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RAY OPTICS AND OPTICAL INSTRUMENTS, , 9/65, , SOLVED SZaAMPLSsS., , ecg REFLECTION OF LIGHT, , I, , Formulae used., , 1, Number of images of an object held between two, plane mirrors inclined at 20 is, , 360" | ,, 360, , , , n= ss if “er is even integer, 360 .. 360., n= =. if isan odd integer, , 2. When a plane mirror is tumed through an angle, , 9, the reflected ray (of the given incident ray) turns |, , through an angle 26 in the same direction., , 3. On reflection, there is no change in frequency, and wavelength of light., , Units used. 6 is in degrees., , , , Example f] Light incident normally on a, plane mirror attached to a galvanometer coil, retraces backwards as shown in Fig. 9.75. A, current in the coil produces a deflection of 3-5° in, the mirror. What is the displacement of the, reflected spot of light on a screen placed 1-5 m, away ? :, , Solution. In Fig. 9.75, when mirror is tumed, ‘from position M, to position M,j through 20 = 3-5°,, the reflected ray turns through ., 220 = 7° = ZAOB, , , , OA=d=15m, Displacement, AB = x =?, , FIGURE 9.75, Bry., x ~ 10 i, , Te Se emia eet, ‘, ’, =, =, nN, , , , , , , , , , , , Is, , In AAOB — tan20=, x = d tan 20 = 1-5 tan, , © = 1.5 x 01228, = 0-18 metre., , Qk, , Example] A boy 1.5 m tall with his eye, level at 1.38 m stands before a mirror fixed on a, wall. Indicate by means of a ray diagram how the, mirror should be positioned so that he can view, himself fully. What should be the minimum length, of the mirror ? Does the answer depend on the, eye level ?, , Solution. In Fig. 9.76, eye is at 1-38 m from, the foot F. Rays from foot can enter eye after, reflection at Mj, whose height from ground is, 1,38, , , , , , , , , , — =0.69m., 2, FIGURES76.==° ~~ ~*«, WALL FIGURE 9.76 |, 7, M,°, "GROUND, , Again, eye is at 1.5 — 1.38 = 0.12 m from head, H. Rays from head can enter eye after reflection at, , 2, M,, whose height above the eye is a2 =0.06m., , -. Minimum length of the mirror = 0,69 + 0,06, = 0.75 m., Minimum length of the mirror is thus always, half of the length to be seen irrespective of eye level., , However, the position of the top and bottom, edges of the mirror would depend on the eye level., , PRG REFLECTION AT, ‘Tl SPHERICAL MIRRORS, , Formiulae used,, , 1. In a spherical mirror, f = R/2, , Both, f and R are positive for a convex mirror, and both are negative for a concave mirror., , Scanned with CamScanner

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9/66, , al Physics (XI) Mepay, , , , 2. Mirror Formula = Le ae owt, vou R f, The formula is same for both, convex mirror and, , Concave mirror, whether the image is real or, Virtual., , 3. Linear magnification for spherical mirrors, , meltin 2 fot. f, , h u f fru, Sign of m is determined from the sign of hy/h,, Apply the prescribed New Cartesian sign, Conventions to all distances and heights., Units used. f, R, u, v, all are in metre or cm., , hy is size of image and hy is size of object (both, in metre or cm). m has no units., , &xampte R] When an object is placed at a, , Pradecp's Fundament, , =-15cm,, , Solution. Here, R= oF a-2om, “9 2, , , , —+, > fu 15 10 30 30, v=—30cm. ;, +. The image is at 30 cm from the mirror on the, , same side as object. Also,, , . v__ © 30) _, magnification, m= “= C10) =, , .. The image is real, inverted and magnified, , 3 times,, distance of 60 cm froma convex spherical mirror, (ii) u=—-Scm,v=?, the magnification produced is 1/2. Where should, the object be placed to get a magnification of 1/3 ?, , , , , , , , , , From i + i = +, i 1 vou, Solution. Here, uy =~ 60.em, m, == 4o1 1-21-4462, =% m=5 0 fu 15S 5°30 “30, u,=?, Mm, 3 v=15 cm., os The image is at 15 cm behi ., As m, =—L 72 5 > Vy =30cm ' hind the mirror., 4% 2° =(-60) Also, m=—2-_ 15) _, Fi 11 ¥ 1 1 1 1 The; u ~6), en fy “30 60 “. The image is yj :, f 4 4 30 60 60 three times, 2 #8 Virtual, erect and magnified, f=60cm _ _ 5 gnifi, xample) Suppose wh . ., - __%_1 i el, parked car, YOU notice a jo ile Sitting