DC Pandey Optics And Modern Physics

438 — Optics and Modern Physics 2. Image distance from plane mirror = object 8. distance. Mi Lateral magnifications = 1 i 3. I q–i R q q–i I 90° + i – q N 75° dTotal = dM + dN 60° 15° = (180° - 2i) + [180° - 2 (q - i)]= 360° - 2q 90° – i 9. Apply 1 + 1 = 1 15° 15° vu f (a) f = -10 cm, u = - 25 cm where, I = incident ray R = reflected ray Solving, we get v = - 16.7 cm Since, v is negative, image is in front of Angle of incidence = 15° mirror. So, it is real. Similarly, we can solve Angle between reflected ray and horizontal = 60° for other parts. 4. Image from one mirror will behave like object for 10. (a) From O to F , image is real. other mirror. From F to P, image is virtual. 5. 3b 3b O I4 I2 O I1 I3 M2 M1 bbbb 2m OI1 = OI2 = 2b C I3 is the image of I2 from mirror M1 similarly I4 is 0.5 m the image of I1 from mirror M2. F OI3 = OI4 = 4b 0.5 m 6. Given in the theory. 7. 0.2 m 30° 30° P 30° At O 1+ 1= 1 vu f d 1.6 m \\ 1 + 1 = 1 v - 3.0 - 0.5 d = tan 30° = 1 0.2 3 v = - 0.6 m \\ d = 0.2 (b) When object is at C and P, image coincides 3 with object. N = 1.6 = 8 3 Using d s = 1 gt2 = 13.85 2 Therefore, actual number of reflections required are 14. or t = 2s g

Chapter 30 Reflection of Light — 439 At C t= 2 ´ 2 = 0.639 s Object Image At P 9.8 From O to - ¥ 11. (a) From O to F or from O to t= 2 ´ 3 = 0.782 s + 0.25 m From - ¥ to C or 9.8 from + ¥ to + 0.50 m From F to C or from From C to F or from 16.5 cm + 0.25 m to + 0.50 m + 0.50 m to + 0.25 m From C to + ¥ or from + 0.50 cm to + ¥ PC P F 15. O is placed at centre of curvature of concave 11 cm mirror (= 42 cm). Therefore, image from this 22 cm mirror I1 will coincide with object O. Q' Q' (b) Apply 1 + 1 = 1 O vu f I1 I2 and m = - v 21 cm 21 cm u 42 cm 12. (a) Now, plane mirror will make its image I2 at the same distance from itself. Q 16. OF P P' F Ix y 10 cm (b) Apply, 1 + 1 = 1 Using 1 + 1 = 1 vu f vu f and m = - v \\ 1 f) + 1 f) = 1 u - (y+ - (x + -f 13. f = R = - 18 cm Solving this equation, we get xy = f 2 2 v Let u = - x cm 17. Then, v = - x cm for real image of 1 th size. A I1 9 9 I2 Using, 1+ 1= 1 x vu f 1 1 1 2R x / 9) (- x) - 18 We have, (- + = For convex mirror Solving we get, x = 180 cm 1 + 1 = + 1 v -x R/2 14. \\ 1= 2+ 1 or v = R Rx vR x + 2x +¥ C F O _¥ Now, applying mirror formula for concave mirror we have 0.25 m 1 x) + - 1 v) = - 1 0.5 m - (2R - (2R + R/2 Solving this equation, we can find value of x.

440 — Optics and Modern Physics 18. (a) Image has to be taken on a screen. So, it should AB = PB - PA = 2(a + L) - 2a be real. Hence, mirror should be concave. = 2L = constant (b) 5 m x 3. m = - v = - (- 20) = - 2 u (- 10) IO | Dv | = m2 | Du | = (- 2)2 (0.1 cm) = 0.4 cm Image is 5 times magnified. Object and image travel in opposite directions (along the axis). Hence, |v |= 5|u| or (5 + x) = 5x 4. 3m Solving we get, I2 O I1 x = 1.25 cm x Now using, 1+ 1= 1 = 2 vu fR We have, 3–x 3–x x - 1 - 1 = 2 I1I2 = (3 - x) + (3) + x 6.25 1.25 R =4m Solving we get, Solving we get, x = 1 m R = - 2.08 m 5. Image of c will coincide with this. LEVEL 2 M Single Correct Option C P 1. 10 cm 10 cm vI 60° For P : \\ 60° 1 + 1 = 1 v0 v - 20 -5 v0 = vI = wA = æèçç ÷ö÷ø v = - 20 cm 3 k M A m = - v = - (- 20/3) = - 1 u (- 20) 3 vr = v02 + vI2 - 2v0 cos 120° Length of image of PM : = 3A k I1 = 10 ´ 1 = 10 cm M 3 3 Length of image of PC : 2. L I2 = 10 - 20 = 10 cm 3 3 h \\ I1 = 1 I2 P a A B 6. 1 + 1 = 1 a a+L v - 15 - 10 a+L \\ v = - 30 cm

m = - v = - (- 30) = - 2 Chapter 30 Reflection of Light — 441 u (- 15) s2 = displacement of mirror | D v | = m2 | Du | = (- 2)2 (2 mm) = - 1 gt2 2 = 8 mm \\ Vertical distance of particle from mirror, 7. Let i is the angle between incident ray and original s = s1 - s2 = 0.5 m reflected ray. Then, initial angle between them will Hence, distance between particle and its image be 180° - 2i. When mirror is rotated by 20° , then = 25 = 1 m reflected ray will rotate by 40°. 13. vOM = vO - vM = (- $i - 3 k$ ) \\ 180° - 2 i ± 40° = 45° \\ vIM = (- i$ + 3 k$ ) = vI - vM Solving we get, or vI = (- i$ + 3 k$ ) + vM i = 47.5° or 87.5° = (3 i$ + 4 $j + 11 k$ ) 8. B M 14. 6 cm C D x=0 x = 2 cm x = 12 cm AP 2 cm 10 cm PM = AC = AD + DC m=+ 5 \\ v should be positive and 5 times | u | . = AD + DB = 164 + 6 = 167 cm 22 9. aLHS = a1 = Net pulling force = 3 mg = 3 g 15. 90° - i q + 2r = 180° Total mass 4m 4 \\ = çèæ 90° +i - qøö÷ r 2 aRHS = a2 = Net pulling force = 2 mg Total mass 3m p = 180° - (90° - r) - 2i = 180° - 90° + çæ 90° +i - q÷ö - 2i è 2 ø =2g 3 qq 90° – q The relative acceleration is therefore : 3 g + 2 g = 17 g 90° – q 4 3 12 53° 90° – i 10. See the hint of Q.No-7 of subjective questions for m i i rr qp Level 1. 11. vI = 5 m /s 90° – r 16° 53° vO = 5 m /s = 180° - 4i + 90° + i - q 2 vI = (5 cos 16° ) i$ + (5 sin 16° ) $j = (4.8 $i + 1.4$j) m/s = çæ 270° - 3i - q÷ö è 2ø 12. uy = 2 sin 45° = 1 m/s m = p - q = 180° - (90° - i) - 2i In vertical direction, s1 = displacement of particle \\ æçè 270° - 3i - q÷øö - q = 90° - i = (1) (0.5) - 1 gt2 = 0.5 - 1 gt2 2 22 Substituting q = 20° , we get i = 30° 16. Ray passing through c is only correct.

