y+z=a, , Sol. Answer (1), Wave front is always perpendicular to direction of propagation of light., 3., , In Young double slit experiment, 12 fringes are obtained in a certain fragment of the screen, when light of, wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in, the same segment of the screen is, (1) 12, , (2), , 18, , (3), , 24, , (4), , 30, , Sol. Answer (2), Width of screen is =, , 12λ1D, d, , For new wavelength =, , 12λ1D n λ 2D, =, d, d, , Number of fringes (n) =, , 12λ1, 6000, = 12 ×, = 18, λ2, 4000, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 152 :
64, 4., , Wave Optics, , Solution of Assignment, , Two points separated by 0.05 mm can just be inspected in a microscope when light of wavelength 6000 Å is, used. If light of wavelength 3000 Å is used then the limit of resolution becomes, (1) 0.05 mm, , (2), , 0.025 mm, , (3), , 0.1 mm, , (4), , 0.15 mm, , Sol. Answer (2), Resolving power of a microscope is the shorter distance between two seperate point in a microscope's field, of view that can be seen directly., Resolving power ∝, , λ, (a = aperture), 2a, , d1 λ1, 0.05 6000, =, ⇒, =, ⇒ 0.025 mm, d2 λ2, d2, 3000, , 5., , A thin film of water ( = 4/3) is 3100 Å thick. If it is illuminated by white light at normal incidence. The colour, of film in the reflected light will be, (1) Blue, , (2), , Black, , (3), , Yellow, , (4), , Red, , Sol. Answer (3), Wave length for which maxima occurs, 2t cos r = (2n – 1), , λ=, , λ, 2, , 4μt cos r, 2n − 1, , r = 0, , , 4 t, 2n 1, , for n = 2, , , 4 3100 , 3, , 4, 3, , , , = 5511.11 Å, (Wavelength of yellow), 6., , The central fringe of the interference pattern produced by light of wavelength 6000 Å is found to shift to the, position of 4th bright fringe after a glass plate of refractive index 1.5 is introduced in front of one slit in Young’s, experiment. The thickness of the glass plate will be, (1) 4.8 μm, , (2), , 8.23 μm, , (3), , 14.98 μm, , (4), , 3.78 μm, , Sol. Answer (1), Fringe shift =, , (µ − 1)tD, d, , 4th bright fringe =, , Equating both =, , 4 λD, d, ( μ − 1)tD 4 λ D, =, d, d, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 153 :
Solution of Assignment, , 7., , Wave Optics, , 65, , In Young’s double slit experiment shown in figure. S1 and S2 are coherent sources and S is the screen having, a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits., Which wavelength passing through the hole has strong intensity?, , 0.5 mm, , centre of, screen, , S1, S2, hole, , 50 cm, , (1) 400 nm, , (2), , 700 nm, , 1.0 mm, , S, (3), , screen, , 500 nm, , (4), , 667 nm, , Sol. Answer (3), Yn =, , nD λ, d, , Yn = 10–3m, D = 0.5 m,, , λ=, 8., , d = 0.5 × 10–3 m, , 500, nm, n, , In Young’s double slit experiment, the intensity of light at a point on the screen where the path difference is, , is, is I0. The intensity of light at a point where the path difference becomes, 3, (2), , (1) I0, , I0, 4, , (3), , I0, 3, , (4), , I0, 2, , (iv), , ⎞, ⎛, y 4 a4 sin ⎜ t ⎟, 3, ⎝, ⎠, , Sol. Answer (2), Where the path difference is , the fringe intensity will be maximum., I0 = 4I, (I= intensity of each slit), Where, ∆ × =, , λ, ,, 3, , ∆φ =, , 2π, 3, , I = 4ICos2 ∆φ, 2, , I = I0 Cos 2, I=, , 9., , π, 3, , I0, 4, , Four different independent waves are represented by, (i) y1 = a1sint, , (ii), , y2 = a2sin2t, , (iii) y3 = a3cost, , With which of two waves interference is possible?, (1) (i) & (ii), , (2), , (i) & (iv), , (3) (iii) & (iv), , (4), , Not possible with any combination, , Sol. Answer (4), (They are independent waves), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 154 :
66, , Wave Optics, , Solution of Assignment, , 10. In YDSE, a thin film (μ = 1.6) of thickness 0.01 mm is introduced in the path of one of the two interfering, beams. The central fringe moves to a position occupied by the 10th bright fringe earlier. The wave length of, wave is, (1) 600 Å, , (2), , 6000 Å, , (3), , 60 Å, , (4), , 660 Å, , Sol. Answer (2), Shift of fringes is given by, , (µ − 1)tD, d, , This is equal to position of 10th bright fringe, , 10 λD, d, , 0.6 × 1 × 10−5 D 10λD, =, d, d, = 0.6 × 10–6, = 6 × 10–7, or = 6000 Å, , 11. In Young's double slit experiment, the intensity at a point where path difference is, , , is I. If I0 denotes the, 6, , I, maximum intensity, I ., 0, , (1), , 1, , (2), , 2, , 3, 2, , (3), , 1, 2, , (4), , 3, 4, , Sol. Answer (4), Where, ∆x =, , , , λ, 6, , ∆φ =, , 2π, ∆x, λ, , ∆φ =, , π, 3, , ⎛ ∆φ ⎞, I = I0 cos2 ⎜ ⎟, ⎝ 2 ⎠, ⎛π⎞, I = I0 cos2 ⎜ ⎟, ⎝6⎠, , or I = I0 ×, , , , 3, 4, , I, 3, =, I0 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 155 :
Solution of Assignment, , Wave Optics, , 67, , 12. The maximum intensity of fringes in Young's experiment is I. If one of the slits is closed, then intensity at, that place becomes I0. Then relation between I and I0 is, (2), , (1) I = I0, , I = 2I0, , (3), , I = 4I0, , (4), , There is no relation, , Sol. Answer (3), When the slits were open the waves interfered constructively, I = 4I (I = Intensity of each slit), If one of the slits is closed then intensity at that place is intensity of each slit I = I0, 13. In Young's double slit interference experiment, the slit separation is made 3 folds. The fringe width becomes, (1), , 1, times, 3, , (2), , 1, times, 9, , (3), , 3 times, , (4), , 9 times, , Sol. Answer (1), Fringe width =, , β' =, , λD, d, , λD β, =, 3d 3, , 14. In Young's double slit experiment, the distance between the slits is reduced to half and the distance between, the slit and the screen is doubled, then fringe width, (1) Will not change, , (2), , Will become half, , (3), , Will be doubled, , (4), , Will become four times, , Sol. Answer (4), β=, , λD, d, , β' =, , λ × 2D, d /2, , , , ' = 4, , 15. In Young's experiment, the separation between 5th maxima and 3rd minima is how many times as that of fringe, width?, (1) 5 times, , (2), , 3 times, , (3), , 2.5 times, , (4), , 2 times, , Sol. Answer (3), 5th maximum =, , 5 λD, d, , 3rd minimum =, , (2n − 1)λD 5 λD, =, 2d, 2d, , Distance =, , 5 λD 5 λD, −, d, 2d, , =, , 5 λD, 2d, , Distance = 2.5 times, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 156 :
68, , Wave Optics, , Solution of Assignment, , 16. Refractive index of material is equal to the tangent of polarising angle. It is called, (1) Brewster's law, , (2), , Lambert's law, , (3), , Malus' law, , (4), , Bragg's law, , Sol. Answer (1), m = tan ip, 17. If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities of maxima and, minima is, (1) 25 : 16, , (2), , 5:3, , (3), , 16 : 1, , (4), , 25 : 9, , Sol. Answer (3), A1 : A2 = 3 : 5 then I1 : I2 = 9 : 25, Imax, =, Imin, , =, , (, (, , I1 + I2, I1 − I2, , ), ), , 2, 2, , ⎛3 + 5⎞, =⎜, ⎝ 3 − 5 ⎟⎠, , 2, , 16, 1, , 18. In a Young's double slit experiment, the source illuminating the slits is changed from blue to violet. The width, of the fringes, (1) Increases, , (2), , Decreases, , (3), , Becomes unequal, , (4), , Remains same, , Sol. Answer (2), =, , λD, d, , If decreases width of the fringes decreases., 19. In Young's double slit experiment, when two light waves form third minimum, they have, (1) Phase difference of 3, , (2), , Path difference of 3, , 5, 2, , (4), , Path difference of, , (1) Should be of the same order as wavelength, , (2), , Should be much smaller than the wavelength, , (3) Has no relation to wavelength, , (4), , Should be exactly, , (3) Phase difference of, , 5, 2, , Sol. Answer (4), Path difference is given by x = (2n – 1), , 5λ, λ, =, 2, 2, , 20. To observerve diffraction the size of an obstacle, , , 2, , Sol. Answer (1), 21. If frequency of light wave propagating in water is halved, its speed, (1) Is halved, , (2), , Is doubled, , (3), , Remains same, , (4), , Becomes four times, , Sol. Answer (3), Speed of light remains same in a medium., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 157 :
Solution of Assignment, , Wave Optics, , 69, , 22. White light is used in Young’s double slit experiment. Separation between slits is y and the screen is at, distance x from slits (x >> y). Which of these wavelengths is missing in front of one of the slits?, , y2, x, , (1), , (2), , y2, 2x, , (3), , y2, 4x, , (4), , All of these, , Sol. Answer (1), , Distance of minima from central maxima y1 =, , y1 =, , λ=, , (2m − 1) λD, 2d, , y, , d = y, D = x, 2, , y2, (2m − 1) x, , m = 1, 2, 3, ...., , 23. Corpuscular theory of light predicts speed of light to be, (1) Independent of medium, , (2), , Greater in water than in vacuum, , (3) Greater in vacuum than in water, , (4), , Dependent on intensity of light, , Sol. Answer (2), 24. Shape of interference fringes formed on the screen due to point source P, in the case shown here, , (1) Parabolic, , (2), , Elliptical, , (3), , Circular, , (4), , Hyperobic, , Sol. Answer (3), 25. In Fraunhoffer diffraction from a single slit, wave front incident on the slit is, (1) Planar, , (2), , Spherical, , (3) Cylindrical, , (4), , Either spherical or cylindrical, , Sol. Answer (1), Fraunhoffer diffraction is for the parallel incidence., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 158 :
70, , Wave Optics, , Solution of Assignment, , 26. Young’s double slit experiment is performed with monochromatic light. A thin film is introduced in front of one, of the slits, (1) Intensity at the position of central maxima must decrease, (2) Intensity at the position of central maxima may increase, (3) Central maxima may remain unshifted, (4) Intensity at position of first maxima may decrease, Sol. Answer (4), The entire fringe will experience a shifts and the only thing we can say for certain is the fourth option., 27. Apparent wavelength of light coming from a star moving away from earth is 0.02% more than its actual, wavelength. Velocity of star is, (1) 30 km/s, , (2), , 60 km/s, , (3), , 90 km/s, , (4), , 120 km/s, , Sol. Answer (2), λs, , C +V, C + V ⎛ 0.02 ⎞, = λ0 ⇒ λs, = ⎜1 +, ⎟λ, 100 ⎠ s, C −V, C −V ⎝, 2, , C + V ⎛ 0.02 ⎞, C +V, 0.04, 2V, 1, = ⎜1 +, ⇒, −1=, ⇒, =, ⎟, C −V ⎝, 100 ⎠, C −V, 100, C − V 2500, , or, , V =, , 3 × 105, ≈ 60 km/s, 5001, , 28. Diffraction is easily noticeable for sound waves than for light waves because sound waves, (1) Are high energy waves, , (2), , Are low intensity waves, , (3) Have longer wavelength, , (4), , Are mechanical in nature, , Sol. Answer (3), For diffraction the obstacle and wavelengths must be of same order. Since they have longer wavelength they, can easily cleaned around everyday absects more easily., 29. In the Young’s arrangement, screen starts moving towards right with constant speed v. Initial distance between, screen and plane of slits is x. At t = 0, 1st order maxima is lying at point A. After how much time first order, minima lies at point A?, x, A, y, Source, , v, , Screen, (t = 0), , (1), , x, 2v, , (2), , x, v, , (3), , x, 3v, , (4), , 2x, 3v, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 159 :
Solution of Assignment, , Wave Optics, , 71, , Sol. Answer (2), Distance of 1st maxima from central fringe =, , Distance of 1st minima =, , λD, d, , [D = x], , λD, 2d, , D at time t = Dt = x + Vt, λx λ( x + Vt ), =, d, 2d, , x = Vt, t=, , or, , x, V, , 30. Unpolarized light of intensity x is incident on a polarising sheet. Intensity of light which does not get transmitted, is, (1) x, , (2), , x, 2, , (3), , x, 4, , (4), , Zero, , Sol. Answer (2), Polariser transmits energy travelling along a single plane., Energy transmitted is, , x, 2, , Hence energy left behind is x −, , x x, =, 2 2, , 31. Light of wavelength l is coming from a star. What is the limit of resolution of a telescope whose objective has, diameter r ?, (1), , 0.305 , r, , (2), , 0.61, r, , (3), , 1.22 , r, , (4), , 2, r, , (4), , cot 1, , Sol. Answer (3), , 32. Brewster angle for air to water transition is (refractive index of water is, 1, (1) sin, , 3, 4, , (2), , cos 1, , 3, 4, , (3), , tan 1, , 4, ), 3, , 3, 4, , 3, 4, , Sol. Answer (4), Brewster's angle is given by tan–1 = tan–1, , 4, 3, which is also written as cot–1, 3, 4, , 33. Approximate thickness of oil film to observe interference of light (due to which it looks coloured) is, (1) 10 mm, , (2), , 10–3 mm, , (3), , 10 pm, , (4), , 1 cm, , Sol. Answer (2), The thickness must be in the order of visible light., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 160 :
72, , Wave Optics, , Solution of Assignment, , 34. Slit widths in a Young’s double slit experiment are in the ratio 9 : 4. Ratio of intensity at minima to that at, maxima is, (1) 4 : 9, , (2), , 16 : 81, , (3), , 1 : 25, , (4), , 1 : 16, , Sol. Answer (3), , W ∝I, I1 9, =, I2 4, , Imin, =, Imax, , (, (, , I1 − I2, I1 + I2, , ), ), , 2, 2, , 35. Width of slit in a single slit diffraction experiment such that 20 maxima of double slit interference pattern are, obtained within central maxima of the diffraction pattern is (Slit separation for double slit arrangement = 2 mm), (1) 0.05 mm, , (2), , 0.1 mm, , (3), , 0.2 mm, , (4), , 0.4 mm, , Sol. Answer (3), 20, , D λ 2D λ, =, d, b, , ,, , d = 2mm, , 36. In Fraunhoffer diffraction, at the angular position of first diffraction minimum, phase difference (in radian) between, wavelets from opposite edges of the slit is, (1), , , 2, , (2), , , , (3), , 2, , (4), , 4, , Sol. Answer (3), For minima in diffraction x = m, For first minima x = , = 2, 37. Choose the correct alternative, (1) When plane polarised light passes through polaroid, it changes its nature to linearly polarised, (2) Refracted light, when light is incident at Brewster angle, is linearly polarised, (3) Polarised light can be produced by scattering through, , , in earth’s atmosphere, 2, , (4) Natural light from sun is polarised, Sol.Answer (3), , SECTION - C, Previous Years Questions, 1., , At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference, between the Huygen's wavelet from the edge of the slit and the wavelet from the mid-point of the slit is, [Re-AIPMT-2015], (1), , , radian, 8, , (2), , , radian, 4, , , radian, 2, , (3), , (4), , radian, , Sol. Answer (4), For 1st minimum d sin ., where = Path difference between the wavelets from edges of the slit., Path difference between edge and mid-point of slit =, Phase difference between edge and mid-point of slit = , , , 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 161 :
Solution of Assignment, , 2., , Wave Optics, , 73, , Two slits in Youngs experiment have widths in he ratio 1 : 25. The ratio of intensity at the maxima and minima, Imax, in the interference pattern, I, is, min, , 4, 9, , (1), , (2), , 9, 4, , [Re-AIPMT-2015], , (3), , 121, 49, , (4), , 49, 121, , Sol. Answer (2), I, w, 1, 1 , w 2 I2 25, 2, , Imax, Imin, , 3., , ⎛ I, ⎞, ⎜ 1 1⎟, 2, ⎜ I, ⎟, ⎛ 5 1⎞, 36 9, ⎜ 2, , , ⎟, ⎜, ⎟ , 5, –, 1, 16, 4, ⎝, ⎠, ⎜ I1, ⎟, – 1⎟, ⎜⎜, ⎟, I, ⎝ 2, ⎠, , In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic, light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double, slit within the central maxima of single slit pattern?, [AIPMT-2015], (1) 0.02 mm, , (2), , 0.2 mm, , (3), , 0.1 mm, , (4), , 0.5 mm, , Sol. Answer (2), 10, , D λ 2D λ, =, d, b, , D = 1m. d = 1 mm, l = 500 nm, b = Width of each slit, , b=, , d, 5, , b = 0.2 mm, 4., , For a parallel beam of monochromatic light of wavelength , diffraction is produced by a single slit whose width, a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of, the central maxima will be, [AIPMT-2015], (1), , 2Da, , , (2), , 2D, a, , (3), , D, a, , (4), , Da, , , Sol. Answer (2), Width of central maxima = Distance between first dark fringes on either side of the central fringe =, 5., , 2D λ, a, , A beam of light of = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction, pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central, bright fringe is, [AIPMT-2014], (1) 1.2 cm, , (2), , 1.2 mm, , (3), , 2.4 cm, , (4), , 2.4 mm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 162 :
74, , Wave Optics, , Solution of Assignment, , Sol. Answer (4), Distance between first dark fringes on either side of the central bright fringe, , 2D λ, b, , D=2m, = 600 nm, b = 1 mm, 6., , In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is , is K, ( being the wave length of light used). The intensity at a point where the path difference is /4, will be:, [AIPMT-2014], (1) K, , (2), , K, 4, , K, 2, , (3), , (4), , Zero, , Sol. Answer (3), Where the path difference is the intensity is maximum, Imax = 4I0, I0 (intensity of each slit), K = 4I0, Path difference =, , λ, 4, , Phase difference () =, , I, , λ ⎛ 2π ⎞, π, ⎜ ⎟ =, 4⎝ λ ⎠, 2, , ⎛ ∆φ ⎞, = 4I0cos2 ⎜⎝ ⎟⎠, 2, = 4I0 cos2, , π, 4, , = 2I0, ⎛K ⎞, = 2 ⎜⎝ ⎟⎠, 4, , =, 7., , K, 2, , In Young's double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths 1, = 12000 Å and 2 = 10000 Å. At what minimum distance from the common central bright fringe on the screen 2m, from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?, [NEET-2013], (1) 6 mm, , (2), , 4 mm, , (3), , 3 mm, , (4), , 8 mm, , Sol. Answer (1), 8., , A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed, at a large distance from the slit. If the speed of the electrons is increased, which of the following statements, is correct?, [NEET-2013], (1) The angular width of the central maximum of the diffraction pattern will increase, (2) The angular width of the central maximum will decrease, (3) The angular width of the central maximum will be unaffected, (4) Diffraction pattern is not observed on the screen in the case of electrons, , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 163 :
Solution of Assignment, , 9., , Wave Optics, , 75, , Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum, of the maximum and minimum intensities is, [AIPMT (Prelims)-2008], (1) 2(I1 + I2), , (2), , (I1 + I2), , (3), , ( I1 I2 )2, , (4), , , , I1 I2, , , , 2, , Sol. Answer (1), 10. The angular resolution of a 10 cm diameter telescope at a wavelength of 5000 Å is of the order of, [AIPMT (Prelims)-2005], (1) 106 rad, , (2), , 10–2 rad, , (3), , 10–4 rad, , (4), , 10–6 rad, , Sol. Answer (4), 11. A star, which is emitting radiation at a wavelength of 5000 Å, is approaching the earth with a velocity of, 1.5 × 105 m/s. The change in wavelength of the radiation as received on the earth is, (1) 25 Å, , (2), , 100 Å, , (3), , Zero, , (4), , 2.5 Å, , Sol. Answer (4), Emitting at a wavelength of 5000 Å, ∆λ V, =, λ, C, , ∆λ =, , V, 1.0 × 10 4, ×λ=, × 5000 × 10 −10 = 0.25 Å, C, 3 × 108, , 12. For a wavelength of light ‘l’ and scattering object of size ‘a’, all wavelengths are scattered nearly equally, if, (1) a = , , (2), , a >> , , (3), , a << , , (4), , a , , Sol. Answer (2), For a >> Rayleigh and scattering not valid., Here all wave lengths scattered nearly equally., 13. If two sources have a randomly varying phase difference (t), the resultant intensity will be given by, (1) I0, , (2), , I0, 2, , (3), , 2I0, , (4), , I0, 2, , Sol. Answer (3), If phase difference varies randomly with time, the wave are incoherent and the intensity of resultant waves, is sum of individual intensities or 2I0., 14. In Young’s double-slit experiment, if the distance between the slits is halved and the distance between the, slits and the screen is doubled, the fringe width becomes, (1) Half, , (2), , Double, , (3), , Four times, , (4), , Eight times, , Sol. Answer (3), =, , λD, d, , =, , 2λD, = 4, d /2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 164 :
76, 15, , Wave Optics, , Solution of Assignment, , In a Fresnel biprism experiment, the two positions of lens give separation between the slits as 16 cm and 9, cm respectively. What is the actual distance of separation?, (1) 13 cm, , (2), , 14 cm, , (3), , 12.5 cm, , (4), , 12 cm, , Sol. Answer (4), d1 = 16 cm d2 = 9 cm, Actual distance 2d =, , d1d2, , 2d = 4 × 3, or 2d = 12 cm, 16. Colours appear on a thin soap film and on soap bubbles due to the phenomenon of, (1) Interference, , (2), , Dispersion, , (3), , Refraction, , (4), , Diffraction, , Sol. Answer (1), The phenomenon of thin film interferens is in effect., 17. On introducing a thin film in the path of one of the two interfering beam, the central fringe will shift by one, fringe width. If = 1.5, the thickness of the film is (wavelength of monochromatic light is l), (1) 4, , (2), , 3, , (3), , 2, , (4), , , , Sol. Answer (3), Shift shown =, , λD, d, , λD (μ − 1)tD, =, d, d, or = (1.5 – 1)t, or t = 2, 18. Two coherent monochromatic light beams of intensities I and 4I are superimposed; the maximum and minimum, possible intensitites in the resulting beam are, (1) 5I and I, , (2), , 5I and 3I, , (3), , 9I and I, , (4), , 9I and 3I, , Sol. Answer (3), I1 = I , I2 = 4I, Imax =, , (, , I1 + I2, , Imin =, , (, , I1 + I2, , ), , 2, , ), , 2, , 19. If two waves, each of intensity I0, having the same frequency but differing by a constant phase angle of 60°,, superimposing at a certain point in space, then the intensity of the resultant wave is, (1) 2I0, , (2), , 3I0, , (3), , 3I0, , (4), , 4I0, , Sol. Answer (2), Resultant intensity is given by, Ir = I0 + I2 + 2 I1I2 cos φ, Ir = I0 + I1 + 2I0 cos60º, Ir = 3I0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 165 :
Solution of Assignment, , Wave Optics, , 77, , 20. In Young’s double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed, ⎛4⎞, in water of refractive index ⎜ ⎟ , without disturbing the geometrical arrangement, the new fringe width will be, ⎝3⎠, , (1) 0.40 mm, , (2), , 0.53 mm, , (3), , 0.20 mm, , (4), , 0.30 mm, , Sol. Answer (4), = 4 × 10–3 m, In water becomes, , λ, μ, , β, μ, , or =, , = 0.30 mm, 21. In an interference experiment monochromatic light is replaced by white light, we will see, (1) Uniform illumination on the screen, (2) Uniform darkness on the screen, (3) Equally spaced white and dark bands, (4) A few coloured bands and then uniform illumination, Sol. Answer (4), 22. In Young’s double slit experiment carried out with light of wavelength 5000 Å, the distance between the slit, is 0.2 mm and the screen is at 200 cm from the plane of slits. The central maximum is at x = 0. The third, maximum will be at x equal to, (1) 1.5 cm, , (2), , 1.67 cm, , (3), , 0.5 cm, , (4), , 5.0 cm, , Sol. Answer (1), d = 0.2 mm; = 5000 × 10–10 m; D = 2 m, 3rd maximum =, , n λD, d, , or 3rd maximum is at =, , 3 λD, d, , 23. In Young’s experiment when sodium light of wave length 5893 Å is used, then 62 fringes are seen in the field, of view. Instead, if violet light of wavelength 4350 Å is used, then the number of fringes that will be seen in, the field of view will be, (1) 54, , (2), , 64, , (3), , 74, , (4), , 84, , Sol. Answer (4), Field of view of sodium light =, nv ⋅, , λs D × ns, d, , λv D, D, = ns × λs, d, d, , nv = v = nss, nv =, , 5893 × 62, = 83.99, 4350, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 166 :
78, , Wave Optics, , Solution of Assignment, , 24. In an interference pattern by two identical slits, the intensity of central maxima is I. What will be the intensity, of the same spot, if one of the slits is closed?, , I, 2, , (1), , (2), , I, 4, , (3), , I, , (4), , 2I, , Sol. Answer (2), I = I0 + I 0 + 2 I 0 I0, , I0 =, , I, 4, , 25. If a thin mica sheet of thickness ‘t’ and refractive index ‘m’ is placed in the path of one of the waves producing, interference, then the whole interference pattern shifts towards the side of the sheet by a distance, (1), , d, ( 1) t, D, , (2), , D, ( 1) t, d, , (3), , Dd ( 1) t, , (4), , ( 1)t, , Sol. Answer (2), 26. In Young’s experiment, the wavelength of red light is 7.8 × 10–5 cm and that of blue light 5.2 × 10–5 cm. The, value of n for which (n + 1)th blue bright band coincides with nth red bright band is, (1) 4, , (2), , 3, , (3), , 2, , (4), , 1, , Sol. Answer (3), r = 7.8 × 10–5 cm, b = 5.2 × 10–5 cm, , nλr D (n + 1)λbD, =, d, d, 27. A slit 5 cm wide is irradiated normally with micro waves of wavelength 1 cm. Then the angular spread of the, central maximum on either side of the incident light is nearly, (1), , 1, radian, 5, , (2), , 4 radian, , (3), , 5 radian, , (4), , 6 radian, , Sol. Answer (1), b = 5 cm, = 1 cm, Angular spread of Central maxima =, , λ, on either side., b, , 28. In Young’s double slit experiment, the 10th maximum of wavelength l1 is at a distance of y1 from the central, maximum. When the wavelength of the source is changed to l2, 5th maximum is at a distance of y2 from its, , ⎛ y1 ⎞, central maximum. The ratio ⎜ ⎟ is, ⎝ y2 ⎠, (1), , 2 2, 1, , (2), , 1, 2 2, , (3), , 2, 21, , (4), , 21, 2, , Sol. Answer (4), , y1 =, , 10 λ1D, 5 λ 2D, , y2 =, d, d, , y1 2λ1, =, y2, λ2, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 167 :
Solution of Assignment, , Wave Optics, , 79, , 29. A beam of light strickes a piece of glass at an angle of incidence of 60° and the reflected beam is completely, plane polarised. The refractive index of the glass is, (1) 1.5, , (2), , 3, , (3), , 2, , (4), , 3, 2, , Sol. Answer (2), Brewster's angle = tan–1 , or tan 60º = , , , µ= 3, , 30. Waves that cannot be polarised are, (1) Light waves, , (2), , Electromagnetic waves, , (3) Transverse waves, , (4), , Longitudinal waves, , Sol. Answer (4), Since the vibrate along the direction of propagation., 31. Two polaroids are kept crossed to each other. Now one of them is rotated through an angle of 45°. The, percentage of incident light now transmitted through the system is, (1) 15%, , (2), , 25%, , (3), , 50%, , (4), , 75%, , Sol. Answer (2), Intensities after first polarises =, , I0, 2, , After that applying law of malus, I = I cos2, I =, , = 45º, , I0, 4, , 32. When the angle of incidence is 60° on the surface of a glass slab, it is found that the reflected ray is, completely polarised. The velocity of light in glass is, 2 108 m/s, , (1), , (2), , 3 108 m/s, , (3), , 2 × 108 m/s, , (4), , 3, 108 m/s, 2, , Sol. Answer (2), tan60º = , , [Brewster's law], , Velocity of light in glass =, , C, µ, , = 3 × 108 m/s, , 33. Light of wavelength l is incident on a slit of width ‘d’. The resulting diffraction pattern is observed on a screen, at a distance D. The linear width of the principal maximum is then equal to the width of the slit if D equals, (1), , d, , , (2), , 2, d, , (3), , d2, 2, , (4), , 2 2, d, , Sol. Answer (3), , 2Dλ, = d,, d, , D=, , d2, 2λ, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 168 :
80, , Wave Optics, , Solution of Assignment, , 34. In a Fraunhofer diffraction at a single slit of width d with incident light of wavelength 5500 Å, the first minimum, is observed at angle of 30°. The first secondary maximum is observed at an angle q, equal to, 1 ⎛ 1 ⎞, (1) sin ⎜, ⎟, ⎝ 2⎠, , (2), , ⎛ 1⎞, sin1 ⎜ ⎟, ⎝4⎠, , ⎛3⎞, sin1 ⎜ ⎟, ⎝4⎠, , (3), , (4), , ⎛ 3⎞, sin1 ⎜, ⎟, ⎝ 2 ⎠, , Sol. Answer (3), For first minima, sin θ =, , λ, b, , sin30° =, , λ, b, , For first secondary maxima, sin θ =, , 3λ, 2b, , sin θ =, , 3, 4, , −1, θ = sin, , 3, 4, , 35. Diameter of human eye lens is 2 mm. What will be the minimum distance between two points to resolve them,, which are situated at a distance of 50 m from eye. The wavelength of light is 5000 Å?, (1) 2.32 m, , (2), , 4.28 mm, , (3), , 1.525 cm, , (4), , 12.48 cm, , Sol. Answer (3), Minimum distance between two points =, , 1.22λD, d, , d = 2 mm, D = 50 m, = 5000 Å, , SECTION - D, Assertion-Reason Type Questions, 1., , A : The speed of light in vacuum doesn't depend on nature of the source, direction of propagation, motion of, the source or observer wavelength and intensity of the wave., R : The speed of light in vacuum is a universal constant independent of all the factors listed and anything else., , Sol. Answer (1), 2., , A : The speed of light, sound waves, water waves in a medium is independent of the nature of the source or, intensity (so long it is low)., R : Speed of the waves in a medium depends on wavelength., , Sol. Answer (2), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 169 :
Solution of Assignment, , 3., , Wave Optics, , 81, , A : Speed of light in a medium is independent of the motion of the source relative to the medium., R : Speed of light in a medium depends on the motion of the observer relative to the medium., , Sol. Answer (2), 4., , A : When monochromatic light is incident on a surface separating two media, the reflected and refracted light, both have the same frequency as the incident frequency., R : At any interface between the two media, the electric (and magnetic) fields must satisfy certain boundary, conditions for all times and frequency determines the time dependence of fields., , Sol. Answer (1), 5., , A : When light travels from a rarer to a denser medium, it loses some speed but it doesn't imply a reduction, in the energy carried by the light wave., R : Energy carried by a wave depends on the amplitude of the wave and not on the speed of wave propagation., , Sol. Answer (1), 6., , A : When a narrow pulse of light is sent through a medium, it doesn't retain its shape as it travels through, the medium., R : Since the speed of propagation in a medium depends on wavelength, different wavelength components of, the pulse travel with different speeds., , Sol. Answer (1), 7., , A : In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave., R : In the photon picture of light, for a given frequency, intensity of light is determined by the number of photons, per unit area., , Sol. Answer (2), 8., , A : The speed of light in still water is not same as that in flowing water., R : The speed of light in water is not independent of the relative motion between the observer and the medium., , Sol. Answer (1), 9., , A : In a single-slit diffraction experiment, if the width of the slit is made double the original width the size of, the central diffraction band reduces by half and intensity increase four fold., R : The intensity of interference fringes in a double slit arrangement is modulated by the diffraction pattern of, each slit., , Sol. Answer (2), 10. A : When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen, at the centre of the shadow of the obstacle., R : Waves diffracted from the edge of the circular obstacle interfere constructively at the centre of the shadow, producing a bright spot., Sol. Answer (1), 11. A : Ray optics assumes that light travels in a straight line which is disapproved by diffraction effects, yet the, ray optics assumption is so commonly used in understanding location and several other properties of, images in optical instruments., R : Typical sizes of apertures involved in ordinary optical instruments are much larger than the wavelength of, light., Sol. Answer (1), Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 170 :
82, , Wave Optics, , Solution of Assignment, , 12. A : The phase difference between any two points on a wavefront is zero., R : Corresponding to a beam of parallel rays of light, the wavefronts are planes parallel to one another., Sol. Answer (2), 13. A : Light waves can be polarised., R : Light waves are transverse in nature., Sol. Answer (1), 14. A : The law of conservation of energy is violated during interference., R : For sustained interference the phase difference between the two waves must change with time., Sol. Answer (4), 15. A : When the apparatus of YDSE is brought in a liquid from air, the fringe width decreases., R : The wavelength of light decreases in the liquid., Sol. Answer (1), 16. A : The resolving power of a telescope decreases on decreasing the aperture of its objective lens., R : The resolving power of a telescope is given as, , D, , where D is aperture of the objective and is the, 1.22 , , wavelength of light., Sol. Answer (1), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 171 :
Chapter, , 24, , https://t.me/NEET_StudyMaterial, , Ray Optics and Optical Instruments, Solutions, SECTION - A, Objective Type Questions, 1., , Digital movie projectors work on the principle of, (1) Reflection from micromirrors, , (2), , Refraction from thin lenses, , (3) Dispersion from thin prisms, , (4), , Total internal reflection from optical fibres, , Sol. Answer (1), Digital movie projectors need parabolic mirrors to converge all incident rays to a point., 2., , Day and night settings for rearview mirrors uses, (1) Thin mirrors, , (2), , Thick wedge shaped mirrors, , (3) Convex mirrors, , (4), , Concave mirrors, , Sol. Answer (2), Day and night settings for rear view mirrors., 3., , When a beam of light is incident on a plane mirror, it is found that a real image is formed. The incident beam, must be, (1) Converging, (2) Diverging, (3) Parallel, (4) Formation of real image by a plane mirror is impossible, , Sol. Answer (1), Real images are images formed from actual intersection of light rays., , Position of, image, , Virtual object, , In plane mirror formation of real images is shown above., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 172 :
2, 4., , Ray Optics and Optical Instruments, , Solution of Assignment, , An object is placed symmetrically between two plane mirrors, inclined at an angle of 72°, then the total number, of images observed is, (1) 5, , (2), , 4, , (3), , 2, , (4), , Infinite, , Sol. Answer (2), For number of images formed from plane mirror, m=, , 360 360, =, =5, θ, 72, , If it was at any other point the number of images (n) = m but this is not so., Since it is placed symmetrically :, n=m–1, or n = 5 – 1, or n = 4 images, 5., , A person 1.6 m tall is standing at the centre between two walls three metre high. What is the minimum size, of a plane mirror fixed on the wall in front of him, if he is to see the full height of the wall behind him?, (1) 0.8 m, , (2), , 1m, , (3), , 1.5 m, , (4), , 2.3 m, , Sol. Answer (2), , h2, , 1.6 − h1, h, tan θ = 1 =, d, d /2, , A, , , , ....(i), , , , M1, B, , h, 1.4 − h2, tan α = 2 =, d, d /2, , , , , 3m, , ....(ii), , 1.6 m, h1 =, , 3.2, 2.8, and h2 =, 3, 3, , M2, C, , h1, , 3 – h1 – h2 = 1 m, 6., , While capturing solar energy for commercial purposes we use, (1) Parabolic mirrors, , (2), , Plane mirrors, , (3), , Convex mirrors, , (4), , Concave mirrors, , Sol. Answer (1), While capturing solar energy for commercial purposes we use parabolic mirrors to converge the rays coming, from infinity to a point., 7., , A convex mirror is used to form an image of a real object. Then mark the wrong statement, (1) The image lies between the pole and focus, , (2), , The image is diminished in size, , (3) The image is erect, , (4), , The image is real, , Sol. Answer (4), A convex mirror always forms a virtual image in the care of a real object., In care of a virtual object reflected rays may intersect really to make a real image., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 173 :
Solution of Assignment, , 8., , Ray Optics and Optical Instruments, , 3, , A concave mirror of focal length f produces an image n times the size of the object. If the image is real then, the distance of the object from the mirror is, (1) (n – 1) f, , (2), , ⎧ (n 1) ⎫, ⎬f, ⎨, ⎩ n ⎭, , (3), , ⎧ (n 1) ⎫, ⎬f, ⎨, ⎩ n ⎭, , (4), , (n + 1) f, , Sol. Answer (3), , f, f −u, Focal real image m = –n, (magnification) m =, , –n =, , −f, −f − u, , u =−, 9., , f (n + 1), u, , A convex mirror has a focal length f. A real object is placed at a distance f in front of it, from the pole. It, produces an image at, (1) Infinity, , (2), , f, , (3), , f/2, , (4), , 2f, , Sol. Answer (3), Mirror formula :, , 1 1 1, + =, v u f, , Here object is real so u is negative, , 1 1 1, − =, v u f, Also (u) = f, 1 1 1, − =, v f f, , v=, , f, 2, , 10. An object placed in front of a concave mirror of focal length 0.15 m produces a virtual image, which is twice, the size of the object. The position of the object with respect to the mirror is, (1) –5.5 cm, , (2), , –6.5 cm, , (3), , –7.5 cm, , (4), , –8.5 cm, , Sol. Answer (3), m =, f, , f, f −u, , = –0.15 m, , m = +2 (virtual image), 2 =, , −0.15, −0.15 − u, , = –.075 m or – 7.5 cm., 11. When a light ray from a rarer medium is refracted into a denser medium, its, (1) Speed increases, wavelength increases, , (2), , Speed decreases, wavelength increases, , (3) Speed increases, wavelength decreases, , (4), , Speed decreases, wavelength decreases, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 174 :
4, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (4), When a light ray goes from a rarer to denser medium by definition of refractive index, its speed creases., Its frequency is found to be invarant., So, if velocity of light in a medium is v :, v = f, , [by wave theory], , or v , So, if v decreases, also decreases., 12. A narrow, paraxial beam of light is converging towards a point I on a screen. A plane parallel plate of glass of, thickness t, and refractive index μ is introduced in the path of the beam. The convergence point is shifted by, (1) t (1–1/) away, , (2), , t (1 + 1/) away, , (3), , t (1 – 1/) nearer, , (4), , t (1 + 1/) nearer, , Sol. Answer (1), Longitudinal shift is given by t –, , t, . Hence, point of convergence shifts by the same amount., µ, , 4⎞, ⎛, 13. The length of a vertical pole at the surface of a lake of water ⎜ ⎟ is 24 cm. Then to an under-water fish, 3⎠, ⎝, just below the water surface the tip of the pole appears to be, , (1) 18 cm above the surface, , (2), , 24 cm above the surface, , (3) 32 cm above the surface, , (4), , 36 cm above the surface, , Sol. Answer (3), , µ=, , Apparent height, Real height, , Now, real height = 24 cm and =, , 4, 3, , 4, × 24 = Apparent height, 3, Apparent height = 32 cm, 14. A ray of light strikes a glass plate at an angle 60o. If the reflected and refracted rays are perpendicular to each, other, the index of refraction of glass is, (1), , 3, , (2), , 3/2, , (3), , (3 2), , (4), , Sol. Answer (1), Angle NOT = 30º this is found by geometry after putting angle between, refleted and transmitted ray equal to 90º, , sin i sin 60º, µ=, =, sin r sin30º, , , , µ=, , 3 2, ×, 2 1, , N, R Reflected ray, , I, 60º 60º, O, , 90º, 30º, , or, , 1/2, , N', , T, Refracted ray, , µ= 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 175 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 5, , 15. A microscope is focussed on a coin lying at the bottom of a beaker. The microscope is now raised by 1 cm., To what depth should water be poured into the beaker so that the coin is again in focus? (The refractive index, of water is, , 4, ), 3, , (1) 1 cm, , (2), , 4/3 cm, , (3), , 3 cm, , (4), , 4 cm, , Sol. Answer (4), Water should be poured such that shift in depth of image is 1 cm, ⎛, d⎞, ⎜⎝ d − μ ⎟⎠ = 1, , ⎛ 3⎞, d ⎜1 − ⎟ = 1, ⎝ 4⎠, , d = 4 cm, 16. Two transparent media A and B are separated by a plane boundary. The speed of light in medium A is 2.0 ×, 108 ms–1 and in medium B is 2.5 × 108 ms–1. The critical angle for which a ray of light going from A to B, suffers total internal reflection is, (1) sin–1 1/2, , (2), , sin–1 2/5, , (3), , sin–1 4/5, , (4), , sin–1 3/4, , Sol. Answer (3), , µA =, , 3 × 108, 2 × 10, , 8, , = 1.5 ; µB =, , 3 × 108, 2.5 × 10, , 8, , =, , 6, = 1.2, 5, , µB, 1.2, = 0.8, R.I. going from A to B = µ or, 1.5, A, , µ AB = 0.8 =, , , , sin C, at critical angle r = 90º, sin r, , sin C, = 0.8, sin90º, , C = sin−1, , 4, 5, , 17. Which of the following phenomenon of light forms a rainbow?, (1) Reflection of light, , (2), , Refraction, , (3) Total internal reflection, , (4), , Reflection as well as refraction, , Sol. Answer (4), Formation of a rainbow involves refraction to break white light into its constituent colours. It is also involves, internal reflection in the drop., 18. Which of the following is possible application of fibre optics?, (1) Endoscopy, , (2), , High speed internet traffic, , (3) Radio, TV & Telephone signals, , (4), , All of the above, , Sol. Answer (4), Total internal reflection is used in all of the above as they involve the use of optics fibres., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 176 :
6, , Ray Optics and Optical Instruments, , Solution of Assignment, , 19. An object is placed at a distance of f/2 from a convex lens. The image will be, (1) At one of the foci, virtual and double its size, , (2), , At 3f/2, real and inverted, , (3) At 2f, virtual and erect, , (4), , At f, real and inverted, , Sol. Answer (1), Lens formula, 1 1 1, − =, v u f, , Object is real and placed at –, , f, 2, , 1 1 2, = +, v f f, , 1 3, =, v f, v=, , f, 3, , 20. The least distance between a point object and its real image formed by a convex lens of focal length F is, (1) 2 F, , (2), , 3F, , (3), , 4F, , (4), , Greater than 4 F, , Sol. Answer (3), (Distance between a point object and its real image) d 4 f, 21. The plane faces of two identical plano-convex lenses, each having focal length of 40 cm, are placed against, each other to form a usual convex lens. The distance from this lens at which an object must be placed to, obtain a real, inverted image with magnification '–1' is, (1) 80 cm, , (2), , 40 cm, , (3), , 20 cm, , (4), , 160 cm, , Sol. Answer (2), It forms an equi-convex lens, f = 0.4 m, , P=, , 1, = 2.5 D, 0.4, , Power of combination = 5 D, , f =, , 1, P, , f =, , 1, m, 5, , or f = 20 cm, Therefore, object must be placed at 2f (= 40 cm), 22. Two thin lenses of focal lengths 20 cm and –20 cm are placed in contact with each other. The combination, has a focal length equal to, (1) Infinite, , (2), , 50 cm, , (3), , 60 cm, , (4), , 10 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 177 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 7, , Sol. Answer (1), , 1, 1, and P2 = –, 0.2, 0.2, , Powers : P1 =, , = 5 D and –5 D, Net power = 0, Focal length =, , 1, =, 0, , 23. If in a plano-convex lens, radius of curvature of convex surface is 10 cm and the focal length of lens is, 30 cm, the refractive index of the material of the lens will be, (1) 1.5, , (2), , 1.66, , (3), , 1.33, , (4), , 3, , Sol. Answer (3), R = 10 cm, f = 30 cm, 1, ⎛ 1 1⎞, = (μ − 1) ⎜ − ⎟, ⎝R ∞⎠, f, , 1, 1, = (µ − 1), 30, 10, µ=, , 4, 3, , 24. A glass concave lens is placed in a liquid in which it behaves like a convergent lens. If the refractive indices, of glass and liquid with respect to air are ag and al respectively, then, (1), , ag, , = 5al, , (2), , ag, , > Il, , (3), , ag, , < al, , (4), , ag, , = 2al, , Sol. Answer (3), The glass lens behaves as divergent in air which has less R.I., It will behave as convergent in a medium of higher R.I., 25. The diameter of aperture of a plano-convex lens is 6 cm and its maximum thickness is 3 mm. If the velocity, of light in the material of the lens is 2 × 108 m/s, its focal length is, (1) 10 cm, , (2), , 15 cm, , (3), , 30 cm, , (4), , 60 cm, , Sol. Answer (3), of medium =, , =, , 3 × 108, 2 × 108, , C, v, , = 1.5, , 6 cm, , 3 mm, , Radius of curvature of lens found using geometry, (R – 3)2 + (30)2 = R2, R2 + 9 – 6R + 900 = R2, 909 = 6R, R = 151.5 mm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 178 :
8, , Ray Optics and Optical Instruments, , Solution of Assignment, , or 15.15 cm, 1, ⎛1 1⎞, = (μ − 1) ⎜ − ⎟, ⎝∞ R⎠, f0, , 30 mm 3 mm, , R–3, , R, , 1, 1, = 0.5 ×, f0, 15.15, f0 = 30.30 cm, , This is distance from optical centre O. Distance from lens is, f = f0 – thickness at principal axis = 30.3 – 0.3, f = 30 cm, 26. Two plano-convex lenses of equal focal lengths are arranged as shown, , The ratio of the combined focal lengths is, (1) 1 : 2 : 1, , (2), , 1:2:3, , (3), , 1:1:1, , (4), , 2:1:2, , Sol. Answer (3), The power of lens remains the same no matter how the lenses are placed. So adding their power, the power, and hence the combined focal length will remain same in all three cases., 27. When an object is at a distance u1 and u2 from a lens, real image and a virtual image is formed respectively, having same magnification. The focal length of the lens is, (1) u1 – u2, , (2), , u1 u 2, 2, , (3), , u1 u 2, 2, , (4), , u1 + u2, , Sol. Answer (3), m=, , f, f +u, , For real image, , For virtual image, , f, –m = f − u ,, 1, , f, +m = f − u ,, 1, , −, , f, f, u + u2, =, f = 1, f − u2 f − u1 , 2, , 28. A concave lens of focal length f produces an image (1/x) of the size of the object. The distance of the object, from the lens is, (1) (x – 1) f, , (2), , (x + 1)f, , (3), , {(x – 1)/x}f, , (4), , {(x + 1)/x}f, , Sol. Answer (1), m=, +, , f, f +u, , −f, 1, =, ,, x −f + u, , For virtual image m = +, , 1, x, , u = –f(x – 1), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 179 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 9, , 29. A thin equiconvex glass lens of refractive index 1.5 has power of 5D. When the lens is immersed in a liquid, of refractive index , it acts as a divergent lens of focal length 100 cm. The value of of liquid is, (1) 4/3, , (2), , 3/4, , (3), , 5/3, , (4), , 8/3, , Sol. Answer (3), g = 1.5, , P=5D, , ⎛ 1 1⎞, 5 = (μg − 1) ⎜ + ⎟, ⎝R R ⎠, , 5×2=, , R=, , 2, R, , 1, = 0.2 m = 20 cm, 5, , fl < 0, , ⎞⎛ 2 ⎞, 1 ⎛ μg, =, − 1⎟ ⎜ ⎟, fl ⎝⎜ μl, ⎠ ⎝R ⎠, , −, , ⎛ μg, ⎞ 2, 1, =⎜, − 1⎟, 100 ⎝ μl, ⎠ 20, , µg, µl, , −1= −, , 1, 10, , 1.5 9, 5, =, , µl =, µl, 10, 3, , 30. In case of displacement method of lenses, the product of magnification in both cases is, (1) 1, , (2), , 2, , (3), , Zero, , (4), , Infinite, , Sol. Answer (1), m1m2 = 1 which is a fact., Here the lens is moved between the object and the screen., v and u interchange values between the two positions a clear image is formed on the screen., If m1 =, , v, u, and m2 =, u, v, , m1m2 = 1, 31. In the displacement method, a convex lens is placed in between an object and a screen. If the magnifications, in the two positions are m1 and m2 and the displacement of the lens between the two positions is x, then, the focal length of the lens is, x, (1) m m, 1, 2, , (2), , x, m1 – m2, , x, , (3), , m1 m2 , , 2, , x, , (4), , m1 – m2 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 180 :
10, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (2), By principle of reversibility where must be symmetry in the two position, , a, d, , u, , u, , v, v, d=u+v, , Hence,, , u=, , d −a, d +a, and v =, 2, 2, , Putting in lens equation, , , , f =, , a=v–u, , 2, 2, 1, +, =, d −a d +a f, , d 2 − a2, 4d, , m1 =, , d +a, d −a, , m2 =, , d −a, d +a, , |m1 – m2| =, , 4 da, d 2 + a2, , |m1 – m2| =, , a, . (a = x), f, , x, f = m –m, 1, 2, , 32. The focal length of a planoconvex glass lens is 20 cm (g =1.5). The plane face of it is silvered. An illuminating, object is placed at a distance of 60 cm from the lens on its axis along the convex side. Then the distance, (in cm) of the image is, (1) 20, , (2), , 30, , (3), , 40, , (4), , 12, , Sol. Answer (4), Net power of combination, P = 2PL + Pm, where PL is power of lens, Pm is power of mirrors, Pm = 0, PL =, , 1, 0.2, , P = 10 D, Focal length of combination (f) =, , 1, = 0.1 m or 10 cm, 10, , 1 1, −1, −, =, v 60 10, , 1, 1, 1, =, −, v 60 10, v = 12 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 181 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 11, , 33. Two thin similar convex glass pieces are joined together front to front with its rear portion silvered such that, a sharp image is formed 20 cm from the mirror. When the air between the glass pieces is replaced by water, (mw = 4/3), then the image formed from the mirror is at a distance of, , (1) 8 cm, , (2), , 10 cm, , (3), , 6 cm, , (4), , 12 cm, , Sol. Answer (4), Initially it is simply a concave mirror with f = –20 cm, R = 40 cm, Power of mirror (Pm) =, , =, , Focal length of lens =, , 1, 8, 1, =5D, 0.2, , R, 2(µ − 1), , 40, = 2 × ⎜⎛ 4 − 1⎟⎞, ⎝3 ⎠, , = 60 cm, Power of lens =, , =, , 1, 0.5, 100, D, 50, , Equivalent power = 2PL + Pm, P=, , 50, 6, , Net focal length =, , =, , 1, P, 6, × 100, 50, , = 12 cm, 34. Yellow light is refracted through a prism producing minimum deviation. If i1 and i2 denote the angle of incidence, and emergence for the prism, then, (1) i1 = i2, , (2), , i1 > i2, , (3), , i1 < i2, , (4), , i1 + i2 = 90, , Sol. Answer (1), At angle of minimum deviation angle of emergence of prism is same as angle of incidence., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 182 :
12, , Ray Optics and Optical Instruments, , Solution of Assignment, , 35. At what angle will a ray of light be incident on one face of an equilateral prism, so that the emergent ray may, graze the second surface of the prism ( = 2)?, (1) 30°, , (2), , 90°, , (3), , Let C by critical angle = 2 =, , sin i, sin r, , 45°, , (4), , 60°, , Sol. Answer (2), , 60º, , ie = 90º, 60º = r + C from geometry, , i, , 1 1 sin C, = =, µ 2 sin ie, C = sin−1, , r, , ie, , C, , [At emergent interface], , 60º, , 1, = 30º, 2, , 60º, ie = 90º, , r = 30º, sin i, =2, sin r, , sin i = 1, , i=, , π, 2, , 36. A prism of refractive index 2 has a refracting angle of 60o. At what angle must a ray be incident on it so that, it suffers a minimum deviation?, (1) 30°, , (2), , 45°, , (3), , 60°, , (4), , 75°, , Sol. Answer (2), , µ = 2, A = 60º, At for minimum deviation :, i = ie and r1 = r2, , i, , Also, r1 + r2 = A, or, , r1, , r2, , ie, , 2r = 60º, r = 30º, µ=, , sin i, sin r, , 2 = 2sini, , sin i =, , 1, 2, , [Putting r = 30º], , i = 45º, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 183 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 13, , 37. A glass prism of refractive index 1.5 is immersed in water of refractive index 4/3. A light ray incident normally, on face AB is totally reflected at face AC if, , B, , , , A, , C, , (1) sin > 8/9, , (2), , sin < 2/3, , (3), , sin =, , 3 2, , (4), , 2/3 < sin 8/9, , Sol. Answer (1), g (glass) = 1.5, l (water) = 4/3, Rays enter and pass undeviated at first interface, For total interface reflection at 2nd interface, , ⎛μ ⎞, ic > sin−1 ⎜ e ⎟, ⎝ μg ⎠, , , ic, , ⎛4 2⎞, ic > sin−1 ⎜ × ⎟, ⎝3 3⎠, ⎛8⎞, ic > sin−1 ⎜ ⎟, ⎝9⎠, , sin ic >, , 8, 9, , Since ic = from geometry, sin θ >, , 8, 9, , 38. A person can see clearly only up to a distance of 25 cm. He wants to read a book placed at a distance of, 50 cm. What kind of lens does he require for this purpose and what must be its power?, (1) Concave, – 1.0 D, , (2), , Convex, + 1.5 D, , (3), , Concave, – 2.0 D, , (4), , Convex, + 2.0 D, , Sol. Answer (3), He needs to bring the image of the object closer to 25 cm, Also the image should be virtual, 1 1 1, − =, v u f, , −, , 1, 1 1, +, =, 25 50 f, , f = –50 cm, , P=, , 1, −0.5, , P=2D, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 184 :
14, , Ray Optics and Optical Instruments, , Solution of Assignment, , 39. An astronomical telescope has an objective of focal length 100 cm and an eye piece of focal length 5 cm., The final image of a star is seen 25 cm from the eyepiece. The magnifying power of the telescope is, (1) 20, , (2), , 22, , (3), , 24, , (4), , 26, , Sol. Answer (3), f0 = 100 cm, fe = 5 cm, To form image at near point, , ⎡ 1 1⎤, m = −f0 ⎢ + ⎥, ⎣ fe D ⎦, ⎡1 1 ⎤, = −100 ⎢ +, ⎥, ⎣ 5 25 ⎦, ⎡6⎤, = −100 ⎢ ⎥, ⎣ 25 ⎦, , m = –24, 40. When a telescope is adjusted for normal vision, the distance of the objective from the eye-piece is found to, be 80 cm. The magnifying power of the telescope is 19. What are the focal lengths of the lenses?, (1) 61 cm, 19 cm, , (2), , 40 cm, 40 cm, , (3), , 76 cm, 4 cm, , (4), , 50 cm, 30 cm, , Sol. Answer (3), Distance between objective and eyepiece when telescope is adjusted for normal vision is given by, f0, 19 = f, e, , ⎛, f0 ⎞, ⎜⎝ M = f ⎟⎠, e, , ....(i), , 80 = f0+ fe. (L = f0 + fe), , ....(ii), , Solving (i) & (ii), f0 = 76 cm, f0 = 4 cm, 41. The focal lengths of the objective and eye lens of a telescope are respectively 200 cm and 5 cm. The maximum, magnifying power of the telescope will be, (1) – 40, , (2), , – 48, , (3), , – 60, , (4), , – 100, , Sol. Answer (2), Maximum magnification is at near point, Magnification at near point, m=−, , f0, fe, , ⎛ fe ⎞, ⎜⎝1 + D ⎟⎠, , m=−, , 200 ⎛, 5 ⎞, 200 6, ×, ⎜1 +, ⎟ =−, 5 ⎝ 25 ⎠, 5, 5, , m = –48, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 185 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 15, , 42. A convex lens forms a real image of a point object at a distance of 50 cm from the convex lens. A concave, lens is placed 10 cm behind the convex lens on the image side. On placing a plane mirror on the image side, and facing the concave lens, it is observed that the final image now coincides with the object itself. The focal, length of the concave lens is, (1) 50 cm, , (2), , 20 cm, , (3), , 40 cm, , (4), , 25 cm, , Sol. Answer (3), Final image coincides with the object when rays fall normal, to the mirror or parallel to principal axis. For this virtual, object must be at the focus of concave lens., , 50 cm, , Distance of virtual object from concave lens = 50 – 10 = 40, Focal length of concave lens = 40 cm., 43. A convex lens of focal length 100 cm and a concave lens of focal length 10 cm are placed coaxially at a, separation of 90 cm. If a parallel beam of light is incident on convex lens, then after passing through the two, lenses the beam, (1) Converges, , (2), , Diverges, , (3), , Remains parallel, , (4), , Disappears, , Sol. Answer (3), f = +100 cm, , f = –10 cm, , 90 cm, 100 cm, Virtual object for concave lens is at its focus., , SECTION - B, Objective Type Questions, 1., , If radii of curvature of both convex surfaces is 20 cm, then focal length of the lens for an object placed in air, in the given arrangement is, , 1=1, 2=1.5, (1) 10 cm, , (2), , 20 cm, , (3), , 40 cm, , (4), , 80 cm, , Sol. Answer (3), µ3 µ2 − µ1 µ3 − µ2, =, +, f, R1, R2, , 1 = 1, 2 = 1.5, 1 =, , 4, 3, , R1 = +20 cm, R2 = – 20 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 186 :
16, 2., , Ray Optics and Optical Instruments, , Solution of Assignment, , A driving mirror consists of a cylindrical mirror of radius of curvature 10 cm and the length over the curved, surface is 10 cm. If the eye of the driver be assumed to be at a great distance from the mirror, then field of, view in radian is, (1) 2.0, , (2), , 4.0, , (3), , 3.0, , (4), , 5.0, , Sol. Answer (1), f=, =, 3., , R, = 5 cm, 2, , 10 cm, , f, , , , 10, = 2 rad, 5, , Which of the following statements is correct?, (1) During hot summer days, the trees and other tall objects seem to be quivering because the density of air, changes in an irregular way, (2) When the moon is near the horizon it appears bigger. This is due to optical illusion, (3) If the critical angle for the medium of a prism is C and the angle of prism is A, there will be no emergent ray, when A > 2C, (4) All of these, , Sol. Answer (4), All the statements are true., 4., , An isosceles prism of angle A = 30° has one of its surfaces silvered. Light rays falling at an angle of incidence, 60° on the other surface retrace their path after reflection from the silvered surface. The refractive index of prism, material is, , 60°, , (1) 1.414, , (2), , 1.5, , 30°, , (3), , 1.732, , (4), , 1.866, , Sol. Answer (3), For light to retrace its path it must reflect normally, in the mirror., When it does so, by geometry r = 30º, µ=, , 5., , 60°, , 30°, , sin i, sin60º, =, = 3 = 1.732, sin30º, sin r, , A short linear object of length l lies along the axis of a concave mirror at a distance u from it. If v is the distance, of image from the mirror then size of the image is, (1) l , , v, u, , (2), , l, , u, v, , (3), , ⎛v ⎞, l ⎜ ⎟, ⎝u ⎠, , 2, , (4), , ⎛u ⎞, l ⎜ ⎟, ⎝v ⎠, , 2, , Sol. Answer (3), Axial magnification of a short object is given by, , m=, , l1 ⎛ v ⎞, =⎜ ⎟, l ⎝u ⎠, , 2, , 2, , ⎛v ⎞, l1 = ⎜ ⎟ l, ⎝u ⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 187 :
Solution of Assignment, , 6., , Ray Optics and Optical Instruments, , 17, , A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index, of water is, , 4, and the fish is 12 cm, below the surface. The radius of the circle is, 3, , r, h, CC, , 16, , (1), , 7, , cm, , (2), , 26, 7, , cm, , 36, , (3), , 7, , cm, , (4), , 46, 7, , cm, , Sol. Answer (3), , r, , h = 12, = iC, , sin ic =, , sin ic =, , h, , 1, µw, , , , 3, 4, , r = h tan iC, , 3, , r =h, , r =, , 7., , 16 − 9, , 36, 7, , In optical fibre, refractive index of inner part is 1.68 and refractive index of outer part is 1.44. The numerical, aperture of the fibre is, (1) 0.5653, , (2), , 0.6653, , (3), , 0.7653, , (4), , 0.8653, , Sol. Answer (4), 0sin =, , µ12 − µ22, , (Numerical aperture), 1 = 1.68, 2 = 1.44, 8., , Compare the dispersive powers of two prisms if one of them deviates the blue and red rays through 10° and, 6° respectively and the second prism through 8° and 4.5°, (1) 0.69, , (2), , 0.79, , (3), , 0.89, , (4), , 0.99, , Sol. Answer (3), Dispersive power =, , δy =, , δv − δ r, δy, , δv + δ r, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 188 :
18, , Ray Optics and Optical Instruments, , w1 =, , 10 − 6 1, =, 8, 2, , w2 =, , 8 − 4.5, 3.5, =, 6.25, 6.25, , Solution of Assignment, , w1, ≈ 0.89, w2, , 9., , A thin prism of angle 6° made of glass of refractive index 1.5 is combined with another prism made of glass, of refractive index 1.75 to produce dispersion without deviation. Then the angle of the second prism is, (1) 7°, , (2), , 4°, , (3), , 9°, , (4), , 5°, , Sol. Answer (2), For dispersion without deviation :, A1(1 – 1) + A2(2 – 1) = 0, 6(0.5) + A2(0.75) = 0, A2 = 4º, 10. In a medium of refractive index 1.6 and having a convex surface has a point object in it at a distance of 12, cm from the pole. The radius of curvature is 6 cm. Locate the image as seen from air, (1) A real image at 30 cm, , (2), , A virtual image at 30 cm, , (3) A real image at 4.28 cm, , (4), , A virtual image at 4.28 cm, , Sol. Answer (2), µ 2 µ1 µ2 − µ1, −, =, v, u, R, , 1 = 1.6, 2 = 1, u = 12 cm, R = – 6 cm, , P, , 12 cm O, , 1 1.6 1 − 1.6, −, =, −6, v −12, , v = – 30 cm. (Virtual image), 11. A point object is situated at a distance of 36 cm from the centre of the sphere of radius 12 cm and refractive, index 1.5. Locate the position of the image due to refraction through sphere., (1) 24 cm from the surface, , (2), , 36 cm from the centre, , (3) 24 cm from the centre, , (4), , Both (1) & (2), , Sol. Answer (4), 1 = 1,, 2 = 1.5, , 12 cm, , u = –24 cm,, R = +12 cm, , O, , 36 cm, , µ 2 µ1 µ2 − µ1, −, =, v, u, R, , 1.5, 1, 1.5 − 1, −, =, v, −24, +12, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 189 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 19, , v becomes parallel to the principal axis., On second face, 1 = 1.5 , 2 = 1, u , R = – 12 cm, , µ 2 µ1 µ 2 − µ1, −, =, v, u, R, 1 1.5 1 − 1.5, −, =, ∞, −12, v, , v = 24 cm, Final image at 24 cm from the surface and from centre of sphere., 12. A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12 cm. An object is, situated in the denser medium at a distance of 9 cm from the pole. Locate the image due to refraction in air., (1) A real image at 8 cm, , (2), , A virtual image at 8 cm, , (3) A real image at 4.8 cm, , (4), , A virtual image at 4.8 cm, , Sol. Answer (4), , µ 2 µ1 µ 2 − µ1, −, =, v, u, R, , 1, , 1 1.5 0.5, +, =, v, −12, 9, , 2, P, , 1 −1 1, =, −, v, 6 24, , O, , v = –4.8, 13. A light ray is travelling from air to glass. The reflected and refracted rays are perpendicular to each other. If, the angle of incidence in air is i the refractive index of glass is, (1) sin i, , (2), , cos i, , (3), , tan i, , (4), , cot i, , Sol. Answer (3), 180º = i + 90 + r, , i i, , r = 90 – i, , r, , sin i, µ=, sin r, , = tan i, 14. A ray is incident on boundary separating glass and water. Refractive index for glass is, for water is, , 4, critical angle for glass-air boundary is, 3, , 1⎛ 3 ⎞, (1) sin ⎜ ⎟, ⎝4⎠, , (2), , ⎛2⎞, sin 1⎜ ⎟, ⎝3⎠, , (3), , ⎛8⎞, sin 1⎜ ⎟, ⎝9⎠, , (4), , 3, and refractive index, 2, , ⎛ 1⎞, sin 1⎜ ⎟, ⎝9⎠, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 191 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 21, , 17. A glass slab ( = 1.5) of thickness 2 cm is placed on a spot. The shift of spot if it is viewed from top, (1), , 2, cm, 3, , (2), , 4, cm, 3, , (3), , 1, cm, 3, , (4), , 5, cm, 3, , Sol. Answer (1), ⎛, 1⎞, Shift = t ⎜⎝1 − ⎟⎠, μ, , 1 ⎞, ⎛, = 2 ⎝⎜1 −, ⎟, 1.5 ⎠, , =, , 2, cm, 3, , 18. Ray diagram for two lenses kept at some distance given in the diagram, which of the following option is correct, (f1, f2 = focal length, d = distance between lenses), , (1) f1+f2 > d, , (2), , f1+f2 < d, , (3) f1+f2 = d, , (4), , Combination behaves like converging lens, , Sol. Answer (3), , f1, , O1, , I, , f2, , P, d, , O2, , II, , For parallel incidence image is at f1 distance from O1. For final emergence to be parallel object for II is at, distance f2 from O2. The distance between I and II would be f1 + f2., 19. A ball is projected from top of the table with initial speed u at an angle of inclination q, motion of image of, ball w.r.t ball, u, , , (1) Must be projectile, , (2), , Must be straight line and vertical, , (3) Must be straight line and horizontal, , (4), , May be straight line, depends upon value of , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 192 :
22, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (3), , u sin, , u sin, , , , V I0 = V I − V 0, , V I0 = −u cos θ iˆ + u sin j − (u cos θ i + u sin θ j ), , O, , , V I0 = −2u cos θiˆ, , u, , , , u cos, , u cos, , I, , Straight line and horizontal, 20. In displacement method, convex lens forms a real image of an object for its two different positions. If heights, of the images in two cases be 24 cm and 6 cm, then the height of the object is, (1) 3 cm, , (2), , 36 cm, , (3), , 6 cm, , (4), , 12 cm, , Sol. Answer (4), In displacement method, m1m2 = 1, h1 h2, ×, =1, h, h, , h = h1h2, h = 24 × 6, , h = 12 cm, 21. Two parallel rays of red and violet colour passed through a glass slab, which of the following is correct?, , Vio Red, let, 1, 2, , t, , 4, , 3, , 6, (1) 3 and 4 are parallel, , (2), , 4 and 5 are parallel, , (3), , Sol. Answer (4), , 6 and 3 are parallel (4), , the emergent ray is parallel to incident ray., , t, , 4, , 3, , because 1 || 2, , , 2 || 5, , 2 and 5 are parallel, , Vio Red, let, 1, 2, , When refraction occurs through parallel glass slab,, , 6 || 2, 5 || 1, , 5, , 6, , 5, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 193 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 23, , 22. A plane glass is kept over a coloured word ‘VIBGYOR’, where colour of letters as same as the colours in white, light start by letter, the letter which appears least raised is, (1) R, , (2), , Y, , (3), , O, , (4), , V, , Sol. Answer (1), , µ=, , Real depth, Apparent depth, , When an image is least raised its apparent depth is highest., R.I. is lowest which happens to be for red light., 23. The near point of a person is 75 cm. In order that he may be able to read book at a distance 30 cm. The, power of spectacles lenses should be, (1) –2 D, , (2), , +3.75 D, , (3), , +2 D, , (4), , +3 D, , Sol. Answer (3), , , 1, 1, 1, −, =, −75 −30 f, , , , 1 1, =, f 50, P = +2D, , 24. If a lens is moved towards the object from a distance of 40 cm to 30 cm, the magnification of the image, remains the same (numarically). The focal length of the lens is, (1) 20 cm, , (2), , 15 cm, , (3), , 35 cm, , (4), , 18 cm, , Sol. Answer (3), f, −f, =, f + u1 f + u2, , u1 =–40 cm, u2 =–30 cm, f, , 0, , –(f – 40) = f – 30, –f + 40 = f – 30, f = +35 cm, 25. A convex lens of power +2.5 D is in contact with a concave lens of focal length 25 cm. The power of, combination is, (1) –1.5 D, , (2), , 0D, , (3), , +1.5 D, , (4), , +6.5 D, , Sol. Answer (1), Power of concave lens = −, , 1, = –4 D, 0.25, , Adding combination = –4 D + 2.5 D = –1.5 D, 26. For a telescope in normal adjustment, the length of telescope is found to be 27 cm. If the magnifying power, of telescope, at normal adjustment is 8, the focal lengths of objective and eye piece are respectively, (1) 24 cm, 3 cm, , (2), , 27 cm, 8 cm, , (3), , 12 cm, 6 cm, , (4), , 27 cm, 9 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 194 :
24, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (1), For normal adjustment m =, , f0, and L = f0 + fe, fe, , f0, 8= f, e, , f0 = 8fe, From 27 = f0 + fe, 27 = 8fe + fe, fe = 3 cm, f0 = 24 cm, 27. A converging lens of focal length 30 cm is placed in contact with another converging lens of unknown focal, length, then possible value for focal length of combination is, (1) 15 cm, , (2), , 60 cm, , (3), , 36 cm, , (4), , –12 cm, , Sol. Answer (1), Total power of combination will be more than power of given lens and focal length will be less., 28. In the diagram the ray passing through prism is parallel to the base. Refractive index of material of prism is, , 60º, 45º, 3, 2, , (1), , (2), , 3, , 45º, (3), , (4), , 2, , 3, 2, , Sol. Answer (4), By geometry r = 45º, , µ=, , [Alternate angles], , 60º, , sin60º, 3, =, × 2, sin 45º, 2, , 45º, , 3, 2, , µ=, , r, 45º, , 45º, , 29. In displacement method we use a lens of focal length f and distance between object and screen is 60 cm., Possible value for focal length is, (1) –15 cm, , (2), , 30 cm, , (3), , 12 cm, , (4), , 20 cm, , Sol. Answer (3), f, , D, 4, , f , , 60, 4, , f 15 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 195 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 25, , 4⎞, ⎛, 30. A red colour in air has wavelength 760 nm when light passes through water of refractive index ⎜ n ⎟ ,, 3, ⎝, ⎠, wavelength becomes 570 nm. (wavelength of yellow colour in air is 570 nm). Then colour of red light in water, is, , (1) Red, , (2), , Green, , (3), , Yellow, , (4), , Blue, , Sol. Answer (1), Colour of a wave depends more on frequency than wavelength as it depicts the amount of energy is carries., Since frequency and energy doesnot change it will simply remain red., 31., , n̂1 is the unit vector along incident ray, n̂2 along normal and n̂3 is the unit vector along reflected ray, then, which of the following must be true?, , (1) n̂1 n̂2 = 0, , (2), , n̂1 n̂3 = 0, , (3), , ( n̂1 × n̂2 ) n̂3 = 0, , (4), , ( n̂1 × n̂2 ) × n̂3 = 0, , Sol. Answer (3), The reflected ray, refracted ray a incident ray and normal all lie on the same plane. Hence (3) is true., 32. A double convex lens has two surfaces of equal radii 15 cm and refractive index = 1.5, its focal length is, equal to, (1) –15 cm, , (2), , –30 cm, , (3), , +15 cm, , (4), , +30 cm, , Sol. Answer (3), 1, 1 ⎞, ⎛1, = (μ − 1) ⎜ −, ⎝ R −R ⎟⎠, f, 1, ⎛ 2⎞, = 0.5 ⎜ ⎟, ⎝ 15 ⎠, f, , f = 15 cm, 33. The distance between a real object and its real image is 56 cm formed by converging lens, focal length of, lens is, (1) f 14, , (2), , f > 14, , (3), , f = 14, , (4), , f = 28, , Sol. Answer (1), f , , D, 4, , f , , 56, 4, , f 14 cm, 34. In displacement method, there are two position of a lens for which we get clear image. First position of the, lens is at 40 cm from object and second is at 80 cm, the focal length of lens is, , (1) 40 cm, , (2), , 40, cm, 3, , (3), , 80 cm, , (4), , 80, cm, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 196 :
26, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (4), Displacement method is based on principle of reversibility. The image distance for 1st position of lens will be, same as object position with IInd position of lens., u = – 40 cm, V = + 80 cm, , 1 1, 1, =, −, f 80 −40, f =, , 80, cm, 3, , 35. Maximum magnification produced by simple micro-scope of focal length f = 5 cm is, (1) 5, , (2), , 7, , (3), , 6, , (4), , 8, , Sol. Answer (3), Maximum magnification is when final image is at near point D., , m = 1+, , D, 25, = 1+, [D = 25 cm], f, 5, , =6, , SECTION - C, Previous Years Questions, 1., , In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of, objective lens. The eyepiece forms a real image of this line. The length of this image is I. The magnification, of the telescope is, [Re-AIPMT-2015], (1), , L, I, , L, 1, I, , (2), , (3), , L, –1, I, , (4), , LI, L–I, , Sol. Answer (1), , Objective, , eyepiece, I, , L, v, , f0 + fe, , f0, At normal adjustment M = f, e, , ...(i), , and distance between lenses = f0 + fe, Lateral magnification, , L f0 fe, , I, v, , Using lens equation, , ...(ii), , 1 1 1, – , v u f, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 197 :
