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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , CBSE NCERT Solutions for Class 12 Physics Chapter 9, Back of Chapter Questions, 9.1., , A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of, radius of curvature 36 cm. At what distance from the mirror should a screen be, placed in order to obtain a sharp image? Describe the nature and size of the image., If the candle is moved closer to the mirror, how would the screen have to be moved?, Solution:, , Given,, The Object distance π’π’ = β27 cm, π
π
, , Focal length ππ = β 2 = β, From the mirror formula,, , 36 cm, 2, , = β18 cm, , 1 1 1, + =, π£π£ π’π’ ππ, β π£π£ =, β π£π£ =, , π’π’π’π’, π’π’ β ππ, , (β27)(β18), = β54 cm, (β27) β (β18), π£π£, , β54, , Magnification ππ = β π’π’ = β β27 = β2, , β΄ image height = ππ Γ height of the oject = β2 Γ 2.5 = β5 cm, , The screen must be placed at a distance of 54 cm from the mirror on the same side, as the object. The image formed is inverted and twice the size of the object., , As the candle is moved close to the mirror, object approaches the focal point. Hence, image will approach infinity. Therefore, the screen has to move away from the, mirror to get the image on it., , Practice more on Ray Optics and Optical Instruments, , Page - 1, , www.embibe.com
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Class- XII-CBSE-Physics, , 9.2., , Ray Optics and Optical Instruments, , A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm., Give the location of the image and the magnification. Describe what happens as the, needle is moved farther from the mirror., Solution:, , Given,, The Object distance π’π’ = β12 cm, Focal length ππ = +15 cm, From the mirror formula,, 1 1 1, + =, π£π£ π’π’ ππ, β π£π£ =, , (β12 ) Γ 15, π’π’π’π’, =, = 6.7 cm, π’π’ β ππ (β12 ) β 15, π£π£, , 6.7, , Magnification ππ = β π’π’ = β β12 = 0.56, , The image will be formed at a distance of 6.7 cm right to the mirror. Magnification, produced is 0.56., , As the needle moved away from the object image approaches the focal point, or, image moves away from the mirror. Size of the image decreases continuously as, the object moves away from the mirror., 9.3., , A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle, lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is, the refractive index of water? If water is replaced by a liquid of refractive index, 1.63 up to the same height, by what distance would the microscope have to be, moved to focus on the needle again?, Solution:, Given,, the apparent depth of the needle ππππ = 9.4 cm, , Height of the water or real depth of the object ππππ = 12.5 cm, , Practice more on Ray Optics and Optical Instruments, , Page - 2, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, ππ, , From the relation ππππ =, β, , 9.4, 1, =, 12.5 πππ€π€, , ππ, , 1, , 1 ππ2, , β πππ€π€ = 1.33, , β΄ the refractive index of the water is 1.33, , If water is replaced by a liquid of refractive index 1.63,, , ππππ, 1, 12.5, =, β ππππ =, = 7.7 cm, 1.63, ππππ, 1 ππ2, 9.4., , The microscope must be moved by a distance of 9.4 cm β 7.7 cm = 1.7 cm to, focus the object again., , Figures 9.34(a) and (b) show refraction of a ray in air incident at 60Β° with the, normal to a glass-air and water-air interface, respectively. Predict the angle of, refraction in glass when the angle of incidence in water is 45Β° with the normal to a, water-glass interface [Fig. 9.34(c)]., , Solution:, From the figure 9.34(a),, ππππ sin 60Β° = ππππ sin 35Β°, β ππππππ =, , sin 60Β°, sin 35Β°, , β ππππππ = 1.51, , From figure 9.34(b),, ππππ sin 60Β° = πππ€π€ sin 47Β°, Practice more on Ray Optics and Optical Instruments, , Page - 3, , www.embibe.com
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Class- XII-CBSE-Physics, , β πππ€π€π€π€ =, , Ray Optics and Optical Instruments, , sin 60Β°, sin 47Β°, , β πππ€π€π€π€ = 1.18, , β ππππππ =, , 1.51, = 1.27, 1.18, , From figure 9.34(c),, , πππ€π€ sin 45Β° = ππππ sin ππ, , sin 45Β° = ππππππ sin ππ, 9.5., , β ππ β 34Β°, , A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm., What is the area of the surface of water through which light from the bulb can, emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point, source.), Solution:, , Light comes out from a circle on the surface of the water, known as the circle of, illuminance, The light ray from the source incident with a critical angle on the border of this, circle., From Snellβs law,, πππ€π€ sin ππππ = ππππ sin 90Β°, , β 1.33 Γ, , π
π
, , βπ
π
2 + 802, , =1, , β π
π
= 91 cm = 91 m, Area= 2.6 m2, , Practice more on Ray Optics and Optical Instruments, , Page - 4, , www.embibe.com