in g, Again, my gr P25 towards you in the 9 Jogger appro aching, ” the jogger is running .. °” MTOr of R = ‘it, Id fast is the i 8 at a speed of 5 ms-i, —t+—=— . mage of the jogger . , how, From UV, uy ff Jogger is (2) 39 m (6) 29 Moving, when the, Ll any © 19 m and (a 9, Lf 1 3 ‘ NCerRy Setveg m, 4 fy Oy op REMI, ran ; Solution. Here, R= 2 my f=R_2 1, : Set =-120cm 5 °4* a, or wy 60°"? From Jia, Example [] An object is placed (i) 10 cm v 4 f, ,; i dius of, i ii t of a concave mirror of ra 14, Oe Se a Find the position, nature and si, , . ‘ image in each case., magnification of the imag (Raj. Board 2012), , Us, st, , NCERT Solved Example - ., , ——, , Scanned with CamScanner

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RAY OPTICS AND OPTICAL INSTRUMENTS, , (a) When jogger is 39 m away, u =~ 39 m,, , fue 39) 39, = 2, u-f ~39-] 40, As the jogger MOves at a constant Speed of, , 5 mvs, after 1 second, the posit; :, , Positi, for U=-3945=~ 34 . = of the image (v), , y= _ 1034) 34, , v=, , , , uf =34-1~35™, Difference in apparent position of ; i, wai Pp n of jogger in |, = 39 _ 34 _ 1365-1360 5S 1, 40 35° 1400 1400 = 2g0™, .. Average speed of Jogger’s image, 1, = — mi a, 280", , Similarly, for u = — 29 m, — 19 m and ~- 9 m,, average speed of jogger’s image is, 1 1 1, iso™® ; mss om respectively., The speed increases as the jogger approaches the, car. This can be experienced by the Person in the car., , Note. We can use eqn. (17) on page 9/11 to, calculate speed of jogger’s image., , LAWS OF REFRACTION, LATERAL SHIFT,, ll REFRACTION THROUGH A COMPOUND, PLATE, REAL AND APPARENT DEPTHS, , Formulae used., , _ Yelocity of lightin vacuum _ c, , - velocity of lightinmedium uv, wavelength in vacuum (A), , 7 wavelength in medium A’), , 2. When light travels from medium a to medium b,, , 1,, , , , M, — sini, oy SS, Be yw, sinr, 1, 3. “Hy =, b on,, tsin (i, — 4), He en de, 4. Lateral shift C08 j, 5. "y,x?h. ="H,, ___realdepth (x), 6. f= “pparent depth (), , , , 9/67, , 1, 7. Displacement of image = x- y =x (! x, Units used. Angles, i, r, i,, 7, in degree ; distances, t,, 3, y in metre ; 1 has no units, velocities c, v in m/s., , , , Example [4 What is the speed of light in, glass of refractive index 1-5 ? Given speed of light, , in water is 2-25 x 108 m/s and refractive index of, water is 1-3,, , Solution. Here, Hy = 155, v,= 7p, = 1-3,, v,, = 2-25 x 108 m/s, , 13 _ 13x2-25x108, 8 45 Ww 15, = 1-95 x 10° m/s, Example] A rectangular glass slab rests, at the bottom of a trough of water. A ray of light, incident on water surface at an angle of 50° passes, through water into glass. What is angle of, , refraction in glass ? Take Ht for water 4/3 and », for glass 3/2., , Solution. Here, “1, = 4/3 and Ap, = 3/2, , , , 4/3 8, , a, , , , sinr, sinr = = sin 50° = $x0766 = 0-5745, Second refraction is in going from water to glass, _Ssinr _9, He = sinr’ 8, , sings sin = $x 05745 = 05107, , w, , , , r= sin! (05107) = 30-7°, , (example f] Whatis the apparent position, of an object below a rectangular block of glass, 6 cm thick, if a layer of water 4 cm thick is on the, top of the glass ? Given y,=3/2 and “1,,=4/3., , Solution. Here, real depth of glass, x, = 6 cm,, real depth of water, x,=4 cm., , “py = 3/2, “y,, = 4/3, , Scanned with CamScanner

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If», yy are the corresponding apparent depths,, , , , Jus, , In wp, , ", , ss he, vole &, , 3 3, y= a = gas 3cm, Apparent Position of the object, =(0, +¥2)=(44+3)em=7 cm, “+ Rise in position of the object, = (x, +2) = (, t+y2)=(6+4)-(443), =10-7=3cm, , Exampte ff] 4 Tay of light is incident at an, angle of 45° on one face of a rectangular glass slab, of thickness 10 cm and refractive index 1:5,, Calculate the lateral shift Produced. _., , Solution. Here, 1 = 45° t=10cm= 0-1m, = 1.5, lateral shift = 7, , sin i,, sin 5, , As, , . sini, sin 45°, six =—+=- =— —(.., , yu 15 15, , r= sin“ (0.4713) = 28.140, , tsin @- n), , lateral shift = 1°, , cos y, , 0-1sin (45° —28-14°), =e ee), cos 28-14°, , _ 