Solution of Assignment, , Ray Optics and Optical Instruments, , , , 1, 1, 1, –, , v – f f, fe, e, 0, , , , , , , , f0, 1, , v f f f, e 0, e, , , , , , , , f0, , fe, , , , f0 fe, , 27, , ...(iii), , v, , Comparing equations (i) (ii) & (iii), f0 L, M= f I, e, , 2., , A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive, index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47,, respectively, , A, , Blue, Green, Red, , B, , 45°, , C, , The prism will, , [Re-AIPMT-2015], , (1) Separate the red colour part from the green and blue colours, (2) Separate the blue colour part from the red and green colours, (3) Separate all the three colours from one another, (4) Not separate the three colours at all, Sol. Answer (1), , A, , Red, 45°, , B, , 45°, , C, , Refractive index of light rays that can just pass through the prism at grazing emergence at 2nd surface is, =, , 1, = 1.414, sin 45, , Light having refractive index < 1.414 takes refraction but light having > 1.414 suffers TIR., Only red colour light will come out of prism., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 198 :
28, 3., , Ray Optics and Optical Instruments, , Solution of Assignment, , The refracting angle of a prism is A, and refractive index of the material of the prism is cot(A/2). The angle of, minimum deviation is, [AIPMT-2015], (1) 180° + 2A, , (2), , 180° – 3A, , (3), , 180° – 2A, , (4), , 90° – A, , Sol. Answer (3), ⎛A, ⎞, A sin ⎛ A ⎞, ⎛A, ⎞, sin ⎜ m ⎟, sin ⎜ m ⎟, cos, ⎜2, m⎟, ⎝, ⎠, A, 2 , ⎝2, ⎠,, ⎝2, ⎠ , , cot , A, A, A, A, 2, sin, sin, sin, sin, 2, 2, 2, 2, , 90 , 4., , A A m, , m (180 2 A), 2 2, 2, , Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm, are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive, index 1.7. The focal length of the combination is, [AIPMT-2015], (1) 50 cm, , (2), , – 20 cm, , (3), , – 25 cm, , (4), , – 50 cm, , Sol. Answer (4), 1 1 1 1, , f f1 f2 f3, , f1, , f2, , 1 1, ⎛ 1 1⎞, (1.5 1) ⎜, ⎟, f1 f2, ⎝ 20 ⎠, , 1 1, 1, , f1 f2 40, , f3, , 1, 1 ⎞, 0.7, 7, ⎛ 1, (1.7 1) ⎜ , , ⎟ 10 100, f3, ⎝ 20 20 ⎠, 1 1, 1, 7, , , , f 40 40 100, 1, 2, , f, 100, , f 50 cm, 5., , If the focal length of objective lens is increased then magnifying power of, , [AIPMT-2014], , (1) Microscope will increase but that of telescope decrease, (2) Microscope and telescope both will increase, (3) Microscope and telescope both will decrease, (4) Microscope will decrease but that of telescope will increase, Sol. Answer (4), MP of microscope , , L, f0, , ⎡ P⎤, ⎢1 ⎥, ⎣ fe ⎦, , MP of telescope f 0 ⎡1 f e ⎤, f e ⎢⎣ D ⎥⎦, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 199 :
Solution of Assignment, , 6., , Ray Optics and Optical Instruments, , 29, , The angle of a prism is A. One of its refracting surface is silvered. Light rays falling at an angle of incidence 2A, on the first surface returns back through the same path after suffering reflection at the silvered surface. The, refractive index , of the prism is:, [AIPMT-2014], (1) 2 sinA, , (2), , 2 cosA, , (3), , 1, cosA, 2, , Sol. Answer (2), , (4), , tanA, , A, , i = 2A, r = A, sin i, , sin r, , 2A, , A, , sin 2 A, 2cos A, sin A, , 7., , A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If, lenses are made of different materials of refractive indices 1 and 2 and R is the radius of curvature of the, curved surface of the lenses, then the focal length of the combination is, [NEET-2013], R, (1) 2 , 1, 2, , (2), , R, 1 2 , , (3), , 2R, 2 1 , , (4), , R, 2 1 2 , , Sol. Answer (2), 8., , For a normal eye, the cornea of eye provides a converging power of 40 D and the least converging power of, the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the corneaeye lens can be estimated to be, [NEET-2013], (1) 2.5 cm, , (2), , 1.67 cm, , (3), , 1.5 cm, , (4), , 5 cm, , Sol. Answer (2), 9., , The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the, objective and eyepiece is 20 cm. The focal length of lenses are, [AIPMT (Prelims)-2012], (1) 18 cm, 2 cm, , (2), , 11 cm, 9 cm, , (3), , 10 cm, 10 cm, , (4), , 15 cm, 5 cm, , Sol. Answer (1), fo + fe = 20, , ....(i), , fo, =9, fe, , ....(ii), , Solving (i) and (ii), fo = 18 cm, , fe = 2 cm, , 10. A ray of light is incident at an angle of incidence i on one face of a prism of angle A (assumed to be small) and, emerges normally from the opposite face. It the refractive index of the prism is , the angle of incidence i, is, nearly equal to, [AIPMT (Prelims)-2012], , (1), , A, , , (2), , A, 2, , (3), , A, , (4), , A, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 200 :
30, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (3), r1 + r 2 = A, For ray to pass normally r2 = 0, r1 = A, , µ=, , sin i, sin A, , sinA = sini, If both A and i are small, i = A, 11. A concave mirror of focal length f1 is placed at a distance of 'd' from a convex lens of focal length f2 A beam of, light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance, d must equal, [AIPMT (Prelims)-2012], (1) 2f1 + f2, , (2), , –2f1 + f2, , (3), , f1 + f2, , (4), , –f1 + f2, , Sol. Answer (1), d = 2f1 + f2, , 2f1, , f2, d, 12. When a biconvex lens of glass having refractive index 1.47 is dipped in a liqud, it acts as a plane sheet of, glass. This implies that the liquid must have refractive index, [AIPMT (Prelims)-2012], (1) Greater than that of glass, , (2), , Less than that of glass, , (3) Equal to that of glass, , (4), , Less than one, , Sol. Answer (3), Power of lens is euqal to zero., ⎞⎛ 1, 1 ⎛ μg, 1 ⎞, =⎜, − 1⎟ ⎜ −, =0, f ⎝ μe, ⎠ ⎝ R1 R2 ⎟⎠, , g = l, 13. For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a, material whose refractive index, [AIPMT (Mains)-2012], (1) Lies between, (3) Is less than 1, , 2 and 1, , (2), , Lies between 2 and, , (4), , Is greater than 2, , 2, , Sol. Answer (2), =i+e–A, min, i = e, min = A, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 201 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 31, , 2A = 2i, A=i, At min,, sin A + δ m, A, A, 2 sin .cos, sin A, 2, 2, 2, =, =, =, A, A, A, sin, sin, sin, 2, 2, 2, , = 2cos, , A, 2, , = 2cos, , i, 2, , (A = i), , imax = 90º, , imin = 0º, , = 2 cos 45º, , = 2 cos 0º, , =, , =2, , 2, , As R.I. lies between 2 and, , 2., , 14. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that, its end closer to the pole is 20 cm away from the mirror. The length of the image is [AIPMT (Mains)-2012], (1) 10 cm, , (2), , 15 cm, , (3), , 2.5 cm, , (4), , 5 cm, , Sol. Answer (4), O1 is at C, image of O1 will form at same postion., For image of O2, u = –30 cm, f = –10 cm, , 1 1 1, − =, v u f, , O2, , f = 10 cm, R = 20 cm, , O1, 10 cm, , 20 cm, , P, , 1, 1, 1, +, =, v −30 −10, v = –15 cm, , Length of image, I1I2 = |PI2 – PI1|, = |15 – 20|, = 5 cm, 15. Which of the following is not due to total internal reflection?, , [AIPMT (Prelims)-2011], , (1) Brilliance of diamond, (2) Working if optical fibre, (3) Difference between apparent and real depth of a pond, (4) Mirage on hot summer days, Sol. Answer (3), Real & apparent depth are explained on the basis of refraction only. TIR not involved here., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 202 :
32, , Ray Optics and Optical Instruments, , Solution of Assignment, , 16. A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describe best, the image formed of an object of height 2 cm placed 30 cm from the lens?, [AIPMT (Prelims)-2011], (1) Real, inverted, height = 1 cm, , (2), , Virtual, upright, height = 1 cm, , (3) Virtual, upright, height = 0.5 cm, , (4), , Real, inverted, height = 4 cm, , Sol. Answer (4), , 1, 1 ⎞, ⎛ 1, = (1.5 − 1) ⎜, +, ⎝ 20 20 ⎟⎠, f, f = 20 cm, m=, , f, f +u, , m=, , 20, = −2 (Real and inversed), 20 − 30, , hi, = −2 , h0 = 2 cm, h0, i = –4 cm, 17. A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at, a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move, 5 cm closer to the lens. The focal length of the lens is, [AIPMT (Mains)-2011], (1) –30 cm, , (2), , 5 cm, , (3), , –10 cm, , (4), , 20 cm, , Sol. Answer (1), For virtual object u = +10 cm, v = +15 cm, The ray diagram is as shown., , 1 1 1, − =, v u f, 1, 1 1, −, =, 15 10 f, , 5 cm, , 2−3 1, =, 30, f, , 15 cm, , f = –30 cm, 18. A thin prism of angle 15° made of glass of refractive index 1 = 1.5 is combined with another prism of glass of, refractive index 2 = 1.75. The combination of the prisms produces dispersion without deviation. The angle of, the second prism should be, [AIPMT (Mains)-2011], (1) 12°, , (2), , 5°, , (3), , 7°, , (4), , 10°, , Sol. Answer (4), Angle = 15º, 1 = 1.5, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 203 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 33, , 2 = 1.75, 15(1 – 1) + A(2 – 1) = 0, 7.5 + 0.75A = 0, 0.75 A = –7.5, , A=−, , 7.5, 0.75, , A = 10º, 19. A ray of light travelling in a transparent medium of refractive index , falls on a surface separating the medium, from air at an angle of incidence of 45°. For which of the following value of the ray can undergo total internal, reflection?, [AIPMT (Prelims)-2010], (1) = 1.25, , (2), , = 1.33, , (3), , = 1.40, , (4), , = 1.50, , Sol. Answer (4), C < 45º, sin C < sin 45º, , 1, < sin 45º, µ, >, , 2, , Only possible with = 1.5, 20. A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d / 2, in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be, respectively, [AIPMT (Prelims)-2010], (1) f and, , I, 4, , (2), , I, 3f, and, 2, 4, , (3), , f and, , 3I, 4, , (4), , f, I, and, 2, 2, , Sol. Answer (3), Focal length will not change as long as curvature of lens does not change., I d2, d = Diameter of aperture, I = Intensity of image, , d, = Aperture is covered by black paper, 2, I , , d2, 4, , I =, , I, 4, , obstructed by paper, , Intensity of image = I −, , I 3I, =, 4 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 204 :
34, , Ray Optics and Optical Instruments, , Solution of Assignment, , 21. The speed of light in media M1 and M2 is 1.5 × 108 m/s and 2 × 108 m/s respectively. A ray of light enters, from medium M1 to M2 at an incidence angle i. If the ray suffers total internal reflection, the value of i is, [AIPMT (Mains)-2010], , ⎛2⎞, (1) Equal to sin–1 ⎜ ⎟, ⎝3⎠, , (2), , ⎛3⎞, Equal to or less than sin–1 ⎜ ⎟, ⎝5⎠, , ⎛3⎞, (3) Equal to or greater than sin–1 ⎜ ⎟, ⎝4⎠, , (4), , ⎛2⎞, Less than sin–1 ⎜ ⎟, ⎝3⎠, , Sol. Answer (3), i > C, sin i > sin C, , sin i >, , sin i >, , µ2, µ1, , sin i >, , v1, sin i > v, 2, , 1.5 × 108, 2 × 108, , 3, 4, , −1 ⎛ 3 ⎞, i > sin ⎜⎝ ⎟⎠, 4, , 22. A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first, face (i.e., incident face) of the prism is, [AIPMT (Mains)-2010], (1) Zero, , (2), , 30°, , (3), , 45°, , (4), , 60°, , Sol. Answer (2), At minimum deviation, r = r, According to geometry of prism, r + r = A, 2r = A, r=, , 60, 2, , r = 30º, 23. Two thin lenses of focal lengths f1 and f2 are in contact and coaxial. The power of the combinations is, [AIPMT (Prelims)-2008], , (1), , f1 f2, f1f2, , (2), , f1, f2, , (3), , f2, f1, , (4), , f1 f2, 2, , Sol. Answer (1), , 1 1 1, = +, f f1 f2, P=, , f1 + f2, f1f2, , ⎡1, ⎤, ⎢f = P⎥, ⎣, ⎦, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 205 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 35, , 24. A boy is trying to start a fire by focusing Sunlight on a piece of paper using an equiconvex lens of focal length, 10 cm. The diameter of the Sun is 1.39 × 109 m and its mean distance from the earth is 1.5 × 1011 m. What, is the diameter of the Sun's image on the paper?, [AIPMT (Prelims)-2008], (1) 12.4 × 10–4 m, , (2), , 9.2 × 10–4 m, , (3), , 6.5 × 10–4 m, , (4), , 6.5 × 10–5 m, , Sol. Answer (2), v = 10 cm, u = 1.5 × 1011 m, Magnification =, 0.1, , Image, , 11, , =, , 12, , × 1.39 × 109 = Image, , 1.5 × 10, 1, , 1.5 × 10, , or, , v, Image diameter, =, u, Sun's diameter, 1.39 × 109, , 1.39, × 10−3 = Image, 1.5, , or 9.2 × 10–4 = Image, Image diameter = 9.2 × 10–4 m, 25. The frequency of a light wave in a material is 2 × 1014 Hz and wavelength is 5000 Å. The refractive index of, material will be, [AIPMT (Prelims)-2007], (1) 1.33, , (2), , 1.40, , (3), , 1.50, , (4), , 3.00, , Sol. Answer (4), C, = v, m, C, = f λ (Medium) fm = fa (air), m m, , C, = f λ, a m, =, , 3 × 108, 2 × 1014, , × 5000 × 10−10 = 3, , 26. A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels upto the, surface of the liquid and moves along its surface. How fast is the light travelling in the liquid?, [AIPMT (Prelims)-2007], 3 cm, , 4 cm, , coin, , (1) 1.2 × 108 m/s, , (2) 1.8 × 108 m/s, , (3), , 1.3 × 108 m/s, , (4), , 3.0 × 108 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 206 :
36, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (2), is critical angle sin =, , l =, , 1, 3, =, µl, 5, , 5, 3, , 3m, , v in medium, , , , 4m, , v in air (C ), µl =, v in medium, , , , =, , 3, × 3 × 108, 5, , =, , 9, × 108, 5, , 5m, , v = 1.8 × 108, 27. A microscope is focussed on a mark on a piece of paper and then a slab of glass of thickness 3 cm and, refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus, again?, [AIPMT (Prelims)-2006], (1) 1 cm upward, , (2), , 4.5 cm downward, , (3), , 1 cm downward, , (4), , 2 cm upward, , Sol. Answer (1), ⎛, 1⎞, Shift = t ⎜⎝1 − ⎟⎠, μ, , 1 ⎞, ⎛, = 3 ⎜⎝1 −, ⎟, 1.5 ⎠, , = 1 cm, So the microscope must be moved by 1 mm upwards, , µ=, , Real depth, Apparent depth, , Apparent depth =, , 2, Real depth, = × 3 mm = 2 mm, 3, µ, , Shift = 3 – 2 = 1 mm, So the microscope must be moved 1 mm upwards., 28. A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a, combination of lenses. The power in diopters of the combination is, [AIPMT (Prelims)-2006], (1) 25, , (2), , 50, , (3), , Infinite, , (4), , Zero, , Sol. Answer (4), Each lens of same power but different is sign., When added P = P1 + P2, P = P1 – P 2, P=0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 207 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 37, , 29. A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirror will be, (1) 12 feet, , (2), , 3 feet, , (3), , 6 feet, , (4), , Any length, , Sol. Answer (2), Height - 6 ft, To see any object in a plane mirror complete, a mirror must be half the height of object., So minimum height of mirror =, , h, = 3 ft, 2, , 30. Images formed by an object placed between two plane mirrors whose reflecting surfaces make an angle of, 90° with one another lie on a, (1) Straight line, , (2), , Parabola, , (3), , Sol. Answer (3), , Circle, , (4), , Ellipse, , m2, I2, , O, , m1, I3, , I1, , 31. An object is placed between two plane mirrors inclined at an angle ‘’ to each other. If the number of images, formed is 7, then the angle of inclination ‘’ is, (1) 15°, , (2), , 30°, , (3), , 45°, , (4), , 60°, , (1) Greater than that of the cladding, , (2), , Equal to that of the cladding, , (3) Smaller than that of the cladding, , (4), , Independent of that of the cladding, , Sol. Answer (3), , 360, −1, θ, = 45º, 7=, , 32. In optical fibres, the refractive index of the core is, , Sol. Answer (1), 33. Light travels through a glass plate of thickness t and having a refractive index . If c is the velocity of light in, vacuum, the time taken by light to travel this thickness of glass is, , t, (1) c, , (2), , t, c, , (3), , tc, , (4), , tc, , , Sol. Answer (2), Time =, =, , c, V, , V=, , c, µ, , Time =, , t, v (Speed in glass), , µt, c, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 208 :
38, , Ray Optics and Optical Instruments, , Solution of Assignment, , 34. A ray of light from a denser medium strikes a rare medium as shown in figure. The reflected and refracted, rays make an angle of 90° with each other. The angles of reflection and refraction are r and r. The critical, angle would be, N, A, C, , i r, DENSER, RARER B, r, D, , N, (1) sin–1 (tan r), , (2), , tan–1 (sin r), , (3), , sin–1 (cot r), , (4), , tan–1 (sin r ), , (4), , 2.25 × 108 m/s, , Sol. Answer (1), i=r, r + r = 90º, i + r = 90º, r = 90º – i, sin c =, , 1, µ (R.I. of denser w.r.t. rarer), , sin c = (R.I. of rarer w.r.t. denser), sin c =, , sin i, sin r ′, , sin c =, , sin i, sin(90º −i ), , sin c = tan i, c=, , sin–1, , (i = r), (tan r), , 35. The refractive index of water is 1.33. What will be the speed of light in water?, (1) 4 × 108 m/s, , (2), , 1.33 × 108 m/s, , (3), , 3 × 108 m/s, , Sol. Answer (4), 1.33 =, , c, V (Speed of light in water), , 36. An electromagnetic radiation of frequency n, wavelength l, travelling with velocity v in air, enters a glass slab, of refractive index . The frequency, wavelength and velocity of light in the glass slab will be respectively, (1) n, 2 and, , v, , , (2), , 2n , , and v, , , (3), , n , v, ,, and, , , , (4), , n,, , v, , and, , , , Sol. Answer (4), By definition of refractive index, Velocity of light becomes, becomes, , v, μ, , λ, μ, , But frequency is constant., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 209 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 39, , 37. A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed, 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out is, (1) , , (2), , 3m, , (3), , 6m, , (4), , 4m, , Sol. Answer (2), ic = , , r, , 1, sin iC =, µd, , iC, , 4m, , , , 3, sin iC =, 5, 3, , tan ic =, , 25 − 3, , 2, , =, , 3, 4, , r = 4 tan, r=3m, 38. A ray of light travelling in air have wavelength , frequency n, velocity v and intensity I. If this ray enters into, water then these parameters are , n, v and I respectively. Which relation is correct from following?, (1) = , , (2), , n = n, , (3), , v = v, , (4), , I = I, , (1) Total internal reflection, , (2), , Less scattering, , (3) Refraction, , (4), , Less absorption coefficient, , Sol. Answer (2), When ray enters water, becomes, , λ, v, and velocity becomes, μ, μ, , Intensity also changes, Only frequency n remains same so answer is (2)., 39. Optical fibre are based on, , Sol. Answer (1), Optical fibres depend on total internal reflection., 40. For the given incident ray as shown in figure, the condition of total internal reflection of this ray the required, refractive index of prism will be, 45º, , (1), , 3 1, 2, , (2), , 2 1, 2, , incident, ray, , (3), , 3, 2, , (4), , 7, 6, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 210 :
40, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (3), iC = 90 – r, , 45º, , ⎛ 1 ⎞, sin(90 − r ) = ⎜ ⎟, ⎝ μg ⎠, µg =, , Also, , , , cos r =, , sin 45º, sin r, , r, , ....(ii), , sin r, sin 45º, , tanr = sin45º =, , sin r =, , , , iC, , ....(i), , 1, 2, , , 1, , 1, 3, , sin i, µ=, =, sin r, , r, 2, , 3, 2, , 41. A layer of benzene ( = 1.5) 12 cm thick floats on water layer (m = 4/3) 8 cm thick in a vessel. When viewed, from the top, the apparent depth of bottom of vessel below the surface of benzene will be, (1) 20 cm, , (2), , 14 cm, , (3), , 7 cm, , (4), , 21 cm, , Sol. Answer (2), In case of multiple medium of different R.I., Apparent depth d =, , d=, , t1 t2, +, µ1 µ2, , 12, 8, +, 1.5 4 / 3, , d=8+6, = 14 cm, 42. The critical angle for a light travelling from medium A into medium B is . The speed of light in medium A is, v, the speed of light in medium B is, (1), , v, cos , , (2), , v sin , , (3), , v, sin , , (4), , v cos , , Sol. Answer (3), , sin θ =, , μB, μA, , A sin = B, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 211 :
Solution of Assignment, , µA =, , C, v, , Ray Optics and Optical Instruments, , µB =, , 41, , C, vB, , C, C, sin θ =, v, vB, , , , vB =, , v, sin θ, , 43. A small bulb is placed at a depth of 2 7 m in water and floating opaque disc is placed over the bulb so that, the bulb is not visible from the surface. The minimum diameter of the disc is (water = 4/3), (1) 42 m, , (2), , 6m, , (3), , 2 7 m, , (4), , 12 m, , Sol. Answer (4), , r, , θmax = sin−1, , 1, 3, = sin−1, μ, 4, , 27, sin =, , tan =, , r, 2 7, , =, , 3, 4, , , , , 3, 7, 3, 7, , r=6m, 44. Two optical media of refractive indices 1 and 2 contain x and y number of waves in the same thickness. Their, relative refractive index, , (1) xy, , 2, is equal to, 1, y, x, , (2), , (3), , x, y, , (4), , yx, x, , Sol. Answer (2), let wavelength of light in air is , Wavelength in medium =, t=, , λ, μ, , xλ, yλ, and t =, μ1, μ2, , x µ1, =, y µ2, , , µ2 y, =, µ1 x, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 212 :
42, , Ray Optics and Optical Instruments, , Solution of Assignment, , 45. A light ray is travelling through a ring of an optical fibre which is made of four different glasses (shown below), but each part has the same geometrical thickness. Their respective refractive indices are shown. The light ray, will take the maximum time in crossing the part, , Part-I, , 1.51 1.52, , Part-II, , 1.54 1.53, Part-IV, (1) I, , (2), , II, , Part-III, (3), , IV, , (4), , Same in all, , Sol. Answer (3), R.I. of part IV is highest. So velocity will be lowest. Hence, maximum time is taken in part IV., 46. Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive, index of the material of the rod the light once entered into it will not leave it through its lateral face whatsoever, be the value of angle of incidence?, (1) n = 1.1, , (2), , n=1, , (3), , n 2, , (4), , n = 1.3, , Sol. Answer (3), n2 − 1, , sin <, , , , sin2 θ + 1, Maximum value of = 90º., n>, , n, , n> 2, , 47. If the refractive index of a material of equilateral prism is, is, (1) 60°, , (2), , 45°, , (3), , 3 , then angle of minimum deviation of the prism, 30°, , (4), , 75°, , Sol. Answer (1), 3 ; A = 60º; m = ?, , R.I. =, , sin ( A + δ m ), 2, μ=, ⎛ A⎞, sin ⎜ ⎟, ⎝2⎠, sin(60º + δm ), 2, 3=, sin30º, , sin−1, , 3 60 + δm, =, 2, 2, , 60º = 30º +, 30º =, , δm, 2, , δm, 2, , m = 60º, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 213 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 43, , 48. A beam of light composed of red and green ray is incident obliquely at a point on the face of rectangular glass, slab. When coming out on the opposite parallel face, the red and green ray emerge from, (1) Two points propagating in two different non parallel directions, (2) Two points propagating in two parallel directions, (3) One point propagating in two different directions, (4) One point propagating in the same directions, Sol. Answer (2), Since parallel beams of light will regain their original direction. They will again become parallel after emergence., 49. The refractive index of the material of a prism is 2 and its refracting angle is 30°. One of the refracting, surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the, other face will retrace its path after reflection from the mirrored surface if its angle of incidence on the prism, is, (1) 45°, , (2), , 60°, , (3), , Zero, , (4), , 30°, , Sol. Answer (1), By geometry r = 90º – 60º = 30º, 30°, , sin i, 3=, sin30º, , , i, , r, , 1, , sin i =, , 2, , i = 45º, 50. Four lenses of focal length ±15 cm and ±150 cm are available for making a telescope. To produce the largest, magnification, the focal length of the eyepiece should be, (1) +15 cm, , (2), , +50 cm, , (3), , –150 cm, , (4), , –15 cm, , Sol. Answer (1), 51. A lens is placed between a source of light and a wall. It forms images of area A1 and A2 on the wall, for its, two different positions. The area of the source of light is, , (1), , A1 A2, 2, , (2), , 1, 1, , A1 A2, , (3), , A1 A2, , (4), , A1 A2, 2, , Sol. Answer (3), This is application of the displacement method for finding focal length., Here m1m2 = 1, Let A0 be area of object, , A1 A2, ×, =1, A0 A0, A0 =, , A1A2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 214 :