442 — Optics and Modern Physics More than One Correct Options m = v = - (- 60) = - 2 u ( - 30) 1. For real image, Speed of image (in event -1) is m2 times and m Let u = - x, then v = - 2x times in event-2. \\ 1 + 1 = 1 6. In event -1 -2x -x - 20 Solving, we get x = 30 cm. For concave mirror, For virtual image, 1+ 1 = 1 v -3 f -f Let u = - x, then v = + 2x \\ v = - 1.5 f \\ 1 + 1 = 1 2x -x - 20 For convex mirror, or x = 10 cm 1+ 1 = 1 v -3f +f 2. Inverted and real image is formed by concave v= 3 f = 0.75 f 4 mirror. Let u = - x, then v = + x/2 For plane mirror, \\ 1 + 1 = 1 v=+ 3f +x/2 -x +f In event-2 \\ x=3f For concave mirror, Erect and virtual image is formed by convex 1+ 1 = 1 v - 1.5 f -f mirror. Let u = - x, then v = + x \\ v=-3f 2 1 1 1 For convex mirror, -x/2 + -x = +f 1+ 1 = 1 v - 1.5 f +f x=+ f \\ v = 3 f = 0.6 f 3. vI = v 5 q v0 = v For plane mirror, q v = + 1.5 f Comprehension Based Questions vr = v2 + v2 - 2v × v cos 2 q 1. y x = 2 v sin q 10 cm 4. Ray diagram is as shown in figure. O I O M 10 cm F I C OM = MI O Coordinates of I are (10 cm, - 10 cm). 2. Object is placed at centre at curvature of mirror. Hence, image is at the same point, real, inverted and of same size. Hence, coordinates are (20 cm, 0). 1 + 1 = 1 3. 1 + 1 = 1 v - 30 - 20 v - 20 + 10 5. Solving we get, v = - 60 cm \\ v = + 20 cm 3

Chapter 30 Reflection of Light — 443 m = - v = - (+20/ 3) = + 1 \\ x = 60 cm u (- 20) 3 In the similar manner, other options can be solved. \\ I = 10 èæç 13÷øö = 10 cm 3 Subjective Questions x = 10 - 10 = 20 cm 1. Q 3 3 and y= 20 + 20 = 80 cm B 3 3 4. Plane mirror forms image at equal distance on opposite sides. 20 cm S A Hence, x = 0, y = 40 cm Match the Columns 1. (a) m = - 2 means image is real, inverted and P 2-times magnified. So, mirror should be concave. Same logic can be given for other Insect can see the image of source S in the mirror, so far as it remains in field of view of image options also. overlapping with the road. 2. 1 + 1 = 1 (as u is positive for virtual objects) v +u f Shaded portion is the field of view, which overlaps \\ 1= 1 -1 v fu with the road upto length PQ. By geometry we can see that, PQ = 3AB = 60 cm For plane mirror, f = ¥. So, v is always negative. \\ t = Distance = 60 = 6 s Ans. Hence, image is always real. Speed 10 For concave mirror, f is negative. So, v is again 2. Using mirror formula, èæçç 1 + 1= 1 ÷÷öø negative. Therefore, image is always real. v u f For convex mirror, f is positive. So, v may be 1 - 1 = -1 positive or negative. v 50 25 Hence, image may be virtual or real. 3. (a) Image is inverted, real and diminished. Hence, \\ v = - 50 cm mirror is concave. Same logic can be applied for other options m = - v = -1 Ans. u too. 4. (a) 1 + 1 = 1 M1 v - 20 - 20 Optic axis of M1 I1 \\ v=µ Optic axis of M2 S 0.5 cm M2 0.5 cm m=-v =¥ I2 0.5 cm u 0.5 cm Same formulae can be applied for other options too. 5. (a) See the hint of Q.No-1 of more than one 50 cm correct options section. 3. Using mirror formula 1 + 1 = 1 for concave (b) Half size image is formed only in case of real vu f image. mirror first, we have Let u = - x, then v = - x Now, 2 1 - 1 = 1 çèæQ f = R ÷ö v 60 -40 2ø 1 + 1 = 1 -x/2 -x - 20 or v = - 120 cm

444 — Optics and Modern Physics Time taken to move the boy from G to topmost First image I1 at 120 cm from concave mirror will point and then from topmost point to G will be act as virtual object for plane mirror. Plane mirror will form real image of I1 at S. t2 = 2v = 2.83 s g I1 \\ The required time is Ans. t = t1 - t2 = 1.7 s 5. Applying mirror formula for concave mirror first 60 cm 30 cm 30 cm ççèæ 1 + 1 = 1 øö÷÷ we have, 120 cm v u f 1 - 1 = 1 v 110 -100 Ray diagram is shown in figure. v = - 1100 cm Distance between two mirrors is 90 cm. Ans. AI1 = 100 cm. Therefore, final image will be real and at distance 100 cm below point A at I2. 4. G 1100 cm B I F 100 cm AI2 = 100 cm A H E 45° S D I1 A I2 C 6. In 15 seconds, mirror will rotate 15° in clockwise FG = IH = HS direction. BF BH BH Hence, the reflected ray will rotate 30° in clockwise direction. \\ FG = (BF ) æç HS ö÷ è BH ø A 3m = (5) æç 1.0 ö÷ 30° 30° 60° è 0.5ø = 10 m 3Ö3 m FC = 2 + 10 = 12 m The boy has dropped himself at point F. So, his At t = 15 s velocity is 20 m/s in upward direction. P Let us first find the time to move from F to 3m topmost point and then from topmost point to point C. From s = ut + 1 at2 , we have q y 2 - 12 = (20t) + 1 (- 10) t 2 2 Solving this equation we get, t1 = 4.53 s. Velocity of boy at point G , v = (20)2 - 2 ´ 10 ´ 10 At time t = 14.14 m/s (Q v2 = u2 - 2gh)

Chapter 30 Reflection of Light — 445 y = 3 tan q or 1 = u - 1 = u - f \\ dy = (3 sec2 q). dq …(i) mf f Here, dt dt \\ m = f - dy = vP , dq = 2° per second u f dt dt Differentiating we have, = 2 ´ p = p rad per second 180 90 èæç dm ÷øö = - (u f f )2 . du …(ii) dt - dt At t = 15 s and q = 60° Substituting the values in Eq. (i), we have Using mirror formula to find u with magnification vP = {3 sec2 60°} ìîí9p0ýþü m = 1 we get, 10 =3´4 ´ p 1 -1= 1 90 u/10 u 10 = 2p m/s Ans. or u = 90 m (with sign u = - 90 m) 15 Substituting in Eq. (ii) we have, 7. (a) Differentiating the mirror formula, (with dm = - (- (10) (- 1) dt 90 - 10)2 respect to time) = 10-3 per second Ans. POLICE v Note u is decreasing at a rate of v - 20 or (21 - 20) or 1 m/s 20 m/s Thief \\ du = - 1 m/s dt 8. (a) At t = t, 1+ 1= 1 u = - (2 f + x) vu f = - (2 f + f cos wt) We get velocity of image, Using the mirror formula, 1+ 1= 1 vI = (m)2vO …(i) vu f Here, vO = relative velocity of object with We have, respect to mirror vI = relative velocity of image 1- + 1 wt = -1 m = linear magnification v 2f f cos f Here, vO = (v - 20) m/s \\ v = - æççè 2 + cos wt ÷ö÷ø f vI = 1 cm/s = 0.01 m/s 1 + cos wt m= 1 10 i.e. distance of image from mirror at time t is Substituting in Eq. (i) we have, èæçç 2 + cos wt öø÷÷ f Ans. 1 + cos wt æç 1 ö÷ 2 è 10 ø 0.01 = (v - 20) (b) Ball coincides with its image at centre of curvature, i.e. at x = 0. \\ v = 21 m/s (b) 1 + 1 = 1 (c) At t = T /2 vu f or wt = p, x = - f Multiplying with u we get, i.e. u = - f or ball is at focus. So, its image is u + 1= u at ¥. vf m=¥

446 — Optics and Modern Physics 9. Let the ray is incident at a point P = (x1, y1) on the æççè y - y1 ö÷ø÷ = - tan b x - x1 mirror. Then, slope at P, = - 2q = -2 tan q y 1 - tan2 q tan P \\ y- y1 = 2 cot a (x1 - x) …(ii) q 1 + cot2 a q x Further, x1 = 4by12 …(iii) a bq At F , x=0 …(iv) F From Eq. (i) to Eq. (iv), tan a = æç dy÷ö = 1 we get x = 1 è dx ø (x1y1) 4by1 8b …(i) It shows that the coordinates of F are unique æçè 0÷øö. a = 90° - q 1 , 8b and b = 2q Now, the reflected ray is passing through P(x1 y1) Hence, the reflected ray passing through one focus and has a slope - tan b. Hence, the equation will be and the focal length is 1 . Hence proved. 8b