0-1sin 16-86° _ 0-1x0-2900, , cos 28-14 0.ggig = 9033 m, , Exampte [[] 4 ray PQ is incident normally, on the refracting face of the prism BAC made of, material of refractive index 1-5. Complete the path, of ray through the prism. From which face will, the ray emerge and at what angle ? Justify your, answer., , Solution. Refer to Fig. 9.77. As the ray PQ falls, normally on face AB, it passes undeviated falling on, AC at Zi = 30°. It will be refracted along RS at Zr., The refracted ray emerges from face AC of prism., , tal Physics CX) Reta }, 4% Fundamen, Pradeep d, , FIGURE 9.77, , , , As refraction occurs from denser to rarer, medium., sinr, = anit, = 1-5 sin 30° = 0-75, r= sin! (0-75) = 48°6°, , sinr=psini, , A transparent cube of side, 210 mm contains a small air bubble. Its apparent, distance, when viewed from one face of the cube, is 100 mm, and when viewed through opposite face, is 40 mm. What is the actual distance of the bubble, from the second face and what is the refractive, index of the material of the cube ?, , Solution. Let the bubble be at a distance x from, , real depth, , apparent depth” therefore, for ‘face I,, , , , facel. As p=, , x, .=—, , 100 224 for face 17, y = 210—x, , , , Scanned with CamScanner

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RAY OPTICS AND OPTICAL INSTRUMENTS, aA FATAL INTERNAL REFLECTION, , Formulae used, 1, For TIR, light must travel from, , cone a medium. Angle of incidence (i) in, < tum must be greater than critical angle, (C) for the pair of media in contact., _ 1, hee, , sinC, , Units used. Angles i and C are, , ‘ ind, units, egree, jt has no, , Example [PJ Calculate the speed of light, in a medium whose critical angle is 30°., , (CBSE 2010), Solution. Here, v = 7, C = 30°, , 1, sinC, v=c sin C=3 x 108 x sin 30° = 1:5 x 108 m/s, &xample 8) The critical angle of incidence, , ina glass slab placed in air is 45°. What will be the, critical angle when the glass slab is immersed in, water of refractive index 1-33 ?, , Solution. Here,, , , , _oc, Uv, , , , , , , , 4) ot =) _-=144, § sinC sin 45°, Refractive index of glass w.r.t. water, w He _ 1414, Mga yn, 133, When glass slab is immersed in water,, 1 1:33 :, in C’ =—— =—— = 09432, eT tae OO, , C =sin7! (0.9432) = 70° 36’, ‘exampte[£] Determine the direction in, which a fish under water sees the setting sun., Given, for water } = 1:33. 7, Solution. In Fig. 9.79, the setting sun is in the, direction of water surface. A ray of light from setting, , nters the eye of the fish when apparent position, pribe sun makes critical angle C with the vertical,, , 11, inC =+ =—— =0-7518, sin 133, , C = sin (0-7518) = 48-7°, , 9/69, , , , , , , FIGURE 9.79, , Apparent, os. of sun, as, /7 ~seen by, the fish, , , , Hence angle between apparent position of the, sun and horizontal is @ = 90° - 48-7° = 41:3°, , Example [lH] The refractive index of water, is 4/3. Determine the angle of the cone within, , which the entire outside view will be confined for, a fish under water. (CBSE 2003), , Solution. As is clear from Fig. 9.80, the fish, can see the entire outside view within cone angle, , <o075, , 8 =2C, where snC ==, . : H 4/3, , , , FIGURE 9.80, , , , , , ish=, , , , , , C =sin“! (0-75) = 48-7°, 0=2 C=2x 48.7 = 97-4", , ‘example f{J A point source of light S is, placed at the bottom of a vessel containing a liquid, of refractive index 5/3. A person is viewing the, source from above the surface. There is an opaque, disc of radius 1 cm floating on the surface. The, centre of disc lies vertically above the source O., The liquid from the vessel is gradually drained, out through a tap. What is the maximum height, of the liquid for which the source cannot be seen, at all, (UT 2004), , Solution. In Fig. 9.81, let OS = h be the max., height of liquid above the source for which the source, cannot be seen at all., , & ZOSA=C, , Any other ray from S will be totally internally, reflected because in that case, angle of incidence will, be greater them the critical angle., , Scanned with CamScanner