44, , Ray Optics and Optical Instruments, , Solution of Assignment, , 52. If fV and fR are the focal lengths of a convex lens for violet and red light respectively and FV and FR are the, focal lengths of a concave lens for violet and red light respectively, then we must have, (1) fV > fR and FV > FR, , (2), , fV < fR and FV > FR, , (3), , fV > fR and FV < FR (4), , fV < fR and FV < FR, , Sol. Answer (2), v > r, A converging lens with higher refractive index will converge rays more hence value of fv < fr, Same is true for concave lenses but since values for focal length are taken as negative Fv > Fr, 53. If a convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together, what, will be their resulting power?, (1) + 7.5 D, , (2), , – 0.75 D, , (3), , + 6.5 D, , (4), , – 6.5 D, , Sol. Answer (2), P = P1 + P2, =, , 1, 1, −, 0.8 0.5, , =, , 5, −2, 4, , = –0.75 D, 54. The focal length of a converging lens is measured for violet, green and red colours. It is respectively fv, fg, fr., We will get, (1) fv < fr, , (2), , fg > fr, , (3), , fv = fg, , (4), , fg = fr, , Sol. Answer (1), v > g > r, A converging lens with greater refractive index will bend rays more converging them closer., fv < fg < fr, 55. A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On the other, side of the lens, at what distance from the lens a convex mirror of radius of curvature 10 cm be placed in order, to have an upright image of the object coincident with it?, (1), , 50 cm, , (2), , 30 cm, , (3), , 12 cm, , Sol. Answer (1), , 60 cm, , 20 cm, , For convex lens, , 1, 1, 1, = −, +20 v −30, , (4), , O, , C, d = 60 – 10, d = 50 cm, , v = 60 cm, Virtual object for convex mirror should be at its C., , 30 cm, , 10 cm, , d, 60 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 215 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 45, , 56. A plano-convex lens is made of refractive index 1.6. The radius of curvature of the curved surface is 60 cm., The focal length of the lens is, (1) 200 cm, , (2), , 100 cm, , (3), , 50 cm, , (4), , 400 cm, , Sol. Answer (2), , 1, ⎛ 1 1⎞, = (μ − 1) ⎜ − ⎟, ⎝R ∞⎠, f, 1 0.6, =, f, 60, , f = 100 cm, 57. A planoconvex lens (m = 1.5) has radius of curvature 10 cm. It is silvered on its plane surface. Find focal length, after silvering, (1) 10 cm, , (2), , 20 cm, , (3), , 15 cm, , (4), , 25 cm, , Sol. Answer (1), = 1.5, R = 10 cm, P = 2PL + Pm, Pm = Power of mirror = 0, P = 2PL, 1⎞, ⎛ 1, PL = (1.5 − 1) ⎜, +, ⎝ 10 ∞ ⎟⎠, =, , PL =, , 0.5, 10, , 1, cm, 20, , PL = 5 D, P = 2PL = 10 D, , f =, , 1, m = 10 cm, 10, , 58. A bubble in glass slab (m = 1.5) when viewed from one side appears at 5 cm and 2 cm from other side, then, thickness of slab is, (1) 3.75 cm, , (2), , 3 cm, , (3), , 10.5 cm, , (4), , 2.5 cm, , Sol. Answer (3), = 1.5, , µ=, , Real depth, Apparent depth, , Real depth = Apparent depth 1, , t1, t, t2, , t = t1 + t2, t = 5 × 1.5 + 2 × 1.5, = 10.5 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 216 :
46, , Ray Optics and Optical Instruments, , Solution of Assignment, , 59. A bulb is located on a wall. Its image is to be obtained on a parallel wall with the help of a convex lens. If, the distance between the two walls is d, then required focal length will be, (1) Only, , d, 4, , (3) More than, , d, d, but less than, 4, 2, , d, 2, , (2), , Only, , (4), , Less than or equal to, , d, 4, , Sol. Answer (4), f, , d, 4, , 60. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its, focal length will, (1) Become zero, , (2), , Become infinite, , (3) Become small, but non-zero, , (4), , Remain unchanged, , Sol. Answer (2), When this happens the lens will behave as a glass slab and parallel rays will converge at infinity., f=, 61. An equiconvex lens is cut into two halves along (i) XOX and (ii) YOY as shown in the figure. Let f, f , f , be the focal lengths of the complete lens of each half in case (i), and of each half in case (ii), respectively., Choose the correct statement from the following, Y, , X, , X, , O, , Y, , (1) f = f, f = 2f, , (2), , f = 2f, f = f, , (3), , f = f, f = f, , f = 2f, f = 2f, , (4), , Sol. Answer (1), f = f as radius of curvature of both surfaces is same., Y, , f = 2f, by lens maker's formula :, 1, ⎛ 1 1⎞, = (μ − 1) ⎜ − ⎟, ⎝R ∞⎠, f '', , ....(i), , 1, ⎛ 1 1⎞, = (μ − 1) ⎜ + ⎟, ⎝R R ⎠, f, , ....(ii), , f, 1, =, f '' 2, , [Dividing (i) and (ii)], , X, , O, , X, , Y, , f = 2f, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 217 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 47, , 62. Which of the following is incorrect?, (1) A thin convex lens of focal length f 1 is placed in contact with a thin concave lens of focal, length f2. The combination will act as convex lens if f1 < f2, (2) Light on reflection at water-glass boundary will undergo a phase change of , (3) Spherical aberration is minimized by achromatic lens, (4) If the image of distant object is formed in front of the retina then defect of vision may be myopia, Sol. Answer (3), Achromatic lenses minimize chromatic aberration., 63. A double concave thin lens made out of glass ( = 1.5) have radii of curvature 500 cm. This lens is used to, rectify the defect in vision of a person. The far point of the person will be at, (1) 5 m, , (2), , 2.5 m, , (3), , 1.25 m, , (4), , 1m, , Sol. Answer (1), , 1 1 1, = −, f v ∞, , ⎛ 2 ⎞ 1, (1.5 − 1) ⎜ −, =, ⎝ 500 ⎟⎠ v, v = – 500 cm, v=–5m, 64. A convex lens forms a real image 16 mm long on a screen. If the lens is shifted to a new position without, disturbing the object or the screen then again a real image of length 81 mm is formed. The length of the object, must be, (1) 48.5 mm, , (2), , 36 mm, , (3), , 6 mm, , (4), , 72 mm, , Sol. Answer (2), This is the application of the displacement method., Size of object (m0) =, , , m1m2, , m0 = 16 × 81, =4×9, = 36 cm, , 65. A point object is moving with speed u0 at a position somewhere between 2F and F in front of a convex lens., The speed of its image is, , u0, 2F, , (1) > u0, , (2), , < u0, , F, , F, , 2F, , (3), , = u0, , (4), , May be (1) or (2), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 218 :
48, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (1), , 1 1 1, − =, v u f, Differentiating w.r.t. t, −, , −, , 1 dv ⎛ 1 ⎞ du, − ⎜− ⎟, =0, v 2 dt ⎝ u 2 ⎠ dt, 1, v, , 2, , vi +, , 1, u2, , v0 = 0, , 2, vi = v v0, u2, When f < u < 2f, v lies beyond 2f., , v > 2f, v2, u2, , >1, , vi > v0., 66. The minimum magnifying power of a telescope is M. If focal length of its eye lens is halved, the magnifying, power will become, (1), , M, 2, , (2), , 2M, , (3), , 3M, , (4), , 4M, , Sol. Answer (2), f0, M =f, e, , 1, M f, e, fe =, , fe, 2, , M = 2M, 67. An object is placed in front of two convex lenses one by one at a distance u from the lens. The focal lengths, of the lenses are 30 cm and 15 cm respectively. If the size of image formed in the two cases is same, then, u is, (1) 15 cm, , (2), , 20 cm, , (3), , 25 cm, , (4), , 30 cm, , Sol. Answer (2), Magnification are same, m1 = – m2, , f ⎞, ⎛, ⎜⎝ m =, ⎟, f +u⎠, , 30, 15, =−, 30 − u, 15 − u, , u = 20 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 219 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 49, , 68. If R1 and R2 are the radii of curvature of the spherical surfaces of a thin lens and R1 > R2, then this lens can, , R2, R1, (1) Correct myopia, , (2), , Correct hypermetropia, , (3) Correct presbiopia, , (4), , Correct astigmatism, , Sol. Answer (1), f < 0 (Diverging), 3, 69. The focal length of a thin lens in vacuum is f. If the material of the lens has , its focal length when, 2, 4, will be, immersed in water of refractive index, 3, , (1) f, , (2), , 4, f, 3, , (3), , 2f, , (4), , 4f, , Sol. Answer (4), 1 ⎞, 1, ⎛3 ⎞ ⎛ 1, −, ⎜⎝ − 1⎟⎠ ⎜, =, 2, fv, ⎝ R1 R2 ⎟⎠, , 1⎛ 1, 1 ⎞, 1, = 2 ⎜R − R ⎟, ⎝ 1, f, 2⎠, ⎛ μg, ⎞ ⎛ 1, 1, 1 ⎞, =, ⎜ μ − 1⎟ ⎜⎝ R − R ⎟⎠, fw, ⎝ w, ⎠, 1, 2, , 1, ⎛9 ⎞ ⎛2⎞, =, ⎜⎝ − 1⎟⎠ ⎜⎝ ⎟⎠, fw, 8, f, fw = 4f, 70. For a telescope having fo as the focal length of the objective and fe as the focal length of the eyepiece, the, length of the telescope tube is, (1) fe, , (2), , fo – fe, , (3), , fo, , (4), , fo + fe, , Sol. Answer (4), By the construction of the telescope f0 + fe = L, 71. An astronomical telescope of tenfold angular magnification has a length of 44 cm. The focal length of the, objective is, (1) 44 cm, , (2), , 440 cm, , (3), , 4 cm, , (4), , 40 cm, , Sol. Answer (4), m = 10, f0, = 10, fe, , ....(i), , f0 + fe = 44 cm, , ....(ii), , Solving (i) and (ii), f0 = 40 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 220 :
50, , Ray Optics and Optical Instruments, , Solution of Assignment, , 72. Ray optics is valid, when characteristic dimensions are, (1) Much smaller than the wavelength of light, , (2), , Of the same order as the wavelength of light, , (3) Of the order of one millimeter, , (4), , Much larger than the wavelength of light, , Sol. Answer (4), Ray optics is valid on a more macro scale compared to wavelength of light. On micro scale wave optics and, wave-particle duality is more prominent., 73. The blue colour of the sky is due to the phenomenon of, (1) Scattering, , (2), , Dispersion, , (3), , Reflection, , (4), , Refraction, , Sol. Answer (1), Blue colour of the sky is due to scattering of light also called the Rayleigh scattering., 74. Rainbow is formed due to, (1) Scattering and refraction, , (2), , Total internal reflection and dispersion, , (3) Reflection only, , (4), , Refraction and dispersion, , Sol. Answer (4), Rainbow's colours appear due to dispersion, which occurs after refraction., , SECTION - D, Assertion - Reason Type Questions, 1., , A : Plane and convex mirrors can produce real images of objects., R : A plane or convex mirror can produce a real image if the object is virtual., , Sol. Answer (1), Real images are formed from real intersection of light rays, which occur in convex mirror when the object is, virtual., 2., , A : A virtual image cannot be caught on a screen but it can be photographed., R : The virtual image here serves as an object for the lens of the camera to produce a real image., , Sol. Answer (1), Virtual image can be observed but real intersection of light rays needs to occur to take objects to screen., 3., , A : When a diver under water looks obliquely at a fisherman standing on the bank of a lake then fisherman, looks taller., R : When a ray of light travelling in air enters water, it bends towards the normal., , Sol. Answer (1), The ray of light bending towards normal makes it appear that light comes from further away., 4., , A : The apparent depth of a tank of water decreases if viewed obliquely., R : Real depth decreases if viewed obliquely., , Sol. Answer (3), Apparent depth of tank decreases because the light ray, travels further through the water bending more in the, process. Real depth remains the same are it was., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 221 :
Solution of Assignment, , 5., , Ray Optics and Optical Instruments, , 51, , A : Proper cutting of diamond makes it sparkle., R : Diamond has very large refractive index., , Sol. Answer (1), Proper cutting of diamond makes it sprakle due to total internal refraction. A high reflective index makes total, internal reflection easier., 6., , A : The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual, image produced by a magnifying glass so the magnification produced is one., R : During image formation through magnifying glass, the object as well as its image are at the same position., , Sol. Answer (4), A magnifying glass magnifies the virtual image and magnification is greater than one. The reason is also false., 7., , A : In viewing through a magnifying glass, angular magnification decreases if the eye is moved back., R : Angle subtended at the eye becomes slightly less than the angle subtended at the lens., , Sol. Answer (1), 8., , A : Magnifying power of a simple microscope cannot be increased beyond a limit., R : Magnifying power is inversely proportional to focal length and there are some practical difficulty of grinding,, aberrations etc. because of which focal length cannot be decreased below a limit., , Sol. Answer (1), There is a limit to how curved a glass piece can be made practically., 9., , A : The objective and the eye-piece of a compound microscope should have short focal lengths., R : Magnifying power of a compound microscope is inversely proportional to the focal lengths of both the lenses., , Sol. Answer (1), , LD, For normal adjustment : M = f ⋅ f ., o e, 10. A : When viewing through a compound microscope, our eyes should be positioned not on the eye piece but, a short distance away from it for best viewing., R : The image of the objective in the eye-piece is known as 'eye-ring' and if we position our eyes on the eyering and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect, all the light refracted by the objective., Sol. Answer (1), 11. A : The peculiar fish Anableps anableps swims with its eyes partially extending above the water surface so, that it can see simultaneously above and below water., R : This fish has egg shaped eye lens and two retina., Sol. Answer (1), 12. A : A virtual image cannot be produced on a screen., R : The light energy does not meet at the point(s) where virtual image is formed., Sol. Answer (1), To take a sharp image on a screen light rays must intersect really. This does not happen in virtual images., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 222 :
52, , Ray Optics and Optical Instruments, , Solution of Assignment, , 13. A : Lenses of large aperture suffer from spherical aberration., R : The curvature of the lens at central and peripheral regions is different., Sol. Answer (3), 14. A : The image formed by a concave lens is always virtual., R : The rays emerging from a concave lens never meet on the principal axis., Sol. Answer (4), Assertion is false as real image is formed by concave lens in care of virtual object. Reason is also false as, real images are formed by real intersection of rays., 15. A : When two lenses in contact form an achromatic doublet, then the materials of the two lenses are always, different., R : The dispersive powers of the materials of the two lenses are of opposite sign., Sol. Answer (3), To achieve achromatic condition in two lenses in contact, refractive indices must be different, dispersive power, cannot be negative., 16. A : A reflecting type of telescope is preferred over refracting type in astronomy., R : A reflecting type of telescope is free from chromatic aberration and spherical aberration., Sol. Answer (1), , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 223 :
Chapter, , 24, , Ray Optics and Optical Instruments, Solutions, SECTION - A, Objective Type Questions, 1., , Digital movie projectors work on the principle of, (1) Reflection from micromirrors, , (2), , Refraction from thin lenses, , (3) Dispersion from thin prisms, , (4), , Total internal reflection from optical fibres, , Sol. Answer (1), Digital movie projectors need parabolic mirrors to converge all incident rays to a point., 2., , Day and night settings for rearview mirrors uses, (1) Thin mirrors, , (2), , Thick wedge shaped mirrors, , (3) Convex mirrors, , (4), , Concave mirrors, , Sol. Answer (2), Day and night settings for rear view mirrors., 3., , When a beam of light is incident on a plane mirror, it is found that a real image is formed. The incident beam, must be, (1) Converging, (2) Diverging, (3) Parallel, (4) Formation of real image by a plane mirror is impossible, , Sol. Answer (1), Real images are images formed from actual intersection of light rays., , Position of, image, , Virtual object, , In plane mirror formation of real images is shown above., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 224 :
2, 4., , Ray Optics and Optical Instruments, , Solution of Assignment, , An object is placed symmetrically between two plane mirrors, inclined at an angle of 72°, then the total number, of images observed is, (1) 5, , (2), , 4, , (3), , 2, , (4), , Infinite, , Sol. Answer (2), For number of images formed from plane mirror, m=, , 360 360, =, =5, θ, 72, , If it was at any other point the number of images (n) = m but this is not so., Since it is placed symmetrically :, n=m–1, or n = 5 – 1, or n = 4 images, 5., , A person 1.6 m tall is standing at the centre between two walls three metre high. What is the minimum size, of a plane mirror fixed on the wall in front of him, if he is to see the full height of the wall behind him?, (1) 0.8 m, , (2), , 1m, , (3), , 1.5 m, , (4), , 2.3 m, , Sol. Answer (2), , h2, , 1.6 − h1, h, tan θ = 1 =, d, d /2, , A, , , , ....(i), , , , M1, B, , h, 1.4 − h2, tan α = 2 =, d, d /2, , , , , 3m, , ....(ii), , 1.6 m, h1 =, , 3.2, 2.8, and h2 =, 3, 3, , M2, C, , h1, , 3 – h1 – h2 = 1 m, 6., , While capturing solar energy for commercial purposes we use, (1) Parabolic mirrors, , (2), , Plane mirrors, , (3), , Convex mirrors, , (4), , Concave mirrors, , Sol. Answer (1), While capturing solar energy for commercial purposes we use parabolic mirrors to converge the rays coming, from infinity to a point., 7., , A convex mirror is used to form an image of a real object. Then mark the wrong statement, (1) The image lies between the pole and focus, , (2), , The image is diminished in size, , (3) The image is erect, , (4), , The image is real, , Sol. Answer (4), A convex mirror always forms a virtual image in the care of a real object., In care of a virtual object reflected rays may intersect really to make a real image., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 225 :
Solution of Assignment, , 8., , Ray Optics and Optical Instruments, , 3, , A concave mirror of focal length f produces an image n times the size of the object. If the image is real then, the distance of the object from the mirror is, (1) (n – 1) f, , (2), , ⎧ (n 1) ⎫, ⎬f, ⎨, ⎩ n ⎭, , (3), , ⎧ (n 1) ⎫, ⎬f, ⎨, ⎩ n ⎭, , (4), , (n + 1) f, , Sol. Answer (3), , f, f −u, Focal real image m = –n, (magnification) m =, , –n =, , −f, −f − u, , u =−, 9., , f (n + 1), u, , A convex mirror has a focal length f. A real object is placed at a distance f in front of it, from the pole. It, produces an image at, (1) Infinity, , (2), , f, , (3), , f/2, , (4), , 2f, , Sol. Answer (3), Mirror formula :, , 1 1 1, + =, v u f, , Here object is real so u is negative, , 1 1 1, − =, v u f, Also (u) = f, 1 1 1, − =, v f f, , v=, , f, 2, , 10. An object placed in front of a concave mirror of focal length 0.15 m produces a virtual image, which is twice, the size of the object. The position of the object with respect to the mirror is, (1) –5.5 cm, , (2), , –6.5 cm, , (3), , –7.5 cm, , (4), , –8.5 cm, , Sol. Answer (3), m =, f, , f, f −u, , = –0.15 m, , m = +2 (virtual image), 2 =, , −0.15, −0.15 − u, , = –.075 m or – 7.5 cm., 11. When a light ray from a rarer medium is refracted into a denser medium, its, (1) Speed increases, wavelength increases, , (2), , Speed decreases, wavelength increases, , (3) Speed increases, wavelength decreases, , (4), , Speed decreases, wavelength decreases, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 226 :
4, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (4), When a light ray goes from a rarer to denser medium by definition of refractive index, its speed creases., Its frequency is found to be invarant., So, if velocity of light in a medium is v :, v = f, , [by wave theory], , or v , So, if v decreases, also decreases., 12. A narrow, paraxial beam of light is converging towards a point I on a screen. A plane parallel plate of glass of, thickness t, and refractive index μ is introduced in the path of the beam. The convergence point is shifted by, (1) t (1–1/) away, , (2), , t (1 + 1/) away, , (3), , t (1 – 1/) nearer, , (4), , t (1 + 1/) nearer, , Sol. Answer (1), Longitudinal shift is given by t –, , t, . Hence, point of convergence shifts by the same amount., µ, , 4⎞, ⎛, 13. The length of a vertical pole at the surface of a lake of water ⎜ ⎟ is 24 cm. Then to an under-water fish, 3⎠, ⎝, just below the water surface the tip of the pole appears to be, , (1) 18 cm above the surface, , (2), , 24 cm above the surface, , (3) 32 cm above the surface, , (4), , 36 cm above the surface, , Sol. Answer (3), , µ=, , Apparent height, Real height, , Now, real height = 24 cm and =, , 4, 3, , 4, × 24 = Apparent height, 3, Apparent height = 32 cm, 14. A ray of light strikes a glass plate at an angle 60o. If the reflected and refracted rays are perpendicular to each, other, the index of refraction of glass is, (1), , 3, , (2), , 3/2, , (3), , (3 2), , (4), , Sol. Answer (1), Angle NOT = 30º this is found by geometry after putting angle between, refleted and transmitted ray equal to 90º, , sin i sin 60º, µ=, =, sin r sin30º, , , , µ=, , 3 2, ×, 2 1, , N, R Reflected ray, , I, 60º 60º, O, , 90º, 30º, , or, , 1/2, , N', , T, Refracted ray, , µ= 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 227 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 5, , 15. A microscope is focussed on a coin lying at the bottom of a beaker. The microscope is now raised by 1 cm., To what depth should water be poured into the beaker so that the coin is again in focus? (The refractive index, of water is, , 4, ), 3, , (1) 1 cm, , (2), , 4/3 cm, , (3), , 3 cm, , (4), , 4 cm, , Sol. Answer (4), Water should be poured such that shift in depth of image is 1 cm, ⎛, d⎞, ⎜⎝ d − μ ⎟⎠ = 1, , ⎛ 3⎞, d ⎜1 − ⎟ = 1, ⎝ 4⎠, , d = 4 cm, 16. Two transparent media A and B are separated by a plane boundary. The speed of light in medium A is 2.0 ×, 108 ms–1 and in medium B is 2.5 × 108 ms–1. The critical angle for which a ray of light going from A to B, suffers total internal reflection is, (1) sin–1 1/2, , (2), , sin–1 2/5, , (3), , sin–1 4/5, , (4), , sin–1 3/4, , Sol. Answer (3), , µA =, , 3 × 108, 2 × 10, , 8, , = 1.5 ; µB =, , 3 × 108, 2.5 × 10, , 8, , =, , 6, = 1.2, 5, , µB, 1.2, = 0.8, R.I. going from A to B = µ or, 1.5, A, , µ AB = 0.8 =, , , , sin C, at critical angle r = 90º, sin r, , sin C, = 0.8, sin90º, , C = sin−1, , 4, 5, , 17. Which of the following phenomenon of light forms a rainbow?, (1) Reflection of light, , (2), , Refraction, , (3) Total internal reflection, , (4), , Reflection as well as refraction, , Sol. Answer (4), Formation of a rainbow involves refraction to break white light into its constituent colours. It is also involves, internal reflection in the drop., 18. Which of the following is possible application of fibre optics?, (1) Endoscopy, , (2), , High speed internet traffic, , (3) Radio, TV & Telephone signals, , (4), , All of the above, , Sol. Answer (4), Total internal reflection is used in all of the above as they involve the use of optics fibres., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 228 :