31. Refraction of Light INTRODUCTORY EXERCISE 31.1 Using m 2 - m 1 = m 2- m 1, we get vu R 1. 1m 2 ´ 2m 3´ 3m 1 = 1 1 - 1.5 = 1 -1.5 v - 10 - 15 \\ 4 ´3 = 1 = 1m 3 or 1m 3 = 2 3 2 3m 1 Solving, we get v = - 8.57 cm 3. m = c= c = 3 ´ 108 = 1.67 3. v fl ´ 1014 ´ 300 6 ´ 10-9 O INTRODUCTORY EXERCISE 31.2 (a) Using m 2 - m 1 = m 2 - m 1, we get m2 sin i1 sin 60° vu R m1 sin i2 sin 30° 1. 1m 2 = = = = 3 1.5 - 1.0 = 1.5 - 1.0 v - 20 +6 l1 2. 1m 2 = 1.5 = l2 Solving, we get v = + 45cm Similarly other parts can be solved. INTRODUCTORY EXERCISE 31.3 4. 10 cm 1. (a) dapp = 10 + 10 = 50 cm O 1.5 3 (b) happ = 10 + (1.5)(10) = 25 cm INTRODUCTORY EXERCISE 31.4 = æççè1 - 1 ø÷÷ö + èæçç1 - 1 ö÷ø÷ Using m 2 - m 1 = m 2 -m 1, we get 1. Total shift m1 m2 vu R t1 t2 1 4/3 1- 4/3 = çæ1 - 2÷ö 10 + çæ1 - 1÷ö 10 v - - 10 = - 15 è 3ø è 2ø Solving, we get v = - 9.0 cm = 25 cm 5. 3 \\ Image distance = 50 - 25 = 125 cm 33 INTRODUCTORY EXERCISE 31.5 Using the equation 1. All rays starting from centre pass undeviated as m 2 - m 1 = m 2 - m 1, we get vu R they fall normal to the surface. 2. +ve 1.44 - 1.0 = 1.44 - 1.0 v ¥ + 1.25 10 cm 0 \\ v = 4.0 cm C INTRODUCTORY EXERCISE 31.6 1. 1-1 = 1 = (m - 1) èçæç 1 - 1 ø÷÷ö vu f R1 R2

448 — Optics and Modern Physics \\ 1 - 1 = (1.65 - 1) çèæç 1 - 1 øö÷÷ - v-2 × dv + u-2 × du = 0 (as f = constant) - 20 - 60 -R +R \\ dv = æççè - v2 ÷öø÷ × du Solving we get, R = 39 cm u2 2. 1 - 1 = 1 8. It is just like a concave mirror. - 50 u + 30 | f | = 0.2 m Solving, we get u = - 18.75 cm \\ | R | = 0.4 m m = v = (- 50) = 2.67 Focal length of this equivalent mirror is Air u (- 18.75) I = m (O ) = 2.67 ´ 2 Water = 5.33cm 3. + ve + ve 1 = 2 (m 2/m 1) - 2 (m 2/m 1 - 1) (extra points) F R2 R1 = 2 (4 /3) - 2 (4 /3 - 1) - 0.4 + 0.4 R1 R2 R1 R2 f1 f2 or F = - 0.12 m or - 12 cm 1 = (m - 1) çæèç 1 - 1 öø÷÷ …(i) 9. | R | = 0.5 m (from first case) f1 R1 R2 + - In the shown figure, object appears at distance 1 = (m - 1) èæçç + 1 - - 1 ÷ö÷ø …(ii) d = m e(0.2) + 0.2 f2 R2 R1 Now, for image to further coincide with the object, Solving these two equations, we can see that d =|R| f1 = f2. Solving we get, m e =1.5 4. 10. O = I1I2 (Displacement method) – ¥ F1 O +¥ = 6 ´ 2 = 2 cm 3 When object is moved from O to F1, its virtual, 11. Virtual, magnified and erect image is formed by erect and magnified image should vary from O to - ¥. convex lens. Let u = - x 5. (a) 1 = çæ 1.3 - 1÷ö æççè 1 - 1 ö÷ø÷ Then, v = - 3x f è 1.8 ø - 20 20 + Now, 1 - 1 = 1 - 3x -x + 12 \\ f = + 36cm (b) Between O and F1 image is virtual. Hence, for \\ x = 8 cm real image. Distance between object and image | m | < f or 36 cm = 3x - x = 2x = 16 cm 6. f = 10 cm f = 10 cm 12. Diminished erect image is formed by concave lens. 2F 2F 2F Let u = - x, then v =- x 2 Now, | u | - | v | = 20 cm 20 cm 20 cm 20 cm 20 cm \\ x = 20 cm or x = 40 cm 2 7. Q 1 - 1 = 1 \\ u = - 40 cm and v = - 20 cm vuf Differentiating this equation, we get 1 = 1 - 1 f -20 - 40 or f = - 40 cm

13. If an object is placed at focus of lens (= 10 cm), Chapter 31 Refraction of Light — 449 rays become parallel and fall normal on plane = (1.6)çèæ 21÷öø = 0.8 mirror. So, rays retrace their path. \\ i = 53° P ray deviates from its original path by an angle, INTRODUCTORY EXERCISE 31.7 d = i - 30° = 23° 1. 30° 60° = i Pd 60° q Critical angle = i = 60° = qC d mR sin 60° = m Q sin qC = mD Þ 3 \\ Angle between two rays, q = 2d = 46° 2. q Solving we get, m = 1.5 2. m= c = 3 ´108 = 1.3 iB v 2.3 ´ 108 q Ai qC = sin-1 æç 1 ö÷ = sin -1 èçæ 11.3÷øö = sin -1 (0.77) è m ø 3. (a) m 1 sin i1 = m 2 sin i2 = m 2 sin qC i = qC (1.6)sin q= (1.80) æç 1.30ö÷ \\ sin i = sin qC = 1 = 2 è 1.80ø mg 3 \\ q = sin-1 æç 13÷ö Applying Snell’s law at point A, è 16ø We have (b) If q is decreased, then i2 will decrease from the mvaelduieuqmC-.3H. ence, refraction will take place in m w sin q = m g sin i \\ 4 sin q = 3 ´ 2 3 23 INTRODUCTORY EXERCISE 31.8 \\ sin q = 3 4 1. 30° sin æç A + dm ÷ö è 2 ø 60° 3. m = sin ( A / 2) , dm = 30° i 4. i1 = 0° Þ r1 = 0° or r2 = A 30° Now, r2 = qC = A 1 2 \\ m 3 sin A= sin qC = = or A = sin-1 æç 2ö÷ è 3ø m = sin i 5. d = i1 + i2 - A sin 30° 30° = 60° + i2 - 30° \\ sin i = m sin 30° \\ i2 = 0 or r2 = 0 Hence proved.

450 — Optics and Modern Physics Þ r1 = A = 30° 8. ABC can be treated as a prism with angle of prism Now, m = sin i1 = sin 60° = sin r1 sin 30° 3 A = 90°. Condition of no emergence is 6. 2 = sin i1 AM C sin (i1/2) = 2sin (i1/2)cos (i1/2) sin (i1/2) Solving this, we get i1 = 90° and r1 = i1 = 45° B 2 A ³ 2qC At minimum deviation, r2 = r1 = 45° or sin qC £ sin çæ A ö÷ è 2ø \\ A = r1+ r2 = 90° 1 7. From m = sin çèæ A + dm ö÷ / sin ( A / 2) or m £ sin 45° 2ø We can see that given deviation is the minimum or 1 £ 1 deviation. m 2 A \\ m³ 2 9. qC = sin -1 æç 1 ö÷ = sin-1 çæ 1 ö÷ = 38.7° è m ø è 1.6ø MN BC r2 = qC = 38.7° \\ r1 = A - r2 = 45° - 38.7° = 6.3° At minimum deviation, MN is parallel to BC is Now applying, m = sin i1 ÐB = ÐC. sin r1 we can find i1. Exercises LEVEL 1 5. m1m2 = 1 Assertion and Reason 6. Image is formed at second focus (not the first 1. Shift due to a slab = çæ1 - 1 ÷ö t focus). è m ø 7. A = 60° and dm = 30° is in the direction of ray of R S Substituting the values, we get light. Hence, Ram appears m= 2 nearer to Anoop by that much 8. 1 = 1 + 1 distance. F f1 f2 4. P can be assumed a slab of negligible thickness. f1 = Focal length of convex lens (+ ve) f2 = Focal length of concave lens (- ve) Deviation is almost negligible. If f1 > | f2 | , then F comes out to be negative. So, it becomes a diverging lens. P Negative power of a diverging lens is just for its diverging lens, nothing else.