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Class- XII-CBSE-Physics, , 9.6., , Ray Optics and Optical Instruments, , A prism is made of glass of unknown refractive index. A parallel beam of light is, incident on a face of the prism. The angle of minimum deviation is measured to be, 40Β°. What is the refractive index of the material of the prism? The refracting angle, of the prism is 60Β°. If the prism is placed in water (refractive index 1.33), predict, the new angle of minimum deviation of a parallel beam of light., Solution:, , The angle of minimum deviation is related to the relative refractive index of the, prism as, ππππ,ππππππππππππ =, , π΄π΄ + πΏπΏππ, 2 οΏ½, π΄π΄, sin οΏ½ 2 οΏ½, , sin οΏ½, , If the prism is placed in the air, ππππ,ππ =, , 60Β°+40Β°, οΏ½, 2, 60Β°, sinοΏ½ οΏ½, 2, , sinοΏ½, , β¦β¦β¦β¦β¦β¦(i), , ππππ,ππ β 1.53, , If the prism is placed in water, β ππππ,π€π€, ππππ,ππ, , 60Β° + πΏπΏππ, οΏ½, ππππ,ππ sin οΏ½, 2, =, =, 60Β°, πππ€π€,ππ, sin οΏ½ 2 οΏ½, , β 1.33 =, , 60Β°+πΏπΏππ, οΏ½, 2, 60Β°, sinοΏ½ οΏ½, 2, , sinοΏ½, , β¦β¦β¦β¦β¦β¦β¦ (ii), , From (i) and (ii), , 60Β° + 40Β°, οΏ½, 2, 1.33 =, 60Β° + πΏπΏππ, sin οΏ½, οΏ½, 2, sin οΏ½, , sin οΏ½, , sin 50Β°, 60Β° + πΏπΏππ, οΏ½=, = 0.56, 2, 1.33, , β πΏπΏππ β 10Β°, , 9.7., , β΄ Refractive index of the prism β 1.53 and minimum deviation when placed in, water is 10Β°, , Double-convex lenses are to be manufactured from a glass of refractive index 1.55,, with both faces of the same radius of curvature. What is the radius of curvature, required if the focal length is to be 20 cm?, , Solution:, , Practice more on Ray Optics and Optical Instruments, , Page - 5, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , From the lens makers formula, 1, 1, 1, = οΏ½ππππ,ππππππππππππ β 1οΏ½ οΏ½ β οΏ½, ππ, π
π
1 π
π
2, , For convex lens,, , π
π
1 = π
π
and π
π
2 = βπ
π
, β, , 1, 1, 1, = (1.55 β 1) οΏ½ β, οΏ½, 20, π
π
βπ
π
, , β π
π
= 22 cm, , 9.8., , β΄ The radius of curvature required is 22 cm., , A beam of light converges at a point P. Now a lens is placed in the path of the, convergent beam 12 cm from P. At what point does the beam converge if the lens, is, (a), , (b), , a convex lens of focal length 20 cm, and, a concave lens of focal length 16 cm?, , Solution:, (a), , Here the object is virtual, From the diagram, Object distance π’π’ = 12 cm, , Focal length ππ = 20 cm, , From the thin lens formula,, 1 1 1, β =, π£π£ π’π’ ππ, Practice more on Ray Optics and Optical Instruments, , Page - 6, , www.embibe.com
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Class- XII-CBSE-Physics, , β π£π£ =, , Ray Optics and Optical Instruments, , π’π’π’π’, 12 Γ 20, =, = 7.5 cm, π’π’ + ππ 12 + 20, , The image formed is real and located at a distance of 7.5 cm right side from the, lens, (b), , Here the object is virtual, From the diagram, Object distance π’π’ = 12 cm, , Focal length ππ = β16 cm, , From the thin lens formula,, 1 1 1, β =, π£π£ π’π’ ππ, β π£π£ =, 9.9., , π’π’π’π’, 12 Γ (β16), =, = 48 cm, π’π’ + ππ, 12 β 16, , The image formed is real and located at a distance of 48 cm right side from the lens, An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length, 21 cm. Describe the image produced by the lens. What happens if the object is, moved further away from the lens?, Solution:, , Practice more on Ray Optics and Optical Instruments, , Page - 7, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , Given,, Object distance π’π’ = β14 cm, Focal length ππ = β21 cm, , From the thin lens formula,, 1 1 1, β =, π£π£ π’π’ ππ, β π£π£ =, , (β14) Γ (β21), π’π’π’π’, =, = β8.4 cm, π’π’ + ππ, β14 β 21, π£π£, , Magnification ππ = π’π’ =, , β8.4, β14, , = 0.6, , β size of the image = 0.6 Γ 3.0 = 1.8 cm, , Image is formed at a distance 8.4 cm left to the lens. The image formed is virtual,, erect, diminished to the size 1.8 cm., 9.10., , As the object moves away from the lens, the image moves towards the focal point, (π’π’ β β β π£π£ β ππ) and the magnification approaches zero (ππ β 0), , What is the focal length of a convex lens of focal length 30 cm in contact with a, concave lens of focal length 20 cm? Is the system a converging or a diverging lens?, Ignore thickness of the lenses., , Solution:, Given, , The focal length of the convex lens ππ1 = 30 cm, , The focal length of the concave lens ππ2 = β20 cm, Combined lens focal length ππππππ is related as, 1, 1 1, = +, ππππππ ππ1 ππ2, β, , 1, 1, 1, 1, =, +, =β, ππππππ 30 β20, 60, , β ππππππ = β60 cm, 9.11., , β΄ The combined lens acts as a diverging lens with a focal length of 60 cm., , A compound microscope consists of an objective lens of focal length 2.0 cm and, an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far, from the objective should an object be placed in order to obtain the final image at, (a), , the least distance of distinct vision (25cm), and, , Practice more on Ray Optics and Optical Instruments, , Page - 8, , www.embibe.com
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Class- XII-CBSE-Physics, , (b), , Ray Optics and Optical Instruments, , at infinity? What is the magnifying power of the microscope in each case?, , Solution:, Given,, The focal length of the objective lens ππππ = 2 cm, The focal length of the eyepiece ππππ = 6.25 cm, (a), , To form the final image at 25 cm from the eyepiece, π£π£ππ = β25 cm, From thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , 1, 1, 1, β, =, β25 π’π’ππ 6.25, β π’π’ππ = β5 cm, , Image distance to the objective π£π£ππ = 15 β 5 = 10 cm, From thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , 1, 1, 1, β, =, 10 π’π’ππ 2, , π’π’ππ = β2.5 cm, , Hence object must be placed at a distance of 2.5 cm from the objective, The magnifying power of the compound microscope, , (b), , ππ =, , π£π£ππ, π·π·, 10, 25, οΏ½1 + οΏ½ =, οΏ½1 +, οΏ½ = 20, π’π’ππ, ππππ, 2.5, 6.25, , To form the final image at β, π£π£ππ = β, From thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , 1, 1, 1, β, =, β π’π’ππ 6.25, , β π’π’ππ = β6.25 cm, , Image distance to the objective π£π£ππ = 15 β 6.25 = 8.75 cm, Practice more on Ray Optics and Optical Instruments, , Page - 9, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , From thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , 1, 1, 1, β, =, 8.75 π’π’ππ 2, , π’π’ππ = β2.59 cm, , Hence object must be placed at a distance of 2.59 cm from the objective, π£π£, , π·π·, , The magnifying power of the compound microscope ππ = π’π’ππ οΏ½ππ οΏ½ =, , 9.12., , 8.75, , οΏ½, , 25, , οΏ½ = 13.5, , 2.59 6.25, , ππ, , ππ, , A person with a normal near point (25 cm) using a compound microscope with, objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring, an object placed at 9.0 mm from the objective in sharp focus. What is the separation, between the two lenses? Calculate the magnifying power of the microscope., Solution:, Given,, The focal length of the objective lens ππππ = 8 mm = 0.8 cm, The focal length of the eyepiece ππππ = 2.5 cm, , Object distance from the objective π’π’ππ = β9 mm = β0.9 cm, Let the image formed by the objective is at a distance π£π£ππ, From thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , 1, 1, 1, β, =, π£π£ππ β0.9 0.8, β π£π£ππ = 7.2 cm, , Image distance for the eyepiece π£π£ππ = β25 cm, From thin lens formula, 1, 1, 1, β, =, β25 π’π’ππ 2.5, , β π’π’ππ = β2.27 cm, , β΄ the separation between the two lenses = π£π£ππ + |π’π’ππ | = 7.2 + 2.27 = 9.47 cm, Practice more on Ray Optics and Optical Instruments, , Page - 10, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, π£π£, , 9.13., , π·π·, , 7.2, , 25, , Magnifying power ππ = |π’π’ππ | οΏ½1 + ππ οΏ½ = 0.9 οΏ½1 + 2.5οΏ½ = 88, ππ, , ππ, , A small telescope has an objective lens of focal length 144 cm and an eyepiece of, focal length 6.0 cm. What is the magnifying power of the telescope? What is the, separation between the objective and the eyepiece?, Solution:, Given,, The focal length of the objective ππππ = 144 cm, , The focal length of the eyepiece ππππ = 6.0 cm, , ππ, , The magnifying power of the telescope ππ = ππππ =, 9.14., , ππ, , 144, 6.0, , = 24, , Length of the telescope πΏπΏ = ππππ + ππππ = 144 + 6.0 = 150 cm, , (a), , (b), , A giant refracting telescope at an observatory has an objective lens of focal, length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the, angular magnification of the telescope?, If this telescope is used to view the moon, what is the diameter of the image, of the moon formed by the objective lens? The diameter of the moon is, 3.48 Γ 106 m, and the radius of lunar orbit is 3.8 Γ 108 m., , Solution:, (a), , Given,, The focal length of the objective ππππ = 15 m = 1500 cm, The focal length of the eyepiece ππππ = 1.0 cm, , (b), , Angular magnification ππ =, Given,, , ππππ, ππππ, , =, , 1500, 1, , = 1500, , The diameter of the moon π·π· = 3.48 Γ 106 m, , The radius of the lunar orbit π
π
= 3.8 Γ 108 m, , The angle subtended by the moonβs diameter is equal to the angle subtended, by its image diameter at the objective lens, , Practice more on Ray Optics and Optical Instruments, , Page - 11, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , π·π· ππ, =, π
π
ππππ, , 9.15., , 3.48 Γ 106, Γ 15 m = 13.7 cm, β ππ =, 3.8 Γ 108, , The diameter of the image formed ππ = 13.7 cm, , Use the mirror equation to deduce that:, (a), (b), , an object placed between ππ and 2ππ of a concave mirror produces a real, image beyond 2ππ., , a convex mirror always produces a virtual image independent of the location, of the object., , (c), , the virtual image produced by a convex mirror is always diminished in size, and is located between the focus and the pole., , (d), , an object placed between the pole and focus of a concave mirror produces, a virtual and enlarged image. [Note: This exercise helps you deduce, algebraically properties of images that one obtains from explicit ray, diagrams.], , Solution:, (a), , Given that the object is placed between ππ and 2ππ β 2ππ < π’π’ < ππ and π’π’ <, 0, ππ < 0, From the mirror formula,, 1 1 1, + =, π£π£ π’π’ ππ, β π£π£ =, , π’π’π’π’, π’π’ β ππ, , If π’π’ = βππ, , Practice more on Ray Optics and Optical Instruments, , Page - 12, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , 1 1 1, + =, π£π£ π’π’ ππ, β, , 1 1 1, = β >0, π£π£ ππ π’π’, , Or π£π£ > 0, , π£π£, , ππ, , Magnification ππ = β π’π’ = ππβπ’π’ > 1, , Hence, an object placed between the focus and pole of a concave mirror, produces a virtual and enlarged image., 9.16., , A small pin fixed on a table top is viewed from above from a distance of 50 cm., By what distance would the pin appear to be raised if it is viewed from the same, point through a 15 cm thick glass slab held parallel to the table? Refractive index, of glass = 1.5. Does the answer depend on the location of the slab?, Solution:, Given,, The thickness of the glass slab π‘π‘ = 15 cm, Refractive index of the glass ππ = 1.5, , 1, , Shift in the object position due to the slab π π = π‘π‘ οΏ½1 β πποΏ½, β π π = 15 οΏ½1 β, , 1, οΏ½ = 5 cm, 1.5, , The shift is independent of the location of the slab for a small angle of incidence., 9.17., , (a), , Figure 9.35 shows a cross-section of a βlight pipeβ made of a glass fibre of, refractive index 1.68. The outer covering of the pipe is made of a material, of refractive index 1.44. What is the range of the angles of the incident rays, with the axis of the pipe for which total reflections inside the pipe take place,, as shown in the figure., , (b), , What is the answer if there is no outer covering of the pipe?, , Practice more on Ray Optics and Optical Instruments, , Page - 14, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , Solution:, (a), , For total internal reflection, near the wall ππ β² > ππππ, , sin ππππ =, , 1, 1.44, =, = 0.86, ππ21 1.68, , ππππ = 59Β°, , β ππ β² > 59Β°, β ππ < 31Β°, , From Snellβs law, 1 Γ sin ππ = 1.68 Γ sin ππ, , As ππ < 31Β° β sin ππ < 1.68 Γ sin 31Β°, ππ < 60Β°, , (b), , Therefore, the range of angle of incidence with the axis is 0 < ππ < 60Β°, , If there is no outer covering,, , For total internal reflection, near the wall ππ β² > ππππ, , sin ππππ =, , 1, 1, =, = 0.60, ππ21 1.68, , ππππ = 36.5Β°, , β ππ β² > 36.5Β°, β ππ < 53.5Β°, , From Snellβs law, for the angle of incidence of 90Β°, , 1 Γ sin 90Β° = 1.68 Γ sin ππ, , β ππ = 36.5Β° andππ β² = 53.5Β° Which is greater than ππππ ., Practice more on Ray Optics and Optical Instruments, , Page - 15, , www.embibe.com