6, , Ray Optics and Optical Instruments, , Solution of Assignment, , 19. An object is placed at a distance of f/2 from a convex lens. The image will be, (1) At one of the foci, virtual and double its size, , (2), , At 3f/2, real and inverted, , (3) At 2f, virtual and erect, , (4), , At f, real and inverted, , Sol. Answer (1), Lens formula, 1 1 1, − =, v u f, , Object is real and placed at –, , f, 2, , 1 1 2, = +, v f f, , 1 3, =, v f, v=, , f, 3, , 20. The least distance between a point object and its real image formed by a convex lens of focal length F is, (1) 2 F, , (2), , 3F, , (3), , 4F, , (4), , Greater than 4 F, , Sol. Answer (3), (Distance between a point object and its real image) d 4 f, 21. The plane faces of two identical plano-convex lenses, each having focal length of 40 cm, are placed against, each other to form a usual convex lens. The distance from this lens at which an object must be placed to, obtain a real, inverted image with magnification '–1' is, (1) 80 cm, , (2), , 40 cm, , (3), , 20 cm, , (4), , 160 cm, , Sol. Answer (2), It forms an equi-convex lens, f = 0.4 m, , P=, , 1, = 2.5 D, 0.4, , Power of combination = 5 D, , f =, , 1, P, , f =, , 1, m, 5, , or f = 20 cm, Therefore, object must be placed at 2f (= 40 cm), 22. Two thin lenses of focal lengths 20 cm and –20 cm are placed in contact with each other. The combination, has a focal length equal to, (1) Infinite, , (2), , 50 cm, , (3), , 60 cm, , (4), , 10 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 229 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 7, , Sol. Answer (1), , 1, 1, and P2 = –, 0.2, 0.2, , Powers : P1 =, , = 5 D and –5 D, Net power = 0, Focal length =, , 1, =, 0, , 23. If in a plano-convex lens, radius of curvature of convex surface is 10 cm and the focal length of lens is, 30 cm, the refractive index of the material of the lens will be, (1) 1.5, , (2), , 1.66, , (3), , 1.33, , (4), , 3, , Sol. Answer (3), R = 10 cm, f = 30 cm, 1, ⎛ 1 1⎞, = (μ − 1) ⎜ − ⎟, ⎝R ∞⎠, f, , 1, 1, = (µ − 1), 30, 10, µ=, , 4, 3, , 24. A glass concave lens is placed in a liquid in which it behaves like a convergent lens. If the refractive indices, of glass and liquid with respect to air are ag and al respectively, then, (1), , ag, , = 5al, , (2), , ag, , > Il, , (3), , ag, , < al, , (4), , ag, , = 2al, , Sol. Answer (3), The glass lens behaves as divergent in air which has less R.I., It will behave as convergent in a medium of higher R.I., 25. The diameter of aperture of a plano-convex lens is 6 cm and its maximum thickness is 3 mm. If the velocity, of light in the material of the lens is 2 × 108 m/s, its focal length is, (1) 10 cm, , (2), , 15 cm, , (3), , 30 cm, , (4), , 60 cm, , Sol. Answer (3), of medium =, , =, , 3 × 108, 2 × 108, , C, v, , = 1.5, , 6 cm, , 3 mm, , Radius of curvature of lens found using geometry, (R – 3)2 + (30)2 = R2, R2 + 9 – 6R + 900 = R2, 909 = 6R, R = 151.5 mm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 230 :
8, , Ray Optics and Optical Instruments, , Solution of Assignment, , or 15.15 cm, 1, ⎛1 1⎞, = (μ − 1) ⎜ − ⎟, ⎝∞ R⎠, f0, , 30 mm 3 mm, , R–3, , R, , 1, 1, = 0.5 ×, f0, 15.15, f0 = 30.30 cm, , This is distance from optical centre O. Distance from lens is, f = f0 – thickness at principal axis = 30.3 – 0.3, f = 30 cm, 26. Two plano-convex lenses of equal focal lengths are arranged as shown, , The ratio of the combined focal lengths is, (1) 1 : 2 : 1, , (2), , 1:2:3, , (3), , 1:1:1, , (4), , 2:1:2, , Sol. Answer (3), The power of lens remains the same no matter how the lenses are placed. So adding their power, the power, and hence the combined focal length will remain same in all three cases., 27. When an object is at a distance u1 and u2 from a lens, real image and a virtual image is formed respectively, having same magnification. The focal length of the lens is, (1) u1 – u2, , (2), , u1 u 2, 2, , (3), , u1 u 2, 2, , (4), , u1 + u2, , Sol. Answer (3), m=, , f, f +u, , For real image, , For virtual image, , f, –m = f − u ,, 1, , f, +m = f − u ,, 1, , −, , f, f, u + u2, =, f = 1, f − u2 f − u1 , 2, , 28. A concave lens of focal length f produces an image (1/x) of the size of the object. The distance of the object, from the lens is, (1) (x – 1) f, , (2), , (x + 1)f, , (3), , {(x – 1)/x}f, , (4), , {(x + 1)/x}f, , Sol. Answer (1), m=, +, , f, f +u, , −f, 1, =, ,, x −f + u, , For virtual image m = +, , 1, x, , u = –f(x – 1), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 231 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 9, , 29. A thin equiconvex glass lens of refractive index 1.5 has power of 5D. When the lens is immersed in a liquid, of refractive index , it acts as a divergent lens of focal length 100 cm. The value of of liquid is, (1) 4/3, , (2), , 3/4, , (3), , 5/3, , (4), , 8/3, , Sol. Answer (3), g = 1.5, , P=5D, , ⎛ 1 1⎞, 5 = (μg − 1) ⎜ + ⎟, ⎝R R ⎠, , 5×2=, , R=, , 2, R, , 1, = 0.2 m = 20 cm, 5, , fl < 0, , ⎞⎛ 2 ⎞, 1 ⎛ μg, =, − 1⎟ ⎜ ⎟, fl ⎝⎜ μl, ⎠ ⎝R ⎠, , −, , ⎛ μg, ⎞ 2, 1, =⎜, − 1⎟, 100 ⎝ μl, ⎠ 20, , µg, µl, , −1= −, , 1, 10, , 1.5 9, 5, =, , µl =, µl, 10, 3, , 30. In case of displacement method of lenses, the product of magnification in both cases is, (1) 1, , (2), , 2, , (3), , Zero, , (4), , Infinite, , Sol. Answer (1), m1m2 = 1 which is a fact., Here the lens is moved between the object and the screen., v and u interchange values between the two positions a clear image is formed on the screen., If m1 =, , v, u, and m2 =, u, v, , m1m2 = 1, 31. In the displacement method, a convex lens is placed in between an object and a screen. If the magnifications, in the two positions are m1 and m2 and the displacement of the lens between the two positions is x, then, the focal length of the lens is, x, (1) m m, 1, 2, , (2), , x, m1 – m2, , x, , (3), , m1 m2 , , 2, , x, , (4), , m1 – m2 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 232 :
10, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (2), By principle of reversibility where must be symmetry in the two position, , a, d, , u, , u, , v, v, d=u+v, , Hence,, , u=, , d −a, d +a, and v =, 2, 2, , Putting in lens equation, , , , f =, , a=v–u, , 2, 2, 1, +, =, d −a d +a f, , d 2 − a2, 4d, , m1 =, , d +a, d −a, , m2 =, , d −a, d +a, , |m1 – m2| =, , 4 da, d 2 + a2, , |m1 – m2| =, , a, . (a = x), f, , x, f = m –m, 1, 2, , 32. The focal length of a planoconvex glass lens is 20 cm (g =1.5). The plane face of it is silvered. An illuminating, object is placed at a distance of 60 cm from the lens on its axis along the convex side. Then the distance, (in cm) of the image is, (1) 20, , (2), , 30, , (3), , 40, , (4), , 12, , Sol. Answer (4), Net power of combination, P = 2PL + Pm, where PL is power of lens, Pm is power of mirrors, Pm = 0, PL =, , 1, 0.2, , P = 10 D, Focal length of combination (f) =, , 1, = 0.1 m or 10 cm, 10, , 1 1, −1, −, =, v 60 10, , 1, 1, 1, =, −, v 60 10, v = 12 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 233 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 11, , 33. Two thin similar convex glass pieces are joined together front to front with its rear portion silvered such that, a sharp image is formed 20 cm from the mirror. When the air between the glass pieces is replaced by water, (mw = 4/3), then the image formed from the mirror is at a distance of, , (1) 8 cm, , (2), , 10 cm, , (3), , 6 cm, , (4), , 12 cm, , Sol. Answer (4), Initially it is simply a concave mirror with f = –20 cm, R = 40 cm, Power of mirror (Pm) =, , =, , Focal length of lens =, , 1, 8, 1, =5D, 0.2, , R, 2(µ − 1), , 40, = 2 × ⎜⎛ 4 − 1⎟⎞, ⎝3 ⎠, , = 60 cm, Power of lens =, , =, , 1, 0.5, 100, D, 50, , Equivalent power = 2PL + Pm, P=, , 50, 6, , Net focal length =, , =, , 1, P, 6, × 100, 50, , = 12 cm, 34. Yellow light is refracted through a prism producing minimum deviation. If i1 and i2 denote the angle of incidence, and emergence for the prism, then, (1) i1 = i2, , (2), , i1 > i2, , (3), , i1 < i2, , (4), , i1 + i2 = 90, , Sol. Answer (1), At angle of minimum deviation angle of emergence of prism is same as angle of incidence., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 234 :
12, , Ray Optics and Optical Instruments, , Solution of Assignment, , 35. At what angle will a ray of light be incident on one face of an equilateral prism, so that the emergent ray may, graze the second surface of the prism ( = 2)?, (1) 30°, , (2), , 90°, , (3), , Let C by critical angle = 2 =, , sin i, sin r, , 45°, , (4), , 60°, , Sol. Answer (2), , 60º, , ie = 90º, 60º = r + C from geometry, , i, , 1 1 sin C, = =, µ 2 sin ie, C = sin−1, , r, , ie, , C, , [At emergent interface], , 60º, , 1, = 30º, 2, , 60º, ie = 90º, , r = 30º, sin i, =2, sin r, , sin i = 1, , i=, , π, 2, , 36. A prism of refractive index 2 has a refracting angle of 60o. At what angle must a ray be incident on it so that, it suffers a minimum deviation?, (1) 30°, , (2), , 45°, , (3), , 60°, , (4), , 75°, , Sol. Answer (2), , µ = 2, A = 60º, At for minimum deviation :, i = ie and r1 = r2, , i, , Also, r1 + r2 = A, or, , r1, , r2, , ie, , 2r = 60º, r = 30º, µ=, , sin i, sin r, , 2 = 2sini, , sin i =, , 1, 2, , [Putting r = 30º], , i = 45º, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 235 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 13, , 37. A glass prism of refractive index 1.5 is immersed in water of refractive index 4/3. A light ray incident normally, on face AB is totally reflected at face AC if, , B, , , , A, , C, , (1) sin > 8/9, , (2), , sin < 2/3, , (3), , sin =, , 3 2, , (4), , 2/3 < sin 8/9, , Sol. Answer (1), g (glass) = 1.5, l (water) = 4/3, Rays enter and pass undeviated at first interface, For total interface reflection at 2nd interface, , ⎛μ ⎞, ic > sin−1 ⎜ e ⎟, ⎝ μg ⎠, , , ic, , ⎛4 2⎞, ic > sin−1 ⎜ × ⎟, ⎝3 3⎠, ⎛8⎞, ic > sin−1 ⎜ ⎟, ⎝9⎠, , sin ic >, , 8, 9, , Since ic = from geometry, sin θ >, , 8, 9, , 38. A person can see clearly only up to a distance of 25 cm. He wants to read a book placed at a distance of, 50 cm. What kind of lens does he require for this purpose and what must be its power?, (1) Concave, – 1.0 D, , (2), , Convex, + 1.5 D, , (3), , Concave, – 2.0 D, , (4), , Convex, + 2.0 D, , Sol. Answer (3), He needs to bring the image of the object closer to 25 cm, Also the image should be virtual, 1 1 1, − =, v u f, , −, , 1, 1 1, +, =, 25 50 f, , f = –50 cm, , P=, , 1, −0.5, , P=2D, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 236 :
14, , Ray Optics and Optical Instruments, , Solution of Assignment, , 39. An astronomical telescope has an objective of focal length 100 cm and an eye piece of focal length 5 cm., The final image of a star is seen 25 cm from the eyepiece. The magnifying power of the telescope is, (1) 20, , (2), , 22, , (3), , 24, , (4), , 26, , Sol. Answer (3), f0 = 100 cm, fe = 5 cm, To form image at near point, , ⎡ 1 1⎤, m = −f0 ⎢ + ⎥, ⎣ fe D ⎦, ⎡1 1 ⎤, = −100 ⎢ +, ⎥, ⎣ 5 25 ⎦, ⎡6⎤, = −100 ⎢ ⎥, ⎣ 25 ⎦, , m = –24, 40. When a telescope is adjusted for normal vision, the distance of the objective from the eye-piece is found to, be 80 cm. The magnifying power of the telescope is 19. What are the focal lengths of the lenses?, (1) 61 cm, 19 cm, , (2), , 40 cm, 40 cm, , (3), , 76 cm, 4 cm, , (4), , 50 cm, 30 cm, , Sol. Answer (3), Distance between objective and eyepiece when telescope is adjusted for normal vision is given by, f0, 19 = f, e, , ⎛, f0 ⎞, ⎜⎝ M = f ⎟⎠, e, , ....(i), , 80 = f0+ fe. (L = f0 + fe), , ....(ii), , Solving (i) & (ii), f0 = 76 cm, f0 = 4 cm, 41. The focal lengths of the objective and eye lens of a telescope are respectively 200 cm and 5 cm. The maximum, magnifying power of the telescope will be, (1) – 40, , (2), , – 48, , (3), , – 60, , (4), , – 100, , Sol. Answer (2), Maximum magnification is at near point, Magnification at near point, m=−, , f0, fe, , ⎛ fe ⎞, ⎜⎝1 + D ⎟⎠, , m=−, , 200 ⎛, 5 ⎞, 200 6, ×, ⎜1 +, ⎟ =−, 5 ⎝ 25 ⎠, 5, 5, , m = –48, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 237 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 15, , 42. A convex lens forms a real image of a point object at a distance of 50 cm from the convex lens. A concave, lens is placed 10 cm behind the convex lens on the image side. On placing a plane mirror on the image side, and facing the concave lens, it is observed that the final image now coincides with the object itself. The focal, length of the concave lens is, (1) 50 cm, , (2), , 20 cm, , (3), , 40 cm, , (4), , 25 cm, , Sol. Answer (3), Final image coincides with the object when rays fall normal, to the mirror or parallel to principal axis. For this virtual, object must be at the focus of concave lens., , 50 cm, , Distance of virtual object from concave lens = 50 – 10 = 40, Focal length of concave lens = 40 cm., 43. A convex lens of focal length 100 cm and a concave lens of focal length 10 cm are placed coaxially at a, separation of 90 cm. If a parallel beam of light is incident on convex lens, then after passing through the two, lenses the beam, (1) Converges, , (2), , Diverges, , (3), , Remains parallel, , (4), , Disappears, , Sol. Answer (3), f = +100 cm, , f = –10 cm, , 90 cm, 100 cm, Virtual object for concave lens is at its focus., , SECTION - B, Objective Type Questions, 1., , If radii of curvature of both convex surfaces is 20 cm, then focal length of the lens for an object placed in air, in the given arrangement is, , 1=1, 2=1.5, (1) 10 cm, , (2), , 20 cm, , (3), , 40 cm, , (4), , 80 cm, , Sol. Answer (3), µ3 µ2 − µ1 µ3 − µ2, =, +, f, R1, R2, , 1 = 1, 2 = 1.5, 1 =, , 4, 3, , R1 = +20 cm, R2 = – 20 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 238 :
16, 2., , Ray Optics and Optical Instruments, , Solution of Assignment, , A driving mirror consists of a cylindrical mirror of radius of curvature 10 cm and the length over the curved, surface is 10 cm. If the eye of the driver be assumed to be at a great distance from the mirror, then field of, view in radian is, (1) 2.0, , (2), , 4.0, , (3), , 3.0, , (4), , 5.0, , Sol. Answer (1), f=, =, 3., , R, = 5 cm, 2, , 10 cm, , f, , , , 10, = 2 rad, 5, , Which of the following statements is correct?, (1) During hot summer days, the trees and other tall objects seem to be quivering because the density of air, changes in an irregular way, (2) When the moon is near the horizon it appears bigger. This is due to optical illusion, (3) If the critical angle for the medium of a prism is C and the angle of prism is A, there will be no emergent ray, when A > 2C, (4) All of these, , Sol. Answer (4), All the statements are true., 4., , An isosceles prism of angle A = 30° has one of its surfaces silvered. Light rays falling at an angle of incidence, 60° on the other surface retrace their path after reflection from the silvered surface. The refractive index of prism, material is, , 60°, , (1) 1.414, , (2), , 1.5, , 30°, , (3), , 1.732, , (4), , 1.866, , Sol. Answer (3), For light to retrace its path it must reflect normally, in the mirror., When it does so, by geometry r = 30º, µ=, , 5., , 60°, , 30°, , sin i, sin60º, =, = 3 = 1.732, sin30º, sin r, , A short linear object of length l lies along the axis of a concave mirror at a distance u from it. If v is the distance, of image from the mirror then size of the image is, (1) l , , v, u, , (2), , l, , u, v, , (3), , ⎛v ⎞, l ⎜ ⎟, ⎝u ⎠, , 2, , (4), , ⎛u ⎞, l ⎜ ⎟, ⎝v ⎠, , 2, , Sol. Answer (3), Axial magnification of a short object is given by, , m=, , l1 ⎛ v ⎞, =⎜ ⎟, l ⎝u ⎠, , 2, , 2, , ⎛v ⎞, l1 = ⎜ ⎟ l, ⎝u ⎠, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 239 :
Solution of Assignment, , 6., , Ray Optics and Optical Instruments, , 17, , A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index, of water is, , 4, and the fish is 12 cm, below the surface. The radius of the circle is, 3, , r, h, CC, , 16, , (1), , 7, , cm, , (2), , 26, 7, , cm, , 36, , (3), , 7, , cm, , (4), , 46, 7, , cm, , Sol. Answer (3), , r, , h = 12, = iC, , sin ic =, , sin ic =, , h, , 1, µw, , , , 3, 4, , r = h tan iC, , 3, , r =h, , r =, , 7., , 16 − 9, , 36, 7, , In optical fibre, refractive index of inner part is 1.68 and refractive index of outer part is 1.44. The numerical, aperture of the fibre is, (1) 0.5653, , (2), , 0.6653, , (3), , 0.7653, , (4), , 0.8653, , Sol. Answer (4), 0sin =, , µ12 − µ22, , (Numerical aperture), 1 = 1.68, 2 = 1.44, 8., , Compare the dispersive powers of two prisms if one of them deviates the blue and red rays through 10° and, 6° respectively and the second prism through 8° and 4.5°, (1) 0.69, , (2), , 0.79, , (3), , 0.89, , (4), , 0.99, , Sol. Answer (3), Dispersive power =, , δy =, , δv − δ r, δy, , δv + δ r, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 240 :
18, , Ray Optics and Optical Instruments, , w1 =, , 10 − 6 1, =, 8, 2, , w2 =, , 8 − 4.5, 3.5, =, 6.25, 6.25, , Solution of Assignment, , w1, ≈ 0.89, w2, , 9., , A thin prism of angle 6° made of glass of refractive index 1.5 is combined with another prism made of glass, of refractive index 1.75 to produce dispersion without deviation. Then the angle of the second prism is, (1) 7°, , (2), , 4°, , (3), , 9°, , (4), , 5°, , Sol. Answer (2), For dispersion without deviation :, A1(1 – 1) + A2(2 – 1) = 0, 6(0.5) + A2(0.75) = 0, A2 = 4º, 10. In a medium of refractive index 1.6 and having a convex surface has a point object in it at a distance of 12, cm from the pole. The radius of curvature is 6 cm. Locate the image as seen from air, (1) A real image at 30 cm, , (2), , A virtual image at 30 cm, , (3) A real image at 4.28 cm, , (4), , A virtual image at 4.28 cm, , Sol. Answer (2), µ 2 µ1 µ2 − µ1, −, =, v, u, R, , 1 = 1.6, 2 = 1, u = 12 cm, R = – 6 cm, , P, , 12 cm O, , 1 1.6 1 − 1.6, −, =, −6, v −12, , v = – 30 cm. (Virtual image), 11. A point object is situated at a distance of 36 cm from the centre of the sphere of radius 12 cm and refractive, index 1.5. Locate the position of the image due to refraction through sphere., (1) 24 cm from the surface, , (2), , 36 cm from the centre, , (3) 24 cm from the centre, , (4), , Both (1) & (2), , Sol. Answer (4), 1 = 1,, 2 = 1.5, , 12 cm, , u = –24 cm,, R = +12 cm, , O, , 36 cm, , µ 2 µ1 µ2 − µ1, −, =, v, u, R, , 1.5, 1, 1.5 − 1, −, =, v, −24, +12, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 241 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 19, , v becomes parallel to the principal axis., On second face, 1 = 1.5 , 2 = 1, u , R = – 12 cm, , µ 2 µ1 µ 2 − µ1, −, =, v, u, R, 1 1.5 1 − 1.5, −, =, ∞, −12, v, , v = 24 cm, Final image at 24 cm from the surface and from centre of sphere., 12. A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12 cm. An object is, situated in the denser medium at a distance of 9 cm from the pole. Locate the image due to refraction in air., (1) A real image at 8 cm, , (2), , A virtual image at 8 cm, , (3) A real image at 4.8 cm, , (4), , A virtual image at 4.8 cm, , Sol. Answer (4), , µ 2 µ1 µ 2 − µ1, −, =, v, u, R, , 1, , 1 1.5 0.5, +, =, v, −12, 9, , 2, P, , 1 −1 1, =, −, v, 6 24, , O, , v = –4.8, 13. A light ray is travelling from air to glass. The reflected and refracted rays are perpendicular to each other. If, the angle of incidence in air is i the refractive index of glass is, (1) sin i, , (2), , cos i, , (3), , tan i, , (4), , cot i, , Sol. Answer (3), 180º = i + 90 + r, , i i, , r = 90 – i, , r, , sin i, µ=, sin r, , = tan i, 14. A ray is incident on boundary separating glass and water. Refractive index for glass is, for water is, , 4, critical angle for glass-air boundary is, 3, , 1⎛ 3 ⎞, (1) sin ⎜ ⎟, ⎝4⎠, , (2), , ⎛2⎞, sin 1⎜ ⎟, ⎝3⎠, , (3), , ⎛8⎞, sin 1⎜ ⎟, ⎝9⎠, , (4), , 3, and refractive index, 2, , ⎛ 1⎞, sin 1⎜ ⎟, ⎝9⎠, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 243 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 21, , 17. A glass slab ( = 1.5) of thickness 2 cm is placed on a spot. The shift of spot if it is viewed from top, (1), , 2, cm, 3, , (2), , 4, cm, 3, , (3), , 1, cm, 3, , (4), , 5, cm, 3, , Sol. Answer (1), ⎛, 1⎞, Shift = t ⎜⎝1 − ⎟⎠, μ, , 1 ⎞, ⎛, = 2 ⎝⎜1 −, ⎟, 1.5 ⎠, , =, , 2, cm, 3, , 18. Ray diagram for two lenses kept at some distance given in the diagram, which of the following option is correct, (f1, f2 = focal length, d = distance between lenses), , (1) f1+f2 > d, , (2), , f1+f2 < d, , (3) f1+f2 = d, , (4), , Combination behaves like converging lens, , Sol. Answer (3), , f1, , O1, , I, , f2, , P, d, , O2, , II, , For parallel incidence image is at f1 distance from O1. For final emergence to be parallel object for II is at, distance f2 from O2. The distance between I and II would be f1 + f2., 19. A ball is projected from top of the table with initial speed u at an angle of inclination q, motion of image of, ball w.r.t ball, u, , , (1) Must be projectile, , (2), , Must be straight line and vertical, , (3) Must be straight line and horizontal, , (4), , May be straight line, depends upon value of , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 244 :
22, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (3), , u sin, , u sin, , , , V I0 = V I − V 0, , V I0 = −u cos θ iˆ + u sin j − (u cos θ i + u sin θ j ), , O, , , V I0 = −2u cos θiˆ, , u, , , , u cos, , u cos, , I, , Straight line and horizontal, 20. In displacement method, convex lens forms a real image of an object for its two different positions. If heights, of the images in two cases be 24 cm and 6 cm, then the height of the object is, (1) 3 cm, , (2), , 36 cm, , (3), , 6 cm, , (4), , 12 cm, , Sol. Answer (4), In displacement method, m1m2 = 1, h1 h2, ×, =1, h, h, , h = h1h2, h = 24 × 6, , h = 12 cm, 21. Two parallel rays of red and violet colour passed through a glass slab, which of the following is correct?, , Vio Red, let, 1, 2, , t, , 4, , 3, , 6, (1) 3 and 4 are parallel, , (2), , 4 and 5 are parallel, , (3), , Sol. Answer (4), , 6 and 3 are parallel (4), , the emergent ray is parallel to incident ray., , t, , 4, , 3, , because 1 || 2, , , 2 || 5, , 2 and 5 are parallel, , Vio Red, let, 1, 2, , When refraction occurs through parallel glass slab,, , 6 || 2, 5 || 1, , 5, , 6, , 5, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 245 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 23, , 22. A plane glass is kept over a coloured word ‘VIBGYOR’, where colour of letters as same as the colours in white, light start by letter, the letter which appears least raised is, (1) R, , (2), , Y, , (3), , O, , (4), , V, , Sol. Answer (1), , µ=, , Real depth, Apparent depth, , When an image is least raised its apparent depth is highest., R.I. is lowest which happens to be for red light., 23. The near point of a person is 75 cm. In order that he may be able to read book at a distance 30 cm. The, power of spectacles lenses should be, (1) –2 D, , (2), , +3.75 D, , (3), , +2 D, , (4), , +3 D, , Sol. Answer (3), , , 1, 1, 1, −, =, −75 −30 f, , , , 1 1, =, f 50, P = +2D, , 24. If a lens is moved towards the object from a distance of 40 cm to 30 cm, the magnification of the image, remains the same (numarically). The focal length of the lens is, (1) 20 cm, , (2), , 15 cm, , (3), , 35 cm, , (4), , 18 cm, , Sol. Answer (3), f, −f, =, f + u1 f + u2, , u1 =–40 cm, u2 =–30 cm, f, , 0, , –(f – 40) = f – 30, –f + 40 = f – 30, f = +35 cm, 25. A convex lens of power +2.5 D is in contact with a concave lens of focal length 25 cm. The power of, combination is, (1) –1.5 D, , (2), , 0D, , (3), , +1.5 D, , (4), , +6.5 D, , Sol. Answer (1), Power of concave lens = −, , 1, = –4 D, 0.25, , Adding combination = –4 D + 2.5 D = –1.5 D, 26. For a telescope in normal adjustment, the length of telescope is found to be 27 cm. If the magnifying power, of telescope, at normal adjustment is 8, the focal lengths of objective and eye piece are respectively, (1) 24 cm, 3 cm, , (2), , 27 cm, 8 cm, , (3), , 12 cm, 6 cm, , (4), , 27 cm, 9 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 246 :
24, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (1), For normal adjustment m =, , f0, and L = f0 + fe, fe, , f0, 8= f, e, , f0 = 8fe, From 27 = f0 + fe, 27 = 8fe + fe, fe = 3 cm, f0 = 24 cm, 27. A converging lens of focal length 30 cm is placed in contact with another converging lens of unknown focal, length, then possible value for focal length of combination is, (1) 15 cm, , (2), , 60 cm, , (3), , 36 cm, , (4), , –12 cm, , Sol. Answer (1), Total power of combination will be more than power of given lens and focal length will be less., 28. In the diagram the ray passing through prism is parallel to the base. Refractive index of material of prism is, , 60º, 45º, 3, 2, , (1), , (2), , 3, , 45º, (3), , (4), , 2, , 3, 2, , Sol. Answer (4), By geometry r = 45º, , µ=, , [Alternate angles], , 60º, , sin60º, 3, =, × 2, sin 45º, 2, , 45º, , 3, 2, , µ=, , r, 45º, , 45º, , 29. In displacement method we use a lens of focal length f and distance between object and screen is 60 cm., Possible value for focal length is, (1) –15 cm, , (2), , 30 cm, , (3), , 12 cm, , (4), , 20 cm, , Sol. Answer (3), f, , D, 4, , f , , 60, 4, , f 15 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 247 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 25, , 4⎞, ⎛, 30. A red colour in air has wavelength 760 nm when light passes through water of refractive index ⎜ n ⎟ ,, 3, ⎝, ⎠, wavelength becomes 570 nm. (wavelength of yellow colour in air is 570 nm). Then colour of red light in water, is, , (1) Red, , (2), , Green, , (3), , Yellow, , (4), , Blue, , Sol. Answer (1), Colour of a wave depends more on frequency than wavelength as it depicts the amount of energy is carries., Since frequency and energy doesnot change it will simply remain red., 31., , n̂1 is the unit vector along incident ray, n̂2 along normal and n̂3 is the unit vector along reflected ray, then, which of the following must be true?, , (1) n̂1 n̂2 = 0, , (2), , n̂1 n̂3 = 0, , (3), , ( n̂1 × n̂2 ) n̂3 = 0, , (4), , ( n̂1 × n̂2 ) × n̂3 = 0, , Sol. Answer (3), The reflected ray, refracted ray a incident ray and normal all lie on the same plane. Hence (3) is true., 32. A double convex lens has two surfaces of equal radii 15 cm and refractive index = 1.5, its focal length is, equal to, (1) –15 cm, , (2), , –30 cm, , (3), , +15 cm, , (4), , +30 cm, , Sol. Answer (3), 1, 1 ⎞, ⎛1, = (μ − 1) ⎜ −, ⎝ R −R ⎟⎠, f, 1, ⎛ 2⎞, = 0.5 ⎜ ⎟, ⎝ 15 ⎠, f, , f = 15 cm, 33. The distance between a real object and its real image is 56 cm formed by converging lens, focal length of, lens is, (1) f 14, , (2), , f > 14, , (3), , f = 14, , (4), , f = 28, , Sol. Answer (1), f , , D, 4, , f , , 56, 4, , f 14 cm, 34. In displacement method, there are two position of a lens for which we get clear image. First position of the, lens is at 40 cm from object and second is at 80 cm, the focal length of lens is, , (1) 40 cm, , (2), , 40, cm, 3, , (3), , 80 cm, , (4), , 80, cm, 3, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 248 :
26, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (4), Displacement method is based on principle of reversibility. The image distance for 1st position of lens will be, same as object position with IInd position of lens., u = – 40 cm, V = + 80 cm, , 1 1, 1, =, −, f 80 −40, f =, , 80, cm, 3, , 35. Maximum magnification produced by simple micro-scope of focal length f = 5 cm is, (1) 5, , (2), , 7, , (3), , 6, , (4), , 8, , Sol. Answer (3), Maximum magnification is when final image is at near point D., , m = 1+, , D, 25, = 1+, [D = 25 cm], f, 5, , =6, , SECTION - C, Previous Years Questions, 1., , In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of, objective lens. The eyepiece forms a real image of this line. The length of this image is I. The magnification, of the telescope is, [Re-AIPMT-2015], (1), , L, I, , L, 1, I, , (2), , (3), , L, –1, I, , (4), , LI, L–I, , Sol. Answer (1), , Objective, , eyepiece, I, , L, v, , f0 + fe, , f0, At normal adjustment M = f, e, , ...(i), , and distance between lenses = f0 + fe, Lateral magnification, , L f0 fe, , I, v, , Using lens equation, , ...(ii), , 1 1 1, – , v u f, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 249 :
Solution of Assignment, , Ray Optics and Optical Instruments, , , , 1, 1, 1, –, , v – f f, fe, e, 0, , , , , , , , f0, 1, , v f f f, e 0, e, , , , , , , , f0, , fe, , , , f0 fe, , 27, , ...(iii), , v, , Comparing equations (i) (ii) & (iii), f0 L, M= f I, e, , 2., , A beam of light consisting of red, green and blue colours is incident on a right angled prism. The refractive, index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47,, respectively, , A, , Blue, Green, Red, , B, , 45°, , C, , The prism will, , [Re-AIPMT-2015], , (1) Separate the red colour part from the green and blue colours, (2) Separate the blue colour part from the red and green colours, (3) Separate all the three colours from one another, (4) Not separate the three colours at all, Sol. Answer (1), , A, , Red, 45°, , B, , 45°, , C, , Refractive index of light rays that can just pass through the prism at grazing emergence at 2nd surface is, =, , 1, = 1.414, sin 45, , Light having refractive index < 1.414 takes refraction but light having > 1.414 suffers TIR., Only red colour light will come out of prism., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 250 :
28, 3., , Ray Optics and Optical Instruments, , Solution of Assignment, , The refracting angle of a prism is A, and refractive index of the material of the prism is cot(A/2). The angle of, minimum deviation is, [AIPMT-2015], (1) 180° + 2A, , (2), , 180° – 3A, , (3), , 180° – 2A, , (4), , 90° – A, , Sol. Answer (3), ⎛A, ⎞, A sin ⎛ A ⎞, ⎛A, ⎞, sin ⎜ m ⎟, sin ⎜ m ⎟, cos, ⎜2, m⎟, ⎝, ⎠, A, 2 , ⎝2, ⎠,, ⎝2, ⎠ , , cot , A, A, A, A, 2, sin, sin, sin, sin, 2, 2, 2, 2, , 90 , 4., , A A m, , m (180 2 A), 2 2, 2, , Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm, are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive, index 1.7. The focal length of the combination is, [AIPMT-2015], (1) 50 cm, , (2), , – 20 cm, , (3), , – 25 cm, , (4), , – 50 cm, , Sol. Answer (4), 1 1 1 1, , f f1 f2 f3, , f1, , f2, , 1 1, ⎛ 1 1⎞, (1.5 1) ⎜, ⎟, f1 f2, ⎝ 20 ⎠, , 1 1, 1, , f1 f2 40, , f3, , 1, 1 ⎞, 0.7, 7, ⎛ 1, (1.7 1) ⎜ , , ⎟ 10 100, f3, ⎝ 20 20 ⎠, 1 1, 1, 7, , , , f 40 40 100, 1, 2, , f, 100, , f 50 cm, 5., , If the focal length of objective lens is increased then magnifying power of, , [AIPMT-2014], , (1) Microscope will increase but that of telescope decrease, (2) Microscope and telescope both will increase, (3) Microscope and telescope both will decrease, (4) Microscope will decrease but that of telescope will increase, Sol. Answer (4), MP of microscope , , L, f0, , ⎡ P⎤, ⎢1 ⎥, ⎣ fe ⎦, , MP of telescope f 0 ⎡1 f e ⎤, f e ⎢⎣ D ⎥⎦, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 251 :
Solution of Assignment, , 6., , Ray Optics and Optical Instruments, , 29, , The angle of a prism is A. One of its refracting surface is silvered. Light rays falling at an angle of incidence 2A, on the first surface returns back through the same path after suffering reflection at the silvered surface. The, refractive index , of the prism is:, [AIPMT-2014], (1) 2 sinA, , (2), , 2 cosA, , (3), , 1, cosA, 2, , Sol. Answer (2), , (4), , tanA, , A, , i = 2A, r = A, sin i, , sin r, , 2A, , A, , sin 2 A, 2cos A, sin A, , 7., , A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If, lenses are made of different materials of refractive indices 1 and 2 and R is the radius of curvature of the, curved surface of the lenses, then the focal length of the combination is, [NEET-2013], R, (1) 2 , 1, 2, , (2), , R, 1 2 , , (3), , 2R, 2 1 , , (4), , R, 2 1 2 , , Sol. Answer (2), 8., , For a normal eye, the cornea of eye provides a converging power of 40 D and the least converging power of, the eye lens behind the cornea is 20 D. Using this information, the distance between the retina and the corneaeye lens can be estimated to be, [NEET-2013], (1) 2.5 cm, , (2), , 1.67 cm, , (3), , 1.5 cm, , (4), , 5 cm, , Sol. Answer (2), 9., , The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the, objective and eyepiece is 20 cm. The focal length of lenses are, [AIPMT (Prelims)-2012], (1) 18 cm, 2 cm, , (2), , 11 cm, 9 cm, , (3), , 10 cm, 10 cm, , (4), , 15 cm, 5 cm, , Sol. Answer (1), fo + fe = 20, , ....(i), , fo, =9, fe, , ....(ii), , Solving (i) and (ii), fo = 18 cm, , fe = 2 cm, , 10. A ray of light is incident at an angle of incidence i on one face of a prism of angle A (assumed to be small) and, emerges normally from the opposite face. It the refractive index of the prism is , the angle of incidence i, is, nearly equal to, [AIPMT (Prelims)-2012], , (1), , A, , , (2), , A, 2, , (3), , A, , (4), , A, 2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 252 :
30, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (3), r1 + r 2 = A, For ray to pass normally r2 = 0, r1 = A, , µ=, , sin i, sin A, , sinA = sini, If both A and i are small, i = A, 11. A concave mirror of focal length f1 is placed at a distance of 'd' from a convex lens of focal length f2 A beam of, light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance, d must equal, [AIPMT (Prelims)-2012], (1) 2f1 + f2, , (2), , –2f1 + f2, , (3), , f1 + f2, , (4), , –f1 + f2, , Sol. Answer (1), d = 2f1 + f2, , 2f1, , f2, d, 12. When a biconvex lens of glass having refractive index 1.47 is dipped in a liqud, it acts as a plane sheet of, glass. This implies that the liquid must have refractive index, [AIPMT (Prelims)-2012], (1) Greater than that of glass, , (2), , Less than that of glass, , (3) Equal to that of glass, , (4), , Less than one, , Sol. Answer (3), Power of lens is euqal to zero., ⎞⎛ 1, 1 ⎛ μg, 1 ⎞, =⎜, − 1⎟ ⎜ −, =0, f ⎝ μe, ⎠ ⎝ R1 R2 ⎟⎠, , g = l, 13. For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a, material whose refractive index, [AIPMT (Mains)-2012], (1) Lies between, (3) Is less than 1, , 2 and 1, , (2), , Lies between 2 and, , (4), , Is greater than 2, , 2, , Sol. Answer (2), =i+e–A, min, i = e, min = A, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 253 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 31, , 2A = 2i, A=i, At min,, sin A + δ m, A, A, 2 sin .cos, sin A, 2, 2, 2, =, =, =, A, A, A, sin, sin, sin, 2, 2, 2, , = 2cos, , A, 2, , = 2cos, , i, 2, , (A = i), , imax = 90º, , imin = 0º, , = 2 cos 45º, , = 2 cos 0º, , =, , =2, , 2, , As R.I. lies between 2 and, , 2., , 14. A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that, its end closer to the pole is 20 cm away from the mirror. The length of the image is [AIPMT (Mains)-2012], (1) 10 cm, , (2), , 15 cm, , (3), , 2.5 cm, , (4), , 5 cm, , Sol. Answer (4), O1 is at C, image of O1 will form at same postion., For image of O2, u = –30 cm, f = –10 cm, , 1 1 1, − =, v u f, , O2, , f = 10 cm, R = 20 cm, , O1, 10 cm, , 20 cm, , P, , 1, 1, 1, +, =, v −30 −10, v = –15 cm, , Length of image, I1I2 = |PI2 – PI1|, = |15 – 20|, = 5 cm, 15. Which of the following is not due to total internal reflection?, , [AIPMT (Prelims)-2011], , (1) Brilliance of diamond, (2) Working if optical fibre, (3) Difference between apparent and real depth of a pond, (4) Mirage on hot summer days, Sol. Answer (3), Real & apparent depth are explained on the basis of refraction only. TIR not involved here., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 254 :
32, , Ray Optics and Optical Instruments, , Solution of Assignment, , 16. A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options describe best, the image formed of an object of height 2 cm placed 30 cm from the lens?, [AIPMT (Prelims)-2011], (1) Real, inverted, height = 1 cm, , (2), , Virtual, upright, height = 1 cm, , (3) Virtual, upright, height = 0.5 cm, , (4), , Real, inverted, height = 4 cm, , Sol. Answer (4), , 1, 1 ⎞, ⎛ 1, = (1.5 − 1) ⎜, +, ⎝ 20 20 ⎟⎠, f, f = 20 cm, m=, , f, f +u, , m=, , 20, = −2 (Real and inversed), 20 − 30, , hi, = −2 , h0 = 2 cm, h0, i = –4 cm, 17. A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at, a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move, 5 cm closer to the lens. The focal length of the lens is, [AIPMT (Mains)-2011], (1) –30 cm, , (2), , 5 cm, , (3), , –10 cm, , (4), , 20 cm, , Sol. Answer (1), For virtual object u = +10 cm, v = +15 cm, The ray diagram is as shown., , 1 1 1, − =, v u f, 1, 1 1, −, =, 15 10 f, , 5 cm, , 2−3 1, =, 30, f, , 15 cm, , f = –30 cm, 18. A thin prism of angle 15° made of glass of refractive index 1 = 1.5 is combined with another prism of glass of, refractive index 2 = 1.75. The combination of the prisms produces dispersion without deviation. The angle of, the second prism should be, [AIPMT (Mains)-2011], (1) 12°, , (2), , 5°, , (3), , 7°, , (4), , 10°, , Sol. Answer (4), Angle = 15º, 1 = 1.5, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 255 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 33, , 2 = 1.75, 15(1 – 1) + A(2 – 1) = 0, 7.5 + 0.75A = 0, 0.75 A = –7.5, , A=−, , 7.5, 0.75, , A = 10º, 19. A ray of light travelling in a transparent medium of refractive index , falls on a surface separating the medium, from air at an angle of incidence of 45°. For which of the following value of the ray can undergo total internal, reflection?, [AIPMT (Prelims)-2010], (1) = 1.25, , (2), , = 1.33, , (3), , = 1.40, , (4), , = 1.50, , Sol. Answer (4), C < 45º, sin C < sin 45º, , 1, < sin 45º, µ, >, , 2, , Only possible with = 1.5, 20. A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d / 2, in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be, respectively, [AIPMT (Prelims)-2010], (1) f and, , I, 4, , (2), , I, 3f, and, 2, 4, , (3), , f and, , 3I, 4, , (4), , f, I, and, 2, 2, , Sol. Answer (3), Focal length will not change as long as curvature of lens does not change., I d2, d = Diameter of aperture, I = Intensity of image, , d, = Aperture is covered by black paper, 2, I , , d2, 4, , I =, , I, 4, , obstructed by paper, , Intensity of image = I −, , I 3I, =, 4 4, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 256 :
34, , Ray Optics and Optical Instruments, , Solution of Assignment, , 21. The speed of light in media M1 and M2 is 1.5 × 108 m/s and 2 × 108 m/s respectively. A ray of light enters, from medium M1 to M2 at an incidence angle i. If the ray suffers total internal reflection, the value of i is, [AIPMT (Mains)-2010], , ⎛2⎞, (1) Equal to sin–1 ⎜ ⎟, ⎝3⎠, , (2), , ⎛3⎞, Equal to or less than sin–1 ⎜ ⎟, ⎝5⎠, , ⎛3⎞, (3) Equal to or greater than sin–1 ⎜ ⎟, ⎝4⎠, , (4), , ⎛2⎞, Less than sin–1 ⎜ ⎟, ⎝3⎠, , Sol. Answer (3), i > C, sin i > sin C, , sin i >, , sin i >, , µ2, µ1, , sin i >, , v1, sin i > v, 2, , 1.5 × 108, 2 × 108, , 3, 4, , −1 ⎛ 3 ⎞, i > sin ⎜⎝ ⎟⎠, 4, , 22. A ray of light is incident on a 60° prism at the minimum deviation position. The angle of refraction at the first, face (i.e., incident face) of the prism is, [AIPMT (Mains)-2010], (1) Zero, , (2), , 30°, , (3), , 45°, , (4), , 60°, , Sol. Answer (2), At minimum deviation, r = r, According to geometry of prism, r + r = A, 2r = A, r=, , 60, 2, , r = 30º, 23. Two thin lenses of focal lengths f1 and f2 are in contact and coaxial. The power of the combinations is, [AIPMT (Prelims)-2008], , (1), , f1 f2, f1f2, , (2), , f1, f2, , (3), , f2, f1, , (4), , f1 f2, 2, , Sol. Answer (1), , 1 1 1, = +, f f1 f2, P=, , f1 + f2, f1f2, , ⎡1, ⎤, ⎢f = P⎥, ⎣, ⎦, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 257 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 35, , 24. A boy is trying to start a fire by focusing Sunlight on a piece of paper using an equiconvex lens of focal length, 10 cm. The diameter of the Sun is 1.39 × 109 m and its mean distance from the earth is 1.5 × 1011 m. What, is the diameter of the Sun's image on the paper?, [AIPMT (Prelims)-2008], (1) 12.4 × 10–4 m, , (2), , 9.2 × 10–4 m, , (3), , 6.5 × 10–4 m, , (4), , 6.5 × 10–5 m, , Sol. Answer (2), v = 10 cm, u = 1.5 × 1011 m, Magnification =, 0.1, , Image, , 11, , =, , 12, , × 1.39 × 109 = Image, , 1.5 × 10, 1, , 1.5 × 10, , or, , v, Image diameter, =, u, Sun's diameter, 1.39 × 109, , 1.39, × 10−3 = Image, 1.5, , or 9.2 × 10–4 = Image, Image diameter = 9.2 × 10–4 m, 25. The frequency of a light wave in a material is 2 × 1014 Hz and wavelength is 5000 Å. The refractive index of, material will be, [AIPMT (Prelims)-2007], (1) 1.33, , (2), , 1.40, , (3), , 1.50, , (4), , 3.00, , Sol. Answer (4), C, = v, m, C, = f λ (Medium) fm = fa (air), m m, , C, = f λ, a m, =, , 3 × 108, 2 × 1014, , × 5000 × 10−10 = 3, , 26. A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels upto the, surface of the liquid and moves along its surface. How fast is the light travelling in the liquid?, [AIPMT (Prelims)-2007], 3 cm, , 4 cm, , coin, , (1) 1.2 × 108 m/s, , (2) 1.8 × 108 m/s, , (3), , 1.3 × 108 m/s, , (4), , 3.0 × 108 m/s, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 258 :
36, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (2), is critical angle sin =, , l =, , 1, 3, =, µl, 5, , 5, 3, , 3m, , v in medium, , , , 4m, , v in air (C ), µl =, v in medium, , , , =, , 3, × 3 × 108, 5, , =, , 9, × 108, 5, , 5m, , v = 1.8 × 108, 27. A microscope is focussed on a mark on a piece of paper and then a slab of glass of thickness 3 cm and, refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus, again?, [AIPMT (Prelims)-2006], (1) 1 cm upward, , (2), , 4.5 cm downward, , (3), , 1 cm downward, , (4), , 2 cm upward, , Sol. Answer (1), ⎛, 1⎞, Shift = t ⎜⎝1 − ⎟⎠, μ, , 1 ⎞, ⎛, = 3 ⎜⎝1 −, ⎟, 1.5 ⎠, , = 1 cm, So the microscope must be moved by 1 mm upwards, , µ=, , Real depth, Apparent depth, , Apparent depth =, , 2, Real depth, = × 3 mm = 2 mm, 3, µ, , Shift = 3 – 2 = 1 mm, So the microscope must be moved 1 mm upwards., 28. A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a, combination of lenses. The power in diopters of the combination is, [AIPMT (Prelims)-2006], (1) 25, , (2), , 50, , (3), , Infinite, , (4), , Zero, , Sol. Answer (4), Each lens of same power but different is sign., When added P = P1 + P2, P = P1 – P 2, P=0, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 259 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 37, , 29. A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirror will be, (1) 12 feet, , (2), , 3 feet, , (3), , 6 feet, , (4), , Any length, , Sol. Answer (2), Height - 6 ft, To see any object in a plane mirror complete, a mirror must be half the height of object., So minimum height of mirror =, , h, = 3 ft, 2, , 30. Images formed by an object placed between two plane mirrors whose reflecting surfaces make an angle of, 90° with one another lie on a, (1) Straight line, , (2), , Parabola, , (3), , Sol. Answer (3), , Circle, , (4), , Ellipse, , m2, I2, , O, , m1, I3, , I1, , 31. An object is placed between two plane mirrors inclined at an angle ‘’ to each other. If the number of images, formed is 7, then the angle of inclination ‘’ is, (1) 15°, , (2), , 30°, , (3), , 45°, , (4), , 60°, , (1) Greater than that of the cladding, , (2), , Equal to that of the cladding, , (3) Smaller than that of the cladding, , (4), , Independent of that of the cladding, , Sol. Answer (3), , 360, −1, θ, = 45º, 7=, , 32. In optical fibres, the refractive index of the core is, , Sol. Answer (1), 33. Light travels through a glass plate of thickness t and having a refractive index . If c is the velocity of light in, vacuum, the time taken by light to travel this thickness of glass is, , t, (1) c, , (2), , t, c, , (3), , tc, , (4), , tc, , , Sol. Answer (2), Time =, =, , c, V, , V=, , c, µ, , Time =, , t, v (Speed in glass), , µt, c, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 260 :