Chapter 31 Refraction of Light — 451 9. By inserting a slab between the lens and the 8. Frequency does not change during refraction, but object, effective distance between object and lens wavelength and speed decrease in a denser (here decreases. So, object now comes between F and 2F or between F and O. water) medium. 9. m 1 sin i1 = m 3 sin i3 F 2F \\ m 1 sin i = m 3 sin r or sin i = m 3 2F F O sin r m 1 10. r1 = i and r2 = 90° - r1 = 90° - i d = 2f Now, m R sin iR = m D sin iD or In the first case, image is real and magnified and in mR = sin qC = sin iD = sin sin i the second case it is virtual and magnified. mD sin iR (90° - i) 10. If medium on both sides of the lens is same, then it or sin qC = sin i = tan i cos i doesn’t matter, which side the object is kept. 11. i = 45° Þ qC = sin-1 (tan i) For m > 2 or 1.414 Þ qC < 45° 11. At minimum deviation, \\ They get TIR on face AC. r1 = r2 = A = 30° 12. For virtual object a concave lens can form a real 2 image. Now using, m = sin i1 or 2 = sin i1 sin r1 sin 30° 13. 1 = (m - 1) çèæ 1 - 1 ÷øö f R R We get i1 = 45° f =¥ 12. 1 = (1.5 - 1) ççèæ 1 - 1 ÷öø÷ …(i) \\ P= 1 =0 0.2 R1 R2 f 1 æç 1.5 - 1÷ö çæèç 1 1 ö÷÷ø 0.5 è m ø R1 R2 Objective Questions - = - …(ii) 2. m , A and B are dimensionless. Here, m = refractive index of medium or liquid. l2 Dividing Eq. (i) by Eq. (ii), we get Hence, [ B ] = [l2 ]= [L2 ] - 5 = 1 or m = 15 (1.5/m ) 8 3. dapp. = d - 1 m 13. From Snell’s law, m of red is least. So, dapp. for red is maximum. So, m sin i = m 4 sin x they appear to be raised least. m \\ sin x = m4 sin i çæ 1 ö÷ 4. qC = sin -1 è m ø 14. Focal length of any one part will be 2 f . m for violet is maximum, so qC for violet is least. \\ 1 = 1 + 1 + 1 + 1 or F = f 5. 1 = P = (m - 1) æèçç 1 - 1 ö÷ø÷ F 2f 2f 2f 2f 2 f R1 R2 15. In air, focal length is èççæ - 10.1÷øö÷ = (1.6 - 1) 1 - f = 100 = 100 = 20 cm 0.1 P 5 = + 12 D 1 = - 1) ççæè 1 - 1 øö÷÷ 20 R1 R2 6. = c = 3 ´ 108 = 2.25 ´ 108 m/s (1.5 …(i) m (4 / 3) …(ii) v 7. After a certain angle all colours get total internally 1 = èççæ 1.5 - 1÷öø÷ æèçç 1 - 1 ø÷÷ö - 100 me R1 R2 reflected.

452 — Optics and Modern Physics Dividing Eq. (i) by Eq. (ii), we get Originally, - 5 = (1.5 0.5 - 1 F = ( f1 f1 f2 - d = f1 f2 = f1 f2 /m e) + f2) 3d - d 2d or me = 1.5 = 5 When d is made 4 times. 0.9 3 F¢ = f1 f2 = f1 f2 =- f1 f2 16. 1 = 1 + 1 - d ( f1 + f2) - 4d 3d - 4d d F f1 f2 f1 f2 = - 2F To behave like concave lens, F should be negative. 22. 1 = (1.5 - 1) èçæ 1 - 1 ÷øö 24 R 2R So, d > 1 + 1 f1 f2 f1 f2 R = 6 cm \\ 2R = 12 cm or d > f1 + f2 f1 f2 f1 f2 23. Lens formula: or d > ( f1 + f2) or 30 cm 1 - 1 = 1 v1 -15 30 17. Condition of no emergence from opposite face is A > 2qC Þ v1 = - 30 cm 18. r2 = 0, r1 = A Mirror formula: m = sin i1 » i1 = i1 Þ i1 = mA v2 = - u2 = - (-45) = 45 cm sin r1 r1 A Lens formula: sin æç A + dm ÷ö 1- 1 = 1 v3 - 60 30 19. m = è 2 ø \\ v3 = + 60 cm sin (A/2) Hence, distance of final image from object Given, m = cot çèæ A ø÷ö = 60 - 15 = 45 cm 2 60 cm Solving, we get sin = 180° - 2A 20. At minimum deviation, r1 = r2 = A = 30° I3 I1 O I2 2 Now, m = sin i1 or 2 = sin i1 sin r1 sin 30° 15 cm 15 cm 45 cm Solving, we get i1 = 45° 24. n1 sin i1 = n3 sin i3 21. 1 = 1 + 1 - d \\ (1) (sin 45° ) = 2 sin i3 F f1 f2 f1 f2 1 2 = f1 + f2 - d Þ sin i3 = or i3 = 30° f1 f2 f1 f2 sin çæ A + dm ö÷ 1 = ( f1 + f2) - d F f1 f2 25. m = è 2 ø f1 f2 sin (A/2) + f2) \\ F = ( f1 - d …(i) For small angled prism, sin A » A 22 If d is doubled, focal length is doubled or \\ sin æç A + dm ö÷ = mA denominator becomes half. è 2ø 2 \\ ( f1 + f2) - 2d = 1 [( f1 + f2) - d ] With increase in m , sin æç A + dm ö÷ will increase. 2 è 2ø or ( f1 + f2) = 3d Hence, dm will increase. Substituting in Eq. (i), we have

Chapter 31 Refraction of Light — 453 26. n= sin i = 2 sin (i/2) cos (i/2) 33. m = sin i or 3 = sin i or i = 60° sin (i/2) sin (i/2) sin 45° 2 (1/ 2) Solving we get, i = 2 cos-1 (n/2) 27. If object is placed at focus of convex lens, rays i d 45º become parallel and they are incident normally on plane mirror. So, ray of light will retrace its path. 28. m = sin i Þ 2 = sin 45° or r = 30° sin r sin r 45º 45º \\ d = 60° - 45° = 15° A 30º 30º 30º C d 30º q d B q = 2d = 30° dTotal = dA - dB + dC 34. r2 = 0° and r1 = A = 30° = (45° - 30° ) - [180° - 2(30°)] + (45° - 30°) Applying, m = sin i1 = - 90° or | dTotal | = 90° sin r1 sin çæ A + dm ö÷ 2 = sin i1 è 2ø sin 30° 29. m = sin æç A ö÷ è 2ø Solving, we get i1 = 45° 30. 1 = èçæ 1 - 1ö÷ø ççèæ 1 - - 110 ö÷÷ø 35. 1 = 1 + 1 = 1 + 1 f 1.5 10 F f1 f2 10 10 \\ F = 5 cm Solving, we get f = - 15 cm 1 - 1 = 1 v 7.5 +5 31. Image formed by convex lens I1 should coincide at - C1, the centre of curvature C of convex mirror. \\ v = + 15 cm m = v = + 15 = - 2 u - 7.5 I1 OC I = mO = (- 2) (1 cm) = - 2 cm 12 cm 10 cm 2f 36. Applying, m2 - m1 = m2 -m1 vu R \\ For convex lens, 1 - 3/2 = 1 - 3/2 v + 30 + 20 1 - 1 = 1 + (10 + 2 f) - 12 + 10 or 1 = 1 - 3 Solving this equation we get, f = 25 cm v 20 40 32. For convex lens, \\ v = 40 cm 1 - 1 = 1 Þ v = - 30 cm 37. 1 = çæè 1.5 - 1÷öø æèçç 1 - - 150÷÷öø v - 12 + 20 f 1.4 30 - This image I1 is therefore, (30 + 10) cm or 40 cm \\ f = - 1050 cm towards left of plane mirror. Therefore, second image I2 (by the plane mirror) will be formed 38. Between 1 and 3 there is no deviation. Hence, 40 cm behind the mirror. m1 =m3