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Class- XII-CBSE-Physics, , 9.18., , Ray Optics and Optical Instruments, , This gives the upper limit as 90Β° and all the incident rays suffer total internal, reflection, , Answer the following questions:, (a), , You have learnt that plane and convex mirrors produce virtual images of, objects. Can they produce real images under some circumstances? Explain., , (b), , A virtual image, we always say, cannot be caught on a screen. Yet when we, βseeβ a virtual image, we are obviously bringing it on to the βscreenβ (i.e.,, the retina) of our eye. Is there a contradiction?, , (c), , A diver under water, looks obliquely at a fisherman standing on the bank of, a lake. Would the fisherman look taller or shorter to the diver than what he, actually is?, , (d), , Does the apparent depth of a tank of water change if viewed obliquely? If, so, does the apparent depth increase or decrease?, , (e), , The refractive index of diamond is much greater than that of ordinary glass., Is this fact of some use to a diamond cutter?, , Solution:, (a), , The plane mirror always forms a real image for a virtual object, irrespective, of object position. The convex mirror forms a real image of a virtual object, if it is between pole and focus., , (b), , When we see the virtual image, it acts as an object for our eye lens. Eye lens, forms a real image of it on the retina. Hence there is no contradiction., , (c), , For the diver the fisherman appears taller than he actually is., , (d), , Apparent depth decreases if viewed obliquely. For the near normal view, apparent depth is maximum., , (e), , The diamond refractive index is about 2.42, much bigger than the regular, glass index (about 1.5). Diamond's critical angle is about 24Β°, much lower, than glass. A skilled diamond cutter takes advantage of the wider range of, incidence angles (in the diamond), 24Β° to 90Β°, to guarantee that light, , Practice more on Ray Optics and Optical Instruments, , Page - 16, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , entering the diamond is fully reflected from many faces before getting out, to produce a sparkling effect., 9.19., , The image of a small electric bulb fixed on the wall of a room is to be obtained on, the opposite wall 3 m away by means of a large convex lens. What is the maximum, possible focal length of the lens required for the purpose?, Solution:, , Give that, the separation between object and the image is 3 m, , Let the object distance from the lens be π₯π₯, this implies that the image distance is, 3 β π₯π₯. Let the required focal length be ππ., π’π’ = βπ₯π₯, , π£π£ = 3 β π₯π₯, ππ = ππ, , From the thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , β, , 1, 1, 1, β, =, 3 β π₯π₯ βπ₯π₯ ππ, , π₯π₯ 2, β ππ = π₯π₯ β, 3, , ππππ, , To find the maximum of ππ, ππππ = 0, ππππ, 2π₯π₯, = 1β, =0, ππππ, 3, 3, β π₯π₯ = m, 2, , β, , 9.20., , 3, , And ππ = 2 β, , 3 2, 2, , οΏ½ οΏ½, 3, , = 0.75 m, , A screen is placed 90 cm from an object. The image of the object on the screen is, formed by a convex lens at two different locations separated by 20 cm. Determine, the focal length of the lens., Solution:, , The separation between the object and the image π·π· = 90 cm, , The separation between two locations of the object ππ = 20 cm, , Let the object distance from the lens be π₯π₯, this implies that the Image distance is, π·π· β π₯π₯, , Practice more on Ray Optics and Optical Instruments, , Page - 17, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , π’π’ = βπ₯π₯, , π£π£ = π·π· β π₯π₯, ππ = ππ, , From the thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , β, , 1, 1, 1, β, =, π·π· β π₯π₯ βπ₯π₯ ππ, , β π₯π₯ 2 β π·π·π·π· + π·π·π·π· = 0, , π·π· Β± οΏ½π·π·2 β 4π·π·π·π·, β π₯π₯ =, 2, , βThe two positions of the object are,, , π·π·+οΏ½π·π·2 β4π·π·π·π·, 2, , and, , π·π·βοΏ½π·π· 2 β4π·π·π·π·, , The separation between them = οΏ½π·π·2 β 4π·π·π·π· = ππ, β ππ =, , 9.21., , 2, , π·π·2 β ππ 2 902 β 202, =, = 21.4 cm, 4π·π·, 4 Γ 90, , The focal length of the lens ππ = 21.4 cm, (a), , Determine the βeffective focal lengthβ of the combination of the two lenses, in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes, coincident. Does the answer depend on which side of the combination a, beam of parallel light is incident? Is the notion of effective focal length of, this system useful at all?, , (b), , An object 1.5 cm in size is placed on the side of the convex lens in the, arrangement (a) above. The distance between the object and the convex lens, is 40 cm. Determine the magnification produced by the two-lens system,, and the size of the image., , Solution:, (a), , The focal length of the convex lens = 30 cm, , The focal length of the concave lens = β20 cm, The separation between the lenses ππ = 8.0 