38, , Ray Optics and Optical Instruments, , Solution of Assignment, , 34. A ray of light from a denser medium strikes a rare medium as shown in figure. The reflected and refracted, rays make an angle of 90° with each other. The angles of reflection and refraction are r and r. The critical, angle would be, N, A, C, , i r, DENSER, RARER B, r, D, , N, (1) sin–1 (tan r), , (2), , tan–1 (sin r), , (3), , sin–1 (cot r), , (4), , tan–1 (sin r ), , (4), , 2.25 × 108 m/s, , Sol. Answer (1), i=r, r + r = 90º, i + r = 90º, r = 90º – i, sin c =, , 1, µ (R.I. of denser w.r.t. rarer), , sin c = (R.I. of rarer w.r.t. denser), sin c =, , sin i, sin r ′, , sin c =, , sin i, sin(90º −i ), , sin c = tan i, c=, , sin–1, , (i = r), (tan r), , 35. The refractive index of water is 1.33. What will be the speed of light in water?, (1) 4 × 108 m/s, , (2), , 1.33 × 108 m/s, , (3), , 3 × 108 m/s, , Sol. Answer (4), 1.33 =, , c, V (Speed of light in water), , 36. An electromagnetic radiation of frequency n, wavelength l, travelling with velocity v in air, enters a glass slab, of refractive index . The frequency, wavelength and velocity of light in the glass slab will be respectively, (1) n, 2 and, , v, , , (2), , 2n , , and v, , , (3), , n , v, ,, and, , , , (4), , n,, , v, , and, , , , Sol. Answer (4), By definition of refractive index, Velocity of light becomes, becomes, , v, μ, , λ, μ, , But frequency is constant., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 261 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 39, , 37. A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed, 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out is, (1) , , (2), , 3m, , (3), , 6m, , (4), , 4m, , Sol. Answer (2), ic = , , r, , 1, sin iC =, µd, , iC, , 4m, , , , 3, sin iC =, 5, 3, , tan ic =, , 25 − 3, , 2, , =, , 3, 4, , r = 4 tan, r=3m, 38. A ray of light travelling in air have wavelength , frequency n, velocity v and intensity I. If this ray enters into, water then these parameters are , n, v and I respectively. Which relation is correct from following?, (1) = , , (2), , n = n, , (3), , v = v, , (4), , I = I, , (1) Total internal reflection, , (2), , Less scattering, , (3) Refraction, , (4), , Less absorption coefficient, , Sol. Answer (2), When ray enters water, becomes, , λ, v, and velocity becomes, μ, μ, , Intensity also changes, Only frequency n remains same so answer is (2)., 39. Optical fibre are based on, , Sol. Answer (1), Optical fibres depend on total internal reflection., 40. For the given incident ray as shown in figure, the condition of total internal reflection of this ray the required, refractive index of prism will be, 45º, , (1), , 3 1, 2, , (2), , 2 1, 2, , incident, ray, , (3), , 3, 2, , (4), , 7, 6, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 262 :
40, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (3), iC = 90 – r, , 45º, , ⎛ 1 ⎞, sin(90 − r ) = ⎜ ⎟, ⎝ μg ⎠, µg =, , Also, , , , cos r =, , sin 45º, sin r, , r, , ....(ii), , sin r, sin 45º, , tanr = sin45º =, , sin r =, , , , iC, , ....(i), , 1, 2, , , 1, , 1, 3, , sin i, µ=, =, sin r, , r, 2, , 3, 2, , 41. A layer of benzene ( = 1.5) 12 cm thick floats on water layer (m = 4/3) 8 cm thick in a vessel. When viewed, from the top, the apparent depth of bottom of vessel below the surface of benzene will be, (1) 20 cm, , (2), , 14 cm, , (3), , 7 cm, , (4), , 21 cm, , Sol. Answer (2), In case of multiple medium of different R.I., Apparent depth d =, , d=, , t1 t2, +, µ1 µ2, , 12, 8, +, 1.5 4 / 3, , d=8+6, = 14 cm, 42. The critical angle for a light travelling from medium A into medium B is . The speed of light in medium A is, v, the speed of light in medium B is, (1), , v, cos , , (2), , v sin , , (3), , v, sin , , (4), , v cos , , Sol. Answer (3), , sin θ =, , μB, μA, , A sin = B, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 263 :
Solution of Assignment, , µA =, , C, v, , Ray Optics and Optical Instruments, , µB =, , 41, , C, vB, , C, C, sin θ =, v, vB, , , , vB =, , v, sin θ, , 43. A small bulb is placed at a depth of 2 7 m in water and floating opaque disc is placed over the bulb so that, the bulb is not visible from the surface. The minimum diameter of the disc is (water = 4/3), (1) 42 m, , (2), , 6m, , (3), , 2 7 m, , (4), , 12 m, , Sol. Answer (4), , r, , θmax = sin−1, , 1, 3, = sin−1, μ, 4, , 27, sin =, , tan =, , r, 2 7, , =, , 3, 4, , , , , 3, 7, 3, 7, , r=6m, 44. Two optical media of refractive indices 1 and 2 contain x and y number of waves in the same thickness. Their, relative refractive index, , (1) xy, , 2, is equal to, 1, y, x, , (2), , (3), , x, y, , (4), , yx, x, , Sol. Answer (2), let wavelength of light in air is , Wavelength in medium =, t=, , λ, μ, , xλ, yλ, and t =, μ1, μ2, , x µ1, =, y µ2, , , µ2 y, =, µ1 x, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 264 :
42, , Ray Optics and Optical Instruments, , Solution of Assignment, , 45. A light ray is travelling through a ring of an optical fibre which is made of four different glasses (shown below), but each part has the same geometrical thickness. Their respective refractive indices are shown. The light ray, will take the maximum time in crossing the part, , Part-I, , 1.51 1.52, , Part-II, , 1.54 1.53, Part-IV, (1) I, , (2), , II, , Part-III, (3), , IV, , (4), , Same in all, , Sol. Answer (3), R.I. of part IV is highest. So velocity will be lowest. Hence, maximum time is taken in part IV., 46. Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive, index of the material of the rod the light once entered into it will not leave it through its lateral face whatsoever, be the value of angle of incidence?, (1) n = 1.1, , (2), , n=1, , (3), , n 2, , (4), , n = 1.3, , Sol. Answer (3), n2 − 1, , sin <, , , , sin2 θ + 1, Maximum value of = 90º., n>, , n, , n> 2, , 47. If the refractive index of a material of equilateral prism is, is, (1) 60°, , (2), , 45°, , (3), , 3 , then angle of minimum deviation of the prism, 30°, , (4), , 75°, , Sol. Answer (1), 3 ; A = 60º; m = ?, , R.I. =, , sin ( A + δ m ), 2, μ=, ⎛ A⎞, sin ⎜ ⎟, ⎝2⎠, sin(60º + δm ), 2, 3=, sin30º, , sin−1, , 3 60 + δm, =, 2, 2, , 60º = 30º +, 30º =, , δm, 2, , δm, 2, , m = 60º, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 265 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 43, , 48. A beam of light composed of red and green ray is incident obliquely at a point on the face of rectangular glass, slab. When coming out on the opposite parallel face, the red and green ray emerge from, (1) Two points propagating in two different non parallel directions, (2) Two points propagating in two parallel directions, (3) One point propagating in two different directions, (4) One point propagating in the same directions, Sol. Answer (2), Since parallel beams of light will regain their original direction. They will again become parallel after emergence., 49. The refractive index of the material of a prism is 2 and its refracting angle is 30°. One of the refracting, surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the, other face will retrace its path after reflection from the mirrored surface if its angle of incidence on the prism, is, (1) 45°, , (2), , 60°, , (3), , Zero, , (4), , 30°, , Sol. Answer (1), By geometry r = 90º – 60º = 30º, 30°, , sin i, 3=, sin30º, , , i, , r, , 1, , sin i =, , 2, , i = 45º, 50. Four lenses of focal length ±15 cm and ±150 cm are available for making a telescope. To produce the largest, magnification, the focal length of the eyepiece should be, (1) +15 cm, , (2), , +50 cm, , (3), , –150 cm, , (4), , –15 cm, , Sol. Answer (1), 51. A lens is placed between a source of light and a wall. It forms images of area A1 and A2 on the wall, for its, two different positions. The area of the source of light is, , (1), , A1 A2, 2, , (2), , 1, 1, , A1 A2, , (3), , A1 A2, , (4), , A1 A2, 2, , Sol. Answer (3), This is application of the displacement method for finding focal length., Here m1m2 = 1, Let A0 be area of object, , A1 A2, ×, =1, A0 A0, A0 =, , A1A2, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 266 :
44, , Ray Optics and Optical Instruments, , Solution of Assignment, , 52. If fV and fR are the focal lengths of a convex lens for violet and red light respectively and FV and FR are the, focal lengths of a concave lens for violet and red light respectively, then we must have, (1) fV > fR and FV > FR, , (2), , fV < fR and FV > FR, , (3), , fV > fR and FV < FR (4), , fV < fR and FV < FR, , Sol. Answer (2), v > r, A converging lens with higher refractive index will converge rays more hence value of fv < fr, Same is true for concave lenses but since values for focal length are taken as negative Fv > Fr, 53. If a convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together, what, will be their resulting power?, (1) + 7.5 D, , (2), , – 0.75 D, , (3), , + 6.5 D, , (4), , – 6.5 D, , Sol. Answer (2), P = P1 + P2, =, , 1, 1, −, 0.8 0.5, , =, , 5, −2, 4, , = –0.75 D, 54. The focal length of a converging lens is measured for violet, green and red colours. It is respectively fv, fg, fr., We will get, (1) fv < fr, , (2), , fg > fr, , (3), , fv = fg, , (4), , fg = fr, , Sol. Answer (1), v > g > r, A converging lens with greater refractive index will bend rays more converging them closer., fv < fg < fr, 55. A luminous object is placed at a distance of 30 cm from the convex lens of focal length 20 cm. On the other, side of the lens, at what distance from the lens a convex mirror of radius of curvature 10 cm be placed in order, to have an upright image of the object coincident with it?, (1), , 50 cm, , (2), , 30 cm, , (3), , 12 cm, , Sol. Answer (1), , 60 cm, , 20 cm, , For convex lens, , 1, 1, 1, = −, +20 v −30, , (4), , O, , C, d = 60 – 10, d = 50 cm, , v = 60 cm, Virtual object for convex mirror should be at its C., , 30 cm, , 10 cm, , d, 60 cm, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 267 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 45, , 56. A plano-convex lens is made of refractive index 1.6. The radius of curvature of the curved surface is 60 cm., The focal length of the lens is, (1) 200 cm, , (2), , 100 cm, , (3), , 50 cm, , (4), , 400 cm, , Sol. Answer (2), , 1, ⎛ 1 1⎞, = (μ − 1) ⎜ − ⎟, ⎝R ∞⎠, f, 1 0.6, =, f, 60, , f = 100 cm, 57. A planoconvex lens (m = 1.5) has radius of curvature 10 cm. It is silvered on its plane surface. Find focal length, after silvering, (1) 10 cm, , (2), , 20 cm, , (3), , 15 cm, , (4), , 25 cm, , Sol. Answer (1), = 1.5, R = 10 cm, P = 2PL + Pm, Pm = Power of mirror = 0, P = 2PL, 1⎞, ⎛ 1, PL = (1.5 − 1) ⎜, +, ⎝ 10 ∞ ⎟⎠, =, , PL =, , 0.5, 10, , 1, cm, 20, , PL = 5 D, P = 2PL = 10 D, , f =, , 1, m = 10 cm, 10, , 58. A bubble in glass slab (m = 1.5) when viewed from one side appears at 5 cm and 2 cm from other side, then, thickness of slab is, (1) 3.75 cm, , (2), , 3 cm, , (3), , 10.5 cm, , (4), , 2.5 cm, , Sol. Answer (3), = 1.5, , µ=, , Real depth, Apparent depth, , Real depth = Apparent depth 1, , t1, t, t2, , t = t1 + t2, t = 5 × 1.5 + 2 × 1.5, = 10.5 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 268 :
46, , Ray Optics and Optical Instruments, , Solution of Assignment, , 59. A bulb is located on a wall. Its image is to be obtained on a parallel wall with the help of a convex lens. If, the distance between the two walls is d, then required focal length will be, (1) Only, , d, 4, , (3) More than, , d, d, but less than, 4, 2, , d, 2, , (2), , Only, , (4), , Less than or equal to, , d, 4, , Sol. Answer (4), f, , d, 4, , 60. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its, focal length will, (1) Become zero, , (2), , Become infinite, , (3) Become small, but non-zero, , (4), , Remain unchanged, , Sol. Answer (2), When this happens the lens will behave as a glass slab and parallel rays will converge at infinity., f=, 61. An equiconvex lens is cut into two halves along (i) XOX and (ii) YOY as shown in the figure. Let f, f , f , be the focal lengths of the complete lens of each half in case (i), and of each half in case (ii), respectively., Choose the correct statement from the following, Y, , X, , X, , O, , Y, , (1) f = f, f = 2f, , (2), , f = 2f, f = f, , (3), , f = f, f = f, , f = 2f, f = 2f, , (4), , Sol. Answer (1), f = f as radius of curvature of both surfaces is same., Y, , f = 2f, by lens maker's formula :, 1, ⎛ 1 1⎞, = (μ − 1) ⎜ − ⎟, ⎝R ∞⎠, f '', , ....(i), , 1, ⎛ 1 1⎞, = (μ − 1) ⎜ + ⎟, ⎝R R ⎠, f, , ....(ii), , f, 1, =, f '' 2, , [Dividing (i) and (ii)], , X, , O, , X, , Y, , f = 2f, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 269 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 47, , 62. Which of the following is incorrect?, (1) A thin convex lens of focal length f 1 is placed in contact with a thin concave lens of focal, length f2. The combination will act as convex lens if f1 < f2, (2) Light on reflection at water-glass boundary will undergo a phase change of , (3) Spherical aberration is minimized by achromatic lens, (4) If the image of distant object is formed in front of the retina then defect of vision may be myopia, Sol. Answer (3), Achromatic lenses minimize chromatic aberration., 63. A double concave thin lens made out of glass ( = 1.5) have radii of curvature 500 cm. This lens is used to, rectify the defect in vision of a person. The far point of the person will be at, (1) 5 m, , (2), , 2.5 m, , (3), , 1.25 m, , (4), , 1m, , Sol. Answer (1), , 1 1 1, = −, f v ∞, , ⎛ 2 ⎞ 1, (1.5 − 1) ⎜ −, =, ⎝ 500 ⎟⎠ v, v = – 500 cm, v=–5m, 64. A convex lens forms a real image 16 mm long on a screen. If the lens is shifted to a new position without, disturbing the object or the screen then again a real image of length 81 mm is formed. The length of the object, must be, (1) 48.5 mm, , (2), , 36 mm, , (3), , 6 mm, , (4), , 72 mm, , Sol. Answer (2), This is the application of the displacement method., Size of object (m0) =, , , m1m2, , m0 = 16 × 81, =4×9, = 36 cm, , 65. A point object is moving with speed u0 at a position somewhere between 2F and F in front of a convex lens., The speed of its image is, , u0, 2F, , (1) > u0, , (2), , < u0, , F, , F, , 2F, , (3), , = u0, , (4), , May be (1) or (2), , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 270 :
48, , Ray Optics and Optical Instruments, , Solution of Assignment, , Sol. Answer (1), , 1 1 1, − =, v u f, Differentiating w.r.t. t, −, , −, , 1 dv ⎛ 1 ⎞ du, − ⎜− ⎟, =0, v 2 dt ⎝ u 2 ⎠ dt, 1, v, , 2, , vi +, , 1, u2, , v0 = 0, , 2, vi = v v0, u2, When f < u < 2f, v lies beyond 2f., , v > 2f, v2, u2, , >1, , vi > v0., 66. The minimum magnifying power of a telescope is M. If focal length of its eye lens is halved, the magnifying, power will become, (1), , M, 2, , (2), , 2M, , (3), , 3M, , (4), , 4M, , Sol. Answer (2), f0, M =f, e, , 1, M f, e, fe =, , fe, 2, , M = 2M, 67. An object is placed in front of two convex lenses one by one at a distance u from the lens. The focal lengths, of the lenses are 30 cm and 15 cm respectively. If the size of image formed in the two cases is same, then, u is, (1) 15 cm, , (2), , 20 cm, , (3), , 25 cm, , (4), , 30 cm, , Sol. Answer (2), Magnification are same, m1 = – m2, , f ⎞, ⎛, ⎜⎝ m =, ⎟, f +u⎠, , 30, 15, =−, 30 − u, 15 − u, , u = 20 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 271 :
Solution of Assignment, , Ray Optics and Optical Instruments, , 49, , 68. If R1 and R2 are the radii of curvature of the spherical surfaces of a thin lens and R1 > R2, then this lens can, , R2, R1, (1) Correct myopia, , (2), , Correct hypermetropia, , (3) Correct presbiopia, , (4), , Correct astigmatism, , Sol. Answer (1), f < 0 (Diverging), 3, 69. The focal length of a thin lens in vacuum is f. If the material of the lens has , its focal length when, 2, 4, will be, immersed in water of refractive index, 3, , (1) f, , (2), , 4, f, 3, , (3), , 2f, , (4), , 4f, , Sol. Answer (4), 1 ⎞, 1, ⎛3 ⎞ ⎛ 1, −, ⎜⎝ − 1⎟⎠ ⎜, =, 2, fv, ⎝ R1 R2 ⎟⎠, , 1⎛ 1, 1 ⎞, 1, = 2 ⎜R − R ⎟, ⎝ 1, f, 2⎠, ⎛ μg, ⎞ ⎛ 1, 1, 1 ⎞, =, ⎜ μ − 1⎟ ⎜⎝ R − R ⎟⎠, fw, ⎝ w, ⎠, 1, 2, , 1, ⎛9 ⎞ ⎛2⎞, =, ⎜⎝ − 1⎟⎠ ⎜⎝ ⎟⎠, fw, 8, f, fw = 4f, 70. For a telescope having fo as the focal length of the objective and fe as the focal length of the eyepiece, the, length of the telescope tube is, (1) fe, , (2), , fo – fe, , (3), , fo, , (4), , fo + fe, , Sol. Answer (4), By the construction of the telescope f0 + fe = L, 71. An astronomical telescope of tenfold angular magnification has a length of 44 cm. The focal length of the, objective is, (1) 44 cm, , (2), , 440 cm, , (3), , 4 cm, , (4), , 40 cm, , Sol. Answer (4), m = 10, f0, = 10, fe, , ....(i), , f0 + fe = 44 cm, , ....(ii), , Solving (i) and (ii), f0 = 40 cm, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 272 :
50, , Ray Optics and Optical Instruments, , Solution of Assignment, , 72. Ray optics is valid, when characteristic dimensions are, (1) Much smaller than the wavelength of light, , (2), , Of the same order as the wavelength of light, , (3) Of the order of one millimeter, , (4), , Much larger than the wavelength of light, , Sol. Answer (4), Ray optics is valid on a more macro scale compared to wavelength of light. On micro scale wave optics and, wave-particle duality is more prominent., 73. The blue colour of the sky is due to the phenomenon of, (1) Scattering, , (2), , Dispersion, , (3), , Reflection, , (4), , Refraction, , Sol. Answer (1), Blue colour of the sky is due to scattering of light also called the Rayleigh scattering., 74. Rainbow is formed due to, (1) Scattering and refraction, , (2), , Total internal reflection and dispersion, , (3) Reflection only, , (4), , Refraction and dispersion, , Sol. Answer (4), Rainbow's colours appear due to dispersion, which occurs after refraction., , SECTION - D, Assertion - Reason Type Questions, 1., , A : Plane and convex mirrors can produce real images of objects., R : A plane or convex mirror can produce a real image if the object is virtual., , Sol. Answer (1), Real images are formed from real intersection of light rays, which occur in convex mirror when the object is, virtual., 2., , A : A virtual image cannot be caught on a screen but it can be photographed., R : The virtual image here serves as an object for the lens of the camera to produce a real image., , Sol. Answer (1), Virtual image can be observed but real intersection of light rays needs to occur to take objects to screen., 3., , A : When a diver under water looks obliquely at a fisherman standing on the bank of a lake then fisherman, looks taller., R : When a ray of light travelling in air enters water, it bends towards the normal., , Sol. Answer (1), The ray of light bending towards normal makes it appear that light comes from further away., 4., , A : The apparent depth of a tank of water decreases if viewed obliquely., R : Real depth decreases if viewed obliquely., , Sol. Answer (3), Apparent depth of tank decreases because the light ray, travels further through the water bending more in the, process. Real depth remains the same are it was., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 273 :
Solution of Assignment, , 5., , Ray Optics and Optical Instruments, , 51, , A : Proper cutting of diamond makes it sparkle., R : Diamond has very large refractive index., , Sol. Answer (1), Proper cutting of diamond makes it sprakle due to total internal refraction. A high reflective index makes total, internal reflection easier., 6., , A : The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual, image produced by a magnifying glass so the magnification produced is one., R : During image formation through magnifying glass, the object as well as its image are at the same position., , Sol. Answer (4), A magnifying glass magnifies the virtual image and magnification is greater than one. The reason is also false., 7., , A : In viewing through a magnifying glass, angular magnification decreases if the eye is moved back., R : Angle subtended at the eye becomes slightly less than the angle subtended at the lens., , Sol. Answer (1), 8., , A : Magnifying power of a simple microscope cannot be increased beyond a limit., R : Magnifying power is inversely proportional to focal length and there are some practical difficulty of grinding,, aberrations etc. because of which focal length cannot be decreased below a limit., , Sol. Answer (1), There is a limit to how curved a glass piece can be made practically., 9., , A : The objective and the eye-piece of a compound microscope should have short focal lengths., R : Magnifying power of a compound microscope is inversely proportional to the focal lengths of both the lenses., , Sol. Answer (1), , LD, For normal adjustment : M = f ⋅ f ., o e, 10. A : When viewing through a compound microscope, our eyes should be positioned not on the eye piece but, a short distance away from it for best viewing., R : The image of the objective in the eye-piece is known as 'eye-ring' and if we position our eyes on the eyering and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect, all the light refracted by the objective., Sol. Answer (1), 11. A : The peculiar fish Anableps anableps swims with its eyes partially extending above the water surface so, that it can see simultaneously above and below water., R : This fish has egg shaped eye lens and two retina., Sol. Answer (1), 12. A : A virtual image cannot be produced on a screen., R : The light energy does not meet at the point(s) where virtual image is formed., Sol. Answer (1), To take a sharp image on a screen light rays must intersect really. This does not happen in virtual images., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Page 274 :
52, , Ray Optics and Optical Instruments, , Solution of Assignment, , 13. A : Lenses of large aperture suffer from spherical aberration., R : The curvature of the lens at central and peripheral regions is different., Sol. Answer (3), 14. A : The image formed by a concave lens is always virtual., R : The rays emerging from a concave lens never meet on the principal axis., Sol. Answer (4), Assertion is false as real image is formed by concave lens in care of virtual object. Reason is also false as, real images are formed by real intersection of rays., 15. A : When two lenses in contact form an achromatic doublet, then the materials of the two lenses are always, different., R : The dispersive powers of the materials of the two lenses are of opposite sign., Sol. Answer (3), To achieve achromatic condition in two lenses in contact, refractive indices must be different, dispersive power, cannot be negative., 16. A : A reflecting type of telescope is preferred over refracting type in astronomy., R : A reflecting type of telescope is free from chromatic aberration and spherical aberration., Sol. Answer (1), , , , , , , , https://t.me/NEET_StudyMaterial, , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