454 — Optics and Modern Physics MP = PQ tan r \\ MN = 2 (MP) = 2PQ × tan r Between 3 and 2 = 2´ 6 ´tan 28.8° = 6.6cm m = sin 45° = 4 sin r 3 Ray to light binds towards normal. Hence, + ve m 2 > m 3. 5. I1 O PI1 = 4.3 cm sin çæ A + dm ö÷ PQ è 2ø 39. A = dm = 60° Þ m = sin (A/2) Applying m 2 - m 1 = m 2 - m 1 two times Subjective Questions vu R 1. (a) m = 343 = 0.229 1.5 - 1.0 = 1.5 - 1.0 v1 - 2.5 + 10 1498 (b) qC = sin-1 (0.229) = 13.2° Solving, we get 2. t1 = l = l = m 1l Þ t2 = m 2l v1 » - 4.3 cm v1 c/m 1 e c Again applying the same equation, we get \\ t2 - t1 = (m 2 - m 1) l 1.0 - 1.5 = 1.0 - 1.5 c v2 - 24.3 - 10 = (1.63 - 1.47) (20) Solving, we get 3 ´ 108 v2 » - 85cm = 1.07´10-8 s or QI2 = 85 cm (towards left) 3. (a) t1 = l = l = n1l = minimum 6. + ve v1 c/ n1 c = 1.2 ´ 10-6 3 ´ 108 = 4 ´ 10-15 s Applying m 2 - m 1 = m 2 - m 1 we get, (b) l = l0 vu R n 1.0 - 1.5 = 1.0 - 1.5 Number of wavelengths across any film, - 1.0 -u - 2.0 N = l = ln Solving we get, l l0 u = 1.2 cm 4. m = sin i 7. For plane surface, happ = mh sin r = 3 ´ 10 = 15cm 2 M PN r This first image is at a distance, (15 + 3) cm from the plane mirror. So, mirror will make its second r image at a distance 18 cm below the mirror or 21 cm below the plane surface. Q Now further applying, 1.8 = sin 60° dapp = d = 21 = 14 cm sin r m 1.5 Solving, we get r = 28.8° below the plane surface.

Chapter 31 Refraction of Light — 455 8. CM = R sec qC = R cos qC O PQ I3 5 cm = cos 41.8° = 6.7 cm 8 cm 10 cm \\ PM = CM - CP = (6.7 - 5.0) = 1.7 cm 6 cm Applying happ for first refraction, 11. PI1 = 8m (towards left) QI1 = PQ + PI1 = (6 + 8m ) I II III IV \\ QI2 = QI1 = (6 + 8m ), towards right. PI2 = PQ + QI2 = (12 + 8m ) QP 5 cm 10 cm Applying dapp = d for third and last refraction we m have, PI 3 = çæ 12 + 8m ö÷ = 16 è m ø Apply m 2 - m 1 = m 2 - m 1 four times, \\ m = 1.5 We have vu R 9. 1.5 - 1.0 = 1.5 - 1.0 v1 ¥ + 10 i r \\ v1 = + 30 cm i 1.0 - 1.5 = 1.0 - 1.5 h v2 + 25 - 5.0 v2 = - 25 cm - 1.0 1.5 - 1.0 = 1.5 5.0 v3 - 35 - 2 cm v3 = - 11.67 cm 4 cm 1.0 - - 1.5 = 1.0 - 1.5 v4 16.67 - 10 m = sin i we get Applying, sin r Þ v4 = - 25 cm \\`Final image is at 25 cm to the left of P. 4 = 4 / 16 + h2 3 2/ 4 + h2 12. Applying m 2 = m 1 = m 2 - m 1 we have, vu R Solving this equation we get, h = 2.4 cm P 10. + ve qC 90° 1.0 - 1.6 = 1.0 - 1.6 R v - 3.0 - 5.0 qC Solving we get, v = - 2.42 cm C PM This is distance from P (downwards). qC = sin -1 çæ 1ö÷ = sin -1 æç 2ö÷ = 41.8° \\ Distance from observer = 5+ 2.42 è nø è 3ø = 7.42 cm

456 — Optics and Modern Physics 13. Bird or 1 = 1 - 1 f + 15 + 10 x Solving, we get f = - 30 cm y 17. L1 L2 Fish I1 Given, - dy = 4 cm/s 25 cm 15 cm dt 40 cm Distance of bird as observed by fish I1 is formed at second focus of L1 and first focus of L2. Z = y+mx = y+ 4 18. L1 L2 3x O I2 I1 \\ - dZ = - dy + 4 çèæ - dx öø÷ …(i) dt dt 3 dt 7.5 cm Given, - dZ = 16 cm/s dt Substituting in Eq. (i), we get 30 cm 30 cm 30 cm - dx = 9 cm/s dt 14. f = d2 - x2 (displacement method) For L1, 1 1 1 v - 30 + 20 4d - = f = 16 cm, x = 60 cm Substituting the values we get, \\ v = + 60cm d = 100 cm For L2 , 15. (a) 1 = (m - 1) çæèç 1 - 1 ø÷÷ö 1- 1 = 1 f R1 R2 v + 30 + 10 \\ f µ æèçç m 1 1÷öø÷ \\ v = 7.5 cm - 19. Object is placed at distance 2 f from the lens. or f1 = m2 - 1 = 1.7- 1.0 = 1.4 Hence, image is also formed at distance 2 f on f2 m1 - 1 1.5 - 1.0 other side. For mirror, (b) If refraction index of the liquid (or the 40 cm medium) is greater than the refraction index of 40 cm I2 I1 lens it changes its nature or converging lens O behaves as diverging. 5 cm 16. f = 20 cm f = 10 cm O 30 cm I 1 + + 1 = 1 v 10 - 10 15 cm Ray diagram is as shown below. 10 cm I2 I1 Using 1 = 1 - 1, f vu

Chapter 31 Refraction of Light — 457 20. For second lens, 24. qC = sin -1 çèæç m R ÷÷øö = sin -1 ççæè 43// 23 øö÷÷ = sin-1 (8/9) m D 1 - 1 = 1 Þ v = - 4 cm v -5 - 20 d2 10 cm 5 cm 5 cm Water 90° Glass qc I1 i = 0° i = qC f1 = 10 cm f2 = –20 cm f3 = 9 cm \\ d1 = 0° \\ d2 = 90° - qC 8 For third lens, cos d2 = cos (90° - qC ) = sin qC = 9 1 - - 1 9) = 1 \\ d2 = cos-1 æèç 8 ÷øö v (4 + +9 9 \\ v=¥ sin çæ A + dm ö÷ 21. See the result of sample example 31.30 25. Using m = è2 ø, sin çæ A ö÷ R= h è 2ø m2 -1 h= R m2 -1 Given, dm = A \\ = (1 cm) (5/3)2 - 1 = 4 cm 3= sin A 2sin A × cos A 3 = 22 sin A sin A 22. sin qC = 1 = 2 22 m1 3 Solving this equation, we get A = 60°. Now, m 1 sin i1 = m 2 sin i2 or m 1 sin qC = m 2 sin q 26. Condition of no emergence is, \\ çæè 23ø÷ö çæè 23øö÷ = èçæ 4 öø÷ sin q A > 2qC 3 \\ Amax = 2qC = 2sin-1 æçè 11.5ö÷ø or q = sin-1 æç 3 ö÷ = 84° è4ø 27. (a) At minimum deviation, 23. n1 M r1 = r2 = A = 30° qc 2 N q1 90° – qc Applying m = sin i1 or 1.5 = sin i1 We get sin r1 sin 30° n2 i1 = 48.6° n1 should be less than n2 for TIR at M. 60° sin qC = n1 n2 P At N , n1 sin q1 = n2 sin (90° - qC ) = n2 cos qC 48.6° 30° 30° Q 30° \\ sin q1 = n2 cos qC n1 R 60° sin q1 = n2 1 - sin2 qC 48.6° n1 = n22 - n12 dTotal = dP + dQ + dR n1