cm, , Consider a parallel beam of light incident on the first lens, π’π’1 = β, ππ1 = ππ1, , Practice more on Ray Optics and Optical Instruments, , Page - 18, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , Hence the answer depends on the side of the combination on which a, beam of parallel light is incident. Therefore, the notion of the effective, focal length of this system not useful., (b), , Size of the object βππ = 1.5 cm, , Distance between the object and convex lens is 40 cm, , For the convex lens, π’π’ = β40 cm, ππ = 30 cm, , From the thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , π£π£ =, , ππ Γ π’π’ 30 Γ (β40), =, = 120 cm, ππ + π’π’, 30 β 40, , π£π£, , ππ, , π£π£, , ππ, , 30, , Magnification produced by convex lens ππ1 = π’π’ = ππ+π’π’ = 30β40 = β3, For the concave lens, , π’π’ = 120 β 8 = 112 cm, ππ = β20 cm, , β20, , 20, , Magnification produced by convex lens ππ1 = π’π’ = ππ+π’π’ = β20+112 = β 92, Net magnification produced ππ = 0.652, 9.22., , Size of the image βππ = 0.652 Γ 1.5 = 0.98 cm, , At what angle should a ray of light be incident on the face of a prism of refracting, angle 60Β° so that it just suffers total internal reflection at the other face? The, refractive index of the material of the prism is 1.524., Solution:, , Practice more on Ray Optics and Optical Instruments, , Page - 20, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , Given,, The refracting angle of prism π΄π΄ = 60, , Refractive index of the prism ππ = 1.524, , At the second surface, angle of incidence ππ2 must be greater than the critical angle, ππππ, 1, 1, ππππ = sinβ1 οΏ½ οΏ½ = sinβ1 οΏ½, οΏ½ = 41Β°, ππ, 1.524, , β΄ ππ2 > 41Β° β ππ1 < 19Β° (β΅ ππ1 + ππ2 = π΄π΄ = 60Β°), From Snellβs law,, , sin ππ = ππ sin ππ1 β sin ππ < 1.524 sin 19Β°, , β ππ < 30Β° (approximately), 9.23., , Therefore, the beam will be total internal reflected at the second surface if the angle, of incidence at the first surface is lesser than 30Β°, You are given prisms made of crown glass and flint glass with a wide variety of, angles. Suggest a combination of prisms which will, (a), , deviate a pencil of white light without much dispersion., , (b), , disperse (and displace) a pencil of white light without much deviation., , Solution:, , Practice more on Ray Optics and Optical Instruments, , Page - 21, , www.embibe.com
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Class- XII-CBSE-Physics, , 9.24., , Ray Optics and Optical Instruments, , (a), , Combine two prisms inverted as shown in figure without a gap between, them. Pencil of white light falls on the left side of the first prism which gets, dispersed and deviated. Dispersed light falls on the left side of the second, prism in which, it will recombine and pencil of white light comes out of the, second surface., , (b), , Combine two prisms inverted as shown in figure without a gap between, them. The angle of flint glass prism should increased so that the deviation, produced by both prisms is equal. The combination disperse the pencil of, white light without much deviation., , For a normal eye, the far point is at infinity and the near point of distinct vision is, about 25 cm in front of the eye. The cornea of the eye provides a converging power, of about 40 dioptres, and the least converging power of the eye-lens behind the, cornea is about 20 dioptres. From this rough data estimate the range of, accommodation (i.e., the range of converging power of the eye-lens) of a normal, eye., Solution:, Given,, The near point of distinct vision ππ = 25 cm, , Converging power of the cornea ππ1 = 40 diopters, , Least converging power of the eye-lens ππ2,ππππππ = 20 diopters, , Converging power of the eye-lens is least when the object is at infinity, Total power ππππππππ = 40 + 20 = 60 diopters, , Focal length ππ = ππ, , 1, , ππππππ, , 1, , 5, , Γ 100 cm = 60 Γ 100 cm = 3 cm, , Object distance π’π’ = β, , From the thin lens formula,, , Practice more on Ray Optics and Optical Instruments, , Page - 22, , www.embibe.com
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Class- XII-CBSE-Physics, , 1 1 1, β =, π£π£ π’π’ ππ, , β π£π£ = ππ =, , Ray Optics and Optical Instruments, , 5, cm, 3, , As the image is always formed on the retina, the distance between the lens and the, 5, retina = 3 cm, When the object is at the near point, π’π’ = β25 cm, π£π£ =, , 5, cm, 3, , Let the effective focal length be ππ, From the thin lens formula,, 1 1 1, β =, π£π£ π’π’ ππ, β, , 1 3 1, 16, = +, =, ππ 5 25 25, , 1, , Maximum converging power ππππππππ = ππ(ππππ cm) Γ 100 diopters = 64 diopters, Hence converging power of the eye lens = 64 β 40 = 24 diopters, 9.25., , β΄ Range of accommodation is 20 diopters to 24 diopters., , Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply, necessarily that the eye has partially lost its ability of accommodation? If not, what, might cause these defects of vision?, Solution:, It is not necessary that the eye with short-sightedness or long-sightedness imply a, partial loss in the ability of accommodation. Short-sightedness or myopia may, occur when the eye-balls get elongated from front to back and Hypermetropia may, occur when eye-ball get shortened. The defect of vision in which the eye has, partially lost its ability of accommodation is termed as presbyopia., , 9.26., , A myopic person has been using spectacles of power β 1.0 dioptre for distant, vision. During old age he also needs to use separate reading glass of power, + 