458 — Optics and Modern Physics = (48.6°-30° ) + (180°-2 ´30° ) + (48.6° - 30° ) 31. For blue light, = 157.2° m = sin i1 sin r1 (b) 4 sin (48.6° ) = 1.5 sin r1 3 or 1.68 = sin 65° sin r1 Solving we get, r1 = 41.8° r2 = 60° - r1 = 18.2° Solving this equation, we get r3 = 60° - r2 = 41.8° r1 = 32.6° \\ i3 = 48.6° r2 = A - r1 = 27.4° Hence, Again applying, dTotal = (48.6° - 41.8° ) + (180° - 2 ´ 18.2° ) m = sin i2 or 1.68 = sin i2 + (48.6° - 41.8° ) sin r2 sin 27.4° = 157.2° Solving this equation, we get 28. Dispersion power, i2 = 50.6° Now, dB = i1 + i2 - A w = mv - mr or dB = 65° + 50.6° - 60° = 55.6° …(i) my -1 For red light, 29. (a) w1 + w2 = 0 1.65 = sin 65° f1 f2 sin r1 \\ w2 = æèçç - f2 ø÷ö÷ w1 or r1 = 33.3° f1 \\ Now, r2 = A - r1 = 26.7° (- 30) sin i2 = - (+20) (0.18) = 0.27 1.65 = sin 26.7° (b) 1 = 1 + 1 = 1 - 1 \\ i2 = 47.8° F f1 f2 20 30 \\ dR = i1 + i2 - A or dR = 65° + 47.8° - 60° = 52.8° …(ii) or F = + 60 cm From Eqs. (i) and (ii), we get 30. w1 + w2 = 0 dB - dR = 2.8° f1 f2 \\ w1 = f1 LEVEL 2 w2 f2 Let 1-stands for flint glass and 2-stands for crown Single Correct Option glass. Then, = mh = 4 ´ = 8 f1 3 1. happ 3 2 3 m f2 2 = or | f1 | = 1.5 | f2 | Focal length of flint glass is more. So, its power is Distance from mirror = 1+ happ = 11 m. So, mirror 3 less. Combined focal length (and hence combined will make image at same distance (= 11/3 m from power) is positive. So, convex lens (converging itself). Now in third refraction, depth of second lens) should be made up of crown glass (having image, more positive power). d = 11 + 1 = 14 m 33 Now, 1= 1 + 1 F f1 f2 d 14 3 7 dapp = m = 3 ´ 4 = 2 m 1 1 1 \\ 150 = - 1.5 f + f The desired distance is therefore, Solving this equation, we get (dapp + h) f2 = f = 50 cm and or çæ 7 + 2ö÷ m or 11 m f1 = - 1.5, f = - 75 cm è2 ø 2

Chapter 31 Refraction of Light — 459 2. Applying, m 2 - m1 = m 2 - m 1 we get, 7. i1 = r1 = 0° vu R r2 = 90° - q = 53° 1.5 - 1 = 1.5 - 1.0 qC = sin -1 çæ 1 ÷ö = 37° v ¥ +R è m ø \\ v = + 3R Since, r2 > qC , T /R will take place on the face AB d = dAC + dAB 3. = 0 + 180° - 2r2 = 180° - 2 ´ 53° = 74° 60º 90º 30º 8. B 60º m = sin 60° = 3 x sin 30° 4. Refraction from first surface, y m2 - m1 = m2 -m2 F v1 µ +R = + y Refraction from second surface, h1 x m …(i) …(ii) m3 - m2 = m3 -m2 (as v2 = f ) h2 = y + mx f v1 -R Adding these two equations, we get Eq. (i) can be written as m3 = m2 -m1 - m3 -m2 mx + y = h1 or h2 = h1 or m = h2 fR R m m h1 Lens becomes diverging if f is negative or 9. MC = QC - QM = (R - 0.3) cm m3 -m2 >m2 -m1 PC 2 = MC 2 + PM 2 or m 3 + m 1 > 2 m 2 R2 = (R - 0.3)2 + (3)2 Same result is obtained if parallel beam of light is incident from RHS. P R 5. In first case, u = - 16 cm, then v = (+ 16n) cm \\ 1 - 1 = 1 …(i) Q C 16n - 16 f M In second case, u = - 6 cm, then v = - (6n) cm Solving this equation, we get R = 15 cm \\ - 1 - 1 = 1 …(ii) 6n -6 f Solving these two equations, we get 10. See the hint of Q.No-5 of the same section. In that f = 11 cm example, 6. dapp = d f = 16 + 6 = 11 cm (n1 /n2) 2 \\ d = n2 dapp 11. 1 = 1 + 1 n1 F1 f1 f2 - d (d ) = n2 çæ - d dapp ö÷ = ççèæ n2 ÷÷øö x 1=1+ 1- d dt n1 è dt ø n1 F2 f1 f2 f1 f2 pR2 F2 > F1 n1 dv = A é - d (d )úùû = x n2 So, image of distant object will be formed to the dt êë dt right of P.

460 — Optics and Modern Physics sin çæ A + dm ö÷ 18. (a) Two images are formed in case (i). 12. m = è 2 ø (b) One image is formed in case (ii). sin (A/2) (c) 1 = (m - 1) æçèç 1 - 1 ÷÷öø = - 2 (m - 1) …(i) f2 -R R R m and A for both rays are same. Hence, value of dm is also same for both rays. 1=1+ 1 f3 F1 F2 13. qC = sin -1 çæ 1 ö÷ è m ø 1)æç 1 1 ö÷ 1)èæçç 1 1 øö÷÷ = sin-1 çèæç 3 ÷÷öø = (m - è ¥ - Rø + (m - -R - ¥ …(ii) 7 = 40.9° From Eqs. (i) and (ii), we can see that f2 = f3. r2 = qC = 40.9° 19. 1 = 1 - 1 = 1 + 1 r1 = A - r2 = 19.1° f OB - OA OB OA m = sin i1 f = (OA) (OB) sin r1 AB + OB 7 = sin i1 \\ f = (OA) (OB) …(i) 3 sin r1 Now, AB Solving this equation, we get i1 = 30° AB2 = AC 2 + BC 2 14. If object is placed at focus of plano-convex lens, or (OA + OB)2 = AC 2 + BC 2 then it will make rays parallel. Now, these rays fall or OA2 + OB2 + 2 (OA) (OB) = AC 2 + BC 2 normal on plane mirror. So, they retrace their path. \\ (AC 2 - OC 2) + (BC 2 - OC 2) 15. If object is placed at centre of the sphere, then all + 2 (OA) (OB) = AC 2 + BC 2 rays starting from C fall normal on spherical surface and pass understand. Solving, we get (OA) (OB) = OC 2 16. sin qC = mR = 6/5 = 4 mD 3/2 5 Substituting in Eq. (i), we get r1 = 0° Þ r2 = < B = 90° - q f = OC 2 AB Now, r2 > qc or sin r2 > sin qC 20. Shift = èæçç1 - m medium ö÷ø÷ t \\ sin (90° - q) > 4 m slab 5 ëêé1 4/3ù or cos q > 4 …(i) = - 3/2 úû ´ 36 = 4 cm 5 cos 37° = 4 21. h = 21 or h = 21 5 m 2 (4 /3) 2 From Eq. (i), we see that Solving, we get h = 14 cm. q < 37° 22. m = - 2cm = - 2 17. From the first refraction, rays should become 1cm parallel. Or, image should formed at infinity. Þ |v| = m |u| Applying, Object y m2 - m1 = m2 -m1 uu R x = 50 cm 3/2 - 1.0 = 3/2 - 1.0 x = – 40 cm O ¥ -x + 10 Solving, we get x = –10 cm Image x = 20 cm 30 cm 60 cm

Chapter 31 Refraction of Light — 461 23. Using the equation, YI2 = (5 mm) æèç 1 ÷öø = 2.5 mm 2 1 = 2 (m 2 /m 1) - 2 (m 2 /m 1 - 1) F R2 R1 = 0.25 cm 12 we get 28. Ð i = 0° for all values of q, as the rays fall normal 1 = 2 (1.5) - 1 (1.5 - 1) æç as F = R ÷ö to sphere at all points. è 2ø - 10 ¥ R1 29. Cavity of placed at centre. Hence, image of cavity Solving, we get R1 = 10 cm. is also formed at centre, as all rays fall normal to surface at all points. (so pass undedicated). 24. In air, object lies between F and 2F. In liquid, focal length will become 4 times. So, object will now lie between optical centre and focus. 25. Condition of no emergence, A > 2qC \\ qC < A or 45° Now, m = çèæç m 1 ø÷÷ö çæ v ÷ö 2 m 2 è uø \\ sin qC or 1 < sin 45° or 1 (4 /3) æçèç 1 m öø÷÷ m 2 1 1 m = For no emergence, m > 2 26. t = nl = n l0 çæ as l = l0 ö÷ =4 m è m ø 3 30. 1 = (1.5 - 1) æççè 1 - - 120÷÷øö f 20 Þ f = + 20 cm or t µ 1 1 - 1 = 1 m v1 - 30 + 20 \\ m2 = t1 \\ v1 = + 60 cm m1 t2 + 60 m1 = v1 = - 30 = - 2 u1 or m2 = t1 m1 t2 For second lens, = æèç 5 øö÷ èæç 4 öø÷ = 5 1- 1 = 1 4 3 3 v2 - 60 + 20 27. For concave lens, v2 = + 30 cm (+ 30) 1 y 5 cm m2 = v2 = (- 60) = - 2 u2 I2 5 mm m = m1 m2 = 1 O I1 Final image is 30 cm to the right of second lens or 150 cm to the right of first lens. m =1 20 cm 10 cm Hence, image height = object height 1 1 1 = 3 mm v - 10 10 = = - Þ v = - 5 cm 31. Apply m 2 - m 1 = m 2 - m 1 (- 5 cm) 1 vv R (- 10 cm) 2 m = v = = We get 1 - 2 = 1- 2 u v - 10 - 10 X I2 = 20 + 5 = 25 cm Solving we get, v = - 10 cm

462 — Optics and Modern Physics 32. O I1 If angle of q is less than this value, then angle of 2F incidence at N will be greater than qC. Hence, TIR 2F will take place at N . 20 cm 20 cm 37. Reflection from the concave mirror, I1 I1 2 cm 33. 1 = (1.5 - 1) çæ 1 - 1 ö÷ 15 è R ¥ ø 1 cm \\ R = 7.5 cm O With polished surface, 1 1 1 v -1 -2 1 = 2 (m 2 /m 1) - 2 (m 2 /m 1 - 1) + = Þ v = + 2 cm F R2 R1 = 2 (1.5) - 2 (1.5 - 1) Shift due to slab = çæ1 - 1 ÷ö t - 7.5 ¥ è nø F = - 2.5 cm = (1 - 2/3) (9 cm) = 3 cm Now, using mirror formula, 1 + 1 = 1 Now, slab will make next image at a distance 3 cm v - 20 - 2.5 from I1 in the direction of ray of light, i.e. at O itself but it is virtual, as the ray of light has crossed Solving, we get v = - 20 cm the slab and we are making image behind the slab. 7 38. m2 m1 = æç1 - 1 ö÷ = æç1- 23øö÷ (6) = 2 cm 34. Shift è m ø t è For mirror, object distance = 50 - 2 = 48 cm. So, mirror will make image at a distance 48 cm m2 - m1 = m2 -m1 behind it. In return journey of ray of light, we will v u -R have to farther take 2 cm shift in the direction of ray of light. \\ m2 = m1 -m2 + m1 vR u So, image distance as observed by observer If m 1 > m 2 and u is positive (i.e. virtual object), = (50 + 48) - 2 = 96 cm then v is always positive or image is always real. 35. Using 1 - 1 = 1 , we get 39. 1 = 1 + 1 vu f F f1 f2 + 1 40) - - ( f 1 10) = 1 (f + + +f Solving this equation, we get f = + 20 cm 36. Applying Snell’s law at point M, we get n1 = sin q or sin q = n1 cos qC æèçç 1 1 öø÷÷ çèæç 1 - 120÷øö÷ sin (90° - qC ) ¥ 10 10 = (1.6 - 1) - + (1.5 - 1) - n22 - - = n1 1 - sin2 qC = n1 1- n12 = n12 - n22 Solving we get, F = 28.57 cm 90°– q M q 40. q N

1.5 - 1.0 = 1.5 - 1.0 Chapter 31 Refraction of Light — 463 v -u +R For n = 2m + 1, it is just like an identical prism of larger size. or 1.5 = - 1 + 1 v u 2R For image to real (for negative value of u) v should be positive. Hence, 1 < 1 or u > 2R. u 2R 41. ¥ 2. sin i1 = v1 2F F O sin i2 v2 \\ sin r = slope = 1 = vy sin i 3 vx When object moves towards F to O virtual erect \\ vy = vx and magnified image moves from ¥ to O. 3 42. 1 = 1 + 1 - d2 Speed of light in medium- y is less. So, it is denser. TIR takes place when ray of travels from denser to F f1 f2 f1/ f2 rarer medium. f1 and f2 both are negative. Hence, F is also 3. (a) If vacuum speed of light of all colours is same. negative. Object is real so combined lens (having negative focal length) will always make, its virtual image. 43. Using m 2 - m 1 = m 2 - m 1 we get (b) PF uu R ¥ 3/2 - 1.0 = 3/2 - 1 If object moves from ¥ to P, then its virtual, µ -x + 60 erect and diminished image move from F to P. \\ x = 120 cm d Hence for x = 120 cm, rays of light become (c) d0 parallel to principal axis and fall normal to polished surface. Hence, rays retrace their path. 44. 1.6 - 1.0 = 1.6 - 1.0 i v1 -2 + 1.0 I1 I2 Þ v1 = 16 m 4. Displacement method of finding focal length of 2.0 - 1.0 = 2.0 - 1.0 convex lens. v2 -2 + 1.0 5. Deviation, d = (m - 1) A = (1.5 - 1) 4° = 2° \\ v2 = 4 m (a) 2º I2 I2 = v1 - v2 = 12 m 2º NM More than One Correct Options 1. For n = 2m, it is just like slab. 2º \\ Deviation = 0 P To rotate ray MP by 2° (to make it parallel to MN ) we will have to rotate the mirror only by 1°.

464 — Optics and Modern Physics (b) Without prism, ray of light was falling normal \\ 1=1+ 1 to plane mirror. So, ray of light was retracing vu f its path prism has deviated it 2°. So, if we rotate the mirror by 2°, ray of light further falls Between O and F2 or between F2 and 2F2, u is normal to plan mirror and retraces its path. positive. So, v is also positive. \\ m = u is positive v Comprehension Based Questions Therefore, image is real (as v is positive) and erect (as m is also positive). 1. Using the mirror formula, 1 = 1 + 1 = 1 - 1 f v u - 40 10 1 1 1 2. (a) v - u = -f Solving we get, f = - 8 cm \\ 1=1- 1 uu f 2. Focal length of lenses is 1 = (1.8 - 1) çæçè 1 - - 1 ø÷÷ö Between O and F1, u is positive and less than F ¥ 2R f . So, v is positive (therefore image is real). + (1.2 - 1) ççæè - 1 - 1 ö÷÷ø Further from 2R -R m= v F = 2R u 2R We can see that m is allow positive. So, it is R erect also. (b) 1- 1 = 1 v u -f Now combined power of system, \\ 1=1- 1 vu f P = 2PL + Pm or - 1 = 2PL + Pm Between F1 and 2F1, m positive and greater than f . f So, v negative (therefore image is virtual). or 1 = 2 çæ 1 ÷ö + 2 Further from 8 è 2R ø R m= v u or R = 24 cm \\ Radius of curvature of common surface We can see that m is negative so, image is inverted. = 2R = 48 cm 3. (a) and (b) 3. Combined focal length of lens, v = m2 u 1 çæçè 1 + 148÷öø÷ m1 f = (1.2 - 1) + 24 - Þ |v| < |u| + (1.8 - 1) çèçæ + 1 - 1 ÷ø÷ö as m 1 > m 2 48 ¥ i.e v and u are of same sign. =1 Or they are on same side of plane surface. 48 From plane surface, if object is real, image is Now, combined power of system virtual and vice-versa. \\ - 1 = 2 ççæè 1 ÷öø÷ + ¥ = 2 æç 1 ö÷ (c) and (d) F f è 48ø v = m1 u m2 \\ F = - 24 cm Þ |v| > |u| Match the Columns Other explanations are same. 1. (c) and (d) 1 - 1 = 1 4. (a) m - 1 = m -1 vu f v -u -R