2.0 dioptres. Explain what may have happened., , Solution:, Given, , Practice more on Ray Optics and Optical Instruments, , Page - 23, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , The power of spectacles of the person ππ1 = β1.0 diopters, 1, , The focal length of the lens used ππ1 = ππ Γ 100 = β100 cm, 1, , This implies that, he has a far point of 100 cm and normal near point of 25 cm, In his old age he needed separate reading glasses of power +2.0 diopters, 1, , βThe focal length of the lens ππ2 = 2 Γ 100 = 50 cm, , This implies that in his old age his near-point increased from 25 cm, π’π’ = β25 cm, π£π£ = π£π£, , ππ = 50 cm, , From the thin lens formula,, 1 1 1, β =, π£π£ π’π’ ππ, , 1, 1, 1, β, =, π£π£ β25 50, π’π’ = β50 cm, , 9.27., , Therefore, his near-point has changed from 25 cm to 50 cm. In the old age he, partially lost the ability of accommodation. This defect is known as presbyopia., Due to this he cannot see the objects placed in front of him clearly up to the distance, 50 cm., , A person looking at a person wearing a shirt with a pattern comprising vertical and, horizontal lines is able to see the vertical lines more distinctly than the horizontal, ones. What is this defect due to? How is such a defect of vision corrected?, Solution:, In the given situation the person can see the vertical lines more distinctly than the, horizontal ones. The type of defect is called astigmatism. This occurs when the, cornea is not spherical in shape. In the given situation he is not able to focus the, horizontal lines, hence the curvature of the cornea in the horizontal plane is not, sufficient compared to the curvature in the vertical plane. Astigmatism can be, corrected by using a cylindrical lens of the desired radius of curvature with an, appropriately directed axis, , 9.28., , A man with normal near point (25 cm) reads a book with small print using a, magnifying glass: a thin convex lens of focal length 5 cm., , (a), , What is the closest and the farthest distance at which he should keep the, lens from the page so that he can read the book when viewing through the, magnifying glass?, , Practice more on Ray Optics and Optical Instruments, , Page - 24, , www.embibe.com
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Class- XII-CBSE-Physics, , (b), , Ray Optics and Optical Instruments, , What is the maximum and the minimum angular magnification (magnifying, power) possible using the above simple microscope?, , Solution:, (a), , The focal length of the lens ππ = 5 cm, , The image can be formed between 25 cm and infinity, From the thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , For π£π£ = β25 cm, β π’π’ =, , ππππ, 5(β25), =, = β4.16 cm, ππ β π£π£, 5 + 25, , For π£π£ = β, π’π’ = β5 cm, (b), , Hence the closest distance he should keep the lens from the page is, 4.167 cm, and the farthest distance is 5 cm, Minimum magnifying power is when the image formed at infinity., ππππππππ =, , π·π· 25 cm, =, =5, ππ, 5 cm, , Maximum magnifying power is when the image is formed at the nearpoint, , 9.29., , ππππππππ = 1 +, , π·π·, =6, ππ, , A card sheet divided into squares each of size 1 mm2 is being viewed at a distance, of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held, close to the eye., , (a), , What is the magnification produced by the lens? How much is the area of, each square in the virtual image?, , (b), , What is the angular magnification (magnifying power) of the lens?, , (c), , Is the magnification in (a) equal to the magnifying power in (b)? Explain., , (Note: Data in the question is modified to get the proper answers), Solution:, (a), , Given,, The area of each square π΄π΄ = 1 mm2, , The distance between card sheet and the lens π’π’ = β9 cm, , Practice more on Ray Optics and Optical Instruments, , Page - 25, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , The focal length of the lens ππ = 10 cm, From thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , π£π£ =, , π’π’π’π’, β9 Γ 10, =, = β90 cm, π’π’ + ππ β9 + 10, π£π£, , Magnification produced ππ = π’π’ =, , β90, β9, , = 10, , β length and breadth of the square increases 10 times, , (b), (c), , Area of each square box = 10 mm Γ 10 mm = 100 mm2, π·π·, , Angular magnification ππβ² = |π’π’| =, , 25, 9, , = 2.8, , No. Magnification and angular magnification of an optical instrument are, two separate things. Angular magnification is defined as the ratio of the, angular size of the object to the angular size of the object when placed at, near point (25 cm)., π£π£, , 25, , Magnification ππ = οΏ½π’π’οΏ½ and angular magnification ππβ² = |π’π’|, , 9.30., , (a), , Their magnitudes are equal only when the image of an object is formed at, near point (25 cm), , (b), , What is the magnification in this case?, , (c), , Is the magnification equal to the magnifying power in this case? Explain., , At what distance should the lens be held from the figure in Exercise 9.29 in, order to view the squares distinctly with the maximum possible magnifying, power?, , Solution:, (a), , Given,, The area of each square π΄π΄ = 1 mm2, , The focal length of the lens ππ = 10 cm, , Magnifying power is maximum when the image of the given object is, formed at the near point (25 cm), β π£π£ = β25 cm, , From thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , Practice more on Ray Optics and Optical Instruments, , Page - 26, , www.embibe.com
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Class- XII-CBSE-Physics, , π’π’ =, (b), (c), , 9.31., , Ray Optics and Optical Instruments, , π£π£π£π£, β25 Γ 10, =, = β7.14 cm, ππ β π£π£, 10 + 25, , the magnifying power of the lens is maximum when the sheet is placed at, 7.14 cm in front of the lens, π£π£, , β25, , Magnification ππ = π’π’ = β7.14 = 3.5, 25, , 25, , Magnifying power ππβ² = |π’π’| = 7.14 = 3.5, , Magnitudes of magnifying power and magnification are equal in this case., If the image of an object is formed at the near point (25 cm), magnitudes of, magnifying power and magnification are equal., , What should be the distance between the object in Exercise 9.30 and the magnifying, glass if the virtual image of each square in the figure is to have an area of 6.25 mm2 ., Would you be able to see the squares distinctly with your eyes very close to the, magnifier? [Note: Exercises 9.29 to 9.31 will help you clearly understand the, difference between magnification in absolute size and the angular magnification (or, magnifying power) of an instrument.], Solution:, Given,, The area of each square = 6.25 mm2, , β The side length of the square = 2.5 mm, π£π£, , Magnification required ππ = π’π’ =, β π£π£ = 2.5π’π’, , 2.5 mm, 1 mm, , = 2.5, , From the thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , 1, 1, 1, β =, 2.5π’π’ π’π’ 10, β π’π’ = β6 cm, , And π£π£ = β15 cm, which is less than near-point (25 cm), , Hence, we cannot see the square distinctly with our eyes very close to the magnifier, 9.32., , Answer the following questions:, (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. In what, sense then does a magnifying glass provide angular magnification?, , Practice more on Ray Optics and Optical Instruments, , Page - 27, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , (b), , In viewing through a magnifying glass, one usually positions oneβs eyes, very close to the lens. Does angular magnification change if the eye is, moved back?, , (c), , Magnifying power of a simple microscope is inversely proportional to the, focal length of the lens. What then stops us from using a convex lens of, smaller and smaller focal length and achieving greater and greater, magnifying power?, , (d), , Why must both the objective and the eyepiece of a compound microscope, have short focal lengths?, , (e), , When viewing through a compound microscope, our eyes should be, positioned not on the eyepiece but a short distance away from it for best, viewing. Why? How much should be that short distance between the eye, and eyepiece?, , Solution:, (a), , (b), , The angular size of the object and the image are the same. The advantage, of the magnifying glass is that, the object can be placed closer than near, point (25 cm). Hence it gives a larger angular size than the object placed at, near point. In this sense a magnifying glass provides angular magnification., , As the angle subtended by the image at the eye is slightly less than the angle, subtended at the lens Angular magnification decreases a little. If the image, is at a very large distance away this effect is negligible., , (c), , The focal length of a convex lens cannot be decreased to a greater amount., If we reduce the focal length, spherical and chromatic aberrations will, become more pronounced. So, in practice one cannot get a magnifying, power of more than 3 for a simple convex lens. However, using a system of, lenses which corrects the aberration one can increase this limit by a factor, 10., , (d), , The magnifying power of the eyepiece ππππ = οΏ½1 + ππ οΏ½, , π·π·, , ππ, , Hence magnifying power of eyepiece increases with smaller ππππ, π£π£, , Magnification of the objective can be written as |π’π’ππ |, , (e), , ππ, , Objects are placed close to the focal point of the objective. This gives |π’π’ππ | β, ππππ . Hence to increase the magnification, the focal length of the eyepiece is, maintained small., The image of the objective in the eyepiece is called eye-ring. If we place, our eye too close to the eyepiece, our eye cannot collect much of the light, and also the field of view gets reduced. If we position our eye at the eye ring, which is slightly above the eyepiece, the area of the pupil of our eye is, , Practice more on Ray Optics and Optical Instruments, , Page - 28, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , greater or equal to the area of the eye ring and our eyes will collect all the, light refracted by the objective., The precise location of the eye-ring naturally depends on the separation, between the objective and the eyepiece., 9.33., , An angular magnification (magnifying power) of 30X is desired using an objective, of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up, the compound microscope?, Solution:, Given,, The focal length of the objective ππππ = 1.25 cm, , The focal length of the eyepiece ππππ = 5 cm, , The magnifying power of a compound microscope ππ = ππππ Γ ππππ, Assuming near point adjustment,, ππ =, β, , π£π£ππ, π·π·, οΏ½1 + οΏ½ = 30 cm, βπ’π’ππ, ππππ, , π£π£ππ, 25, οΏ½1 + οΏ½ = 30, βπ’π’ππ, 5, , β π£π£ππ = β5π’π’ππ, , From the thin lens formula, 1 1 1, β =, π£π£ π’π’ ππ, , 5, 1, 1, + =, π£π£ππ π£π£ππ 1.25, , β π£π£ππ = 7.5 cm, and π’π’ππ = β1.5 cm, Similarly, for the eyepiece,, 1 1 1, β =, π£π£ π’π’ ππ, , 1, 1, 1, +, =, β25 π’π’ππ 5, , β π’π’ππ = 4.17 cm, , Therefore, the separation between objective and the eyepiece πΏπΏ = 7.5 + 1.17 =, 11.67 cm, And the object must be placed at 1.5 cm to get the given magnification., , Practice more on Ray Optics and Optical Instruments, , Page - 29, , www.embibe.com
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Class- XII-CBSE-Physics, , 9.34., , Ray Optics and Optical Instruments, , A small telescope has an objective lens of focal length 140 cm and an eyepiece of, focal length 5.0 cm. What is the magnifying power of the telescope for viewing, distant objects when, (a), , the telescope is in normal adjustment (i.e., when the final image is at, infinity)?, , (b), , the final image is formed at the least distance of distinct vision (25cm)?, , Solution:, Given,, The focal length of the objective ππππ = 140 cm, , The focal length of the eyepiece ππππ = 5.0 cm, , 9.35., , (a), , For normal adjustment,, , ππππ, , (b), , Magnifying power ππ =, , ππππ, , (a), , Magnifying power ππ =, , (b), (c), , ππππ, , For near-point adjustment,, ππππ, , =, , 140, 5.0, , = 28, , ππ, , οΏ½1 + π·π·πποΏ½ =, , 140, 5.0, , 5.0, , οΏ½1 + 25 οΏ½ = 33.6, , For the telescope described in Exercise 9.34 (a), what is the separation, between the objective lens and the eyepiece?, If this telescope is used to view a 100 m tall tower 3 km away, what is the, height of the image of the tower formed by the objective lens?, What is the height of the final image of the tower if it is formed at 25 cm?, , Solution:, (a), , From Exercise 9.34 (a),, The focal length of the objective ππππ = 140 cm, , The focal length of the eyepiece ππππ = 5.0 cm, , For normal adjustment, the separation between the objective and the eyepiece, πΏπΏ = ππππ + ππππ, , (b), , πΏπΏ = 140 + 5 = 145 cm, Given,, , The Height of the tower βππ = 100 m, , The distance of the tower π’π’ππ = 3 km = 3000 m, Practice more on Ray Optics and Optical Instruments, , Page - 30, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , The angle subtended by the tower at the objective = Angle subtended by the, image of the objective, β, , β Height of the image formed by the objective β1 = ππππ Γ π’π’ππ = 140 Γ, , (c), , 100, , 3000, , = 4.7 cm, , ππ, , For the eyepiece,, 1 1 1, β =, π’π’ π£π£ ππ, β, , 1, 1, 1, β, =, π’π’ππ β25 5, , β π’π’ππ = 6.5 cm, , The angle subtended by the image of the objective at the eyepiece = Angle, subtended by the image of the eyepiece, β, , 4.7 cm, β2, =, 6.5 cm 25 cm, , β β2 = 18.07 cm, , 9.36., , Height of the final image of the tower = 18.07 cm, , A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is, built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is, 220 mm and the small mirror is 140 mm, where will the final image of an object, at infinity be?, , Fig. 9.33, Solution:, Given,, The radius of the curvature of the large mirror π
π
1 = 220 mm, Its focal length = 110 mm = 11 cm, , The radius of the curvature of the small mirror π
π
1 = β140 mm, Its focal length = 70 mm = 7 cm, , Practice more on Ray Optics and Optical Instruments, , Page - 31, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , The separation between the mirrors = 20 mm = 2 cm, For the large mirror, π’π’1 = β, , ππ1 = β11 cm, , From mirror formula,, 1 1 1, + =, π£π£ π’π’ ππ, , 1, 1, 1, + =, π£π£1 β β11, π£π£1 = β11 cm, , The image formed by the large mirror serves as an object to the small mirror, For the small mirror,, π’π’2 = 11 β 2 = 9 cm, , ππ2 = 7 cm, , From mirror formula,, 1 1 1, + =, π£π£ π’π’ ππ, , 1 1 1, + =, π£π£2 9 7, , π£π£2 = 31.5 cm = 315 mm, , 9.37., , Hence the final image will be formed at a distance 315 mm left to the small mirror, , Light incident normally on a plane mirror attached to a galvanometer coil retraces, backwards as shown in Fig. 9.36. A current in the coil produces a deflection of 3.5Β°, of the mirror. What is the displacement of the reflected spot of light on a screen, placed 1.5 m away?, , Practice more on Ray Optics and Optical Instruments, , Page - 32, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, , Fig. 9.36, Solution:, ππ, , Given that the reflection in the mirror ππ = 3.5Β° = οΏ½180 Γ 3.5Β°οΏ½ rad, , When the mirror is rotated through an angle ππ the reflected light beam from it, rotates through an angle 2ππ, ππ, , Thus, the angle rotated by the reflected light beam = 2 Γ οΏ½180 Γ 3.5Β°οΏ½ rad, ππ, , 9.38., , Distance moved by the spot on the screen ππ = 1.5 Γ 2 Γ οΏ½180 Γ 3.5Β°οΏ½ m =, 18.3 cm, , Figure 9.37 shows an equiconvex lens (of refractive index 1.50) in contact with a, liquid layer on top of a plane mirror. A small needle with its tip on the principal, axis is moved along the axis until its inverted image is found at the position of the, needle. The distance of the needle from the lens is measured to be 45.0 cm. The, liquid is removed and the experiment is repeated. The new distance is measured to, be 30.0 cm. What is the refractive index of the liquid?, , Figure 9.37, Solution:, Given:, The refractive index of the convex lens = 1.50, Let the refractive index of the liquid be ππ, , In both cases, object distance and image distances are the same. This implies, incoming rays are retracing the same path after reflecting from the mirror, or rays, from the pin tip going parallel after refracting from the lens and water., Let the radius of curvature of the equiconvex lens be π
π
, , Practice more on Ray Optics and Optical Instruments, , Page - 33, , www.embibe.com
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Class- XII-CBSE-Physics, , Ray Optics and Optical Instruments, β§«β§«β§«, , Practice more on Ray Optics and Optical Instruments, , Page - 35, , www.embibe.com