\\ m = -1-m -1 Chapter 31 Refraction of Light — 465 vu R 2. The paths of the rays are shown in figure. Since, Therefore, v is always negative. Or image is the virtual image is diminished, the lens is concave. always virtual. (b) m - 1 = m -1 v +u -R A B or m = 1 - m - 1 A¢ C vu R F So, v may be positive or negative. Hence, image may be real or virtual. Same logic can be applied for two options. For them R is positive. In option (c), u is negative 3. (a) OA = 8.0 cm and in option (d) u is positive. \\ AI1 = (ng)(OA) = çæ 8ö÷ (8.0) = 12.8 cm 5. 1 or P1 = (1.5 - 1) èççæ 1 - 1 ÷øö÷ …(i) è 5ø R1 R2 f1 For refraction at EG (R = ¥ ), using 1 or P2 = çæèç 1.5 - 1÷øö÷ çæçè 1 - 1 ÷ö÷ø …(ii) CE f2 me R1 R2 Dividing Eq. (ii) by Eq. (i), we get P2 = æèçç 3 - 2÷øö÷ P1 …(iii) me (a) m e = 1.4, then P2 is positive and less than P1. O AB F Other options can be checked from Eq. (iii). 6. Concave lens can make only virtual, erect and DG +ve diminished images of real objects. Convex lens can make real, inverted and n2 - n1 = n2 - n1 diminished size or real inverted and magnified or vu R virtual, erect and magnified images of real objects. 4/3 8/5 Rest type of cases are possible with virtual, \\ BI 2 - -(12.8 + 3) = 0 objects. Subjective Questions \\ BI2 = - (15.8)(4 /3)(5/8) = - 13.2 cm Ans. \\ FI2 = 13.2 + 6.8 = 20.0 cm 1. First draw a ray AA¢ until it intersects with the (b) For face EF principal optical axis and find the centre of the lens C. Since, the virtual image is magnified, the lens is CE convex. A¢ B O AB F A CF D G +ve Draw a ray AB parallel to the principal optical axis. 8/5 - 4/3 = 0 BI 1 -6.8 It is refracted by the lens so that it passes through its \\ BI1 = - (6.8)(8/5)(3/4) focus and its continuation passes through the virtual = - 8.16 cm image. The ray A¢ B intersects the principal optical axis at point F, the focus of the lens.

466 — Optics and Modern Physics For face CD Therefore, we will have to shift the screen a distance x = v2 - v1 = 60 cm away from lens. Ans. 1.0 - 8/5 = 0 AI 2 -11.16 1 èæçç 1 1 øö÷÷ 7. 40 = (1.5 - 1) 120 + R1 \\ AI2 = - (11.16)(5/8) = - 6.975 cm Solving we get, R2 = 24 cm \\ FI2 = 8 + 6.975 Applying lens formula, for L2 = 14.975 cm Ans. O 4. The system behaves like a mirror of x focal length given by n1 n2 1 = 2(n2 /n1) - 2(n2 /n1 - 1) L2 F R2 R1 Substituting the values with proper 10 cm sign. 1 = 2 ´ 4/3 (Q R1 = ¥ ) L1 R1 F -20 R2 or F = - 7.5 cm Ans. i.e. system behaves as a concave mirror of focal 1 + 1= 1 length 7.5 cm. v1 x 20 …(i) 5. 1 = (n - 1) çæçè 1 - 1 ø÷÷ö Using m 2 - m 1 = m 2 - m 1 for unsilvered side of f R1 R2 vu R L1. 1 = (1.5 - 1) æççè 1 - 1 ÷÷øö R1 n R2 1.5 1.0 1.5 – 1.0 10 R1 -10 12 -120 - v1 - 10 = 24 …(ii) \\ R1 = + 10 cm Solving Eqs. (i) and (ii), we get x = 10 cm Ans. Now using, 8. Case I 1 + 1 = 1 1 + 1 = 2(n2 /n1) - 2(n2 /n1 - 1) vu R2 R1 vu f Substituting the values, \\ 1- 1= 1 v1 30 -10 1 - 1 = 2(1.5) - 2(1.5 - 1) v -15 -10 +10 \\ v1 = - 15 cm \\ v = - 2.14 cm Ans. Case II Shift = çæ1 - 1 ÷ö t = èçæ1 - 11.5 ÷øö 6 = 2 cm è m ø 6. Using lens formula, 1 - 1 = 1 \\ 1 - 1 = 1 vu f v2 28 -10 1 - 1 = 1 \\ v2 = - 15.55 cm v1 -40 30 \\ Dv = 0.55 cm Ans. \\ v1 = 120 cm 9. Using m 2 - m 1 = m 2 - m 1, twice with u = ¥, Shift due to the slab, vu R Dx = æç1 - 1 ÷ö d = çæ1 - 11.8 ö÷ø 9 = 4 cm we have è m ø è 1.5 1.5 – 1.4 \\ u¢ = - (40 - Dx) v1 = +20 …(i) = - 36 cm 1.6 1.5 1.6 – 1.5 v2 v1 -20 \\ 1 - 1 = 1 - = …(ii) v2 -36 30 \\ v2 = 180 cm Solving Eqs. (i) and (ii), we get f = v2 = ¥ ,

Chapter 31 Refraction of Light — 467 i.e. the system behaves like a glass plate. qC = sin -1 æç 1 ÷ö = 45° è m ø 10. First image will be formed by direct rays 1 and 2, etc. Applying sine law in DCPM, PI 1 = PO =5 = 3.33 cm Ans. CP = CM + m 1.5 sin qC (90° sin r) Second image will be formed by reflected rays \\ CP = R [R = radius] 3 and 4, etc. (1/ 2) cos r \\ CP = 2 R 3 As we move away from C, angle PMC will P increase. Therefore, CP >| 32R. Same is the case 5cm 2 1 on left side of C. O 3 5cm 3 13. 1 = (1.5 - 1) æç 1 - 1 ö÷ 20 è R -R ø Object is placed at the focus of the mirror. Hence, \\ R = 20 cm I2 is formed at infinity. Applying m 2 - m 1 = m 2 - m 1 twice with the 11. (a) Applying Snell's law at D, vu R 4 sin = 3 sin 30° condition that rays must fall normally on the 3 2 i concave mirror. \\ i = 34.2° Ans. 1.5 - 1.2 = 1.5 – 1.2 …(i) v1 - 40 + 20 …(ii) A 2.0 - 1.5 = 2.0 – 1.5 d - 80 v1 –20 30° DE Solving Eqs. (i) and (ii), we get i 30° 30° i d = 30 cm Ans. and v1 = - 100 cm The ray diagram is as shown in figure. B C 90° (b) d = dD + dE = 2dD I1 I2 = 2(i - 30°) = 8.4° Ans. 100 cm 40 cm 12. 2 = sin 45° 30 cm 80 cm sin r 14. Using lens formula, 45° CP 1- 1 = 1 36 -45 f r qC \\ f = 20 cm M In the second case, let m be the refractive index of the liquid, then r = 30°

468 — Optics and Modern Physics 1 - 1 = 1 Note A ray passing through O and then O¢ goes 48 - çæ 5 + 20 40 ÷ö undeviated. è m ø Therefore I1 and I2 both should be on this line, Solving this, we get m = 1.37 which is also the x-axis. That’s why for final Ans. image we have taken projection of O ¢I2¢ on x-axis. 15. As the angles are small we can take, 17. Since, the vessel is cubical, ÐGDE = 45° A and GE = ED = h (say ) then 90° EF = ED - FD = (h - 10) Further, 4 = sin 45° 3 sin r A A A + 1°15¢ \\ r = 32° 6°30¢ 2A Eye A C B C 45° G Now, sin q » q r m = A + 1°15¢ h A = 6°30¢ B E 45° D 2A F Solving this equation, we get Now, EF = tan r = tan 32° \\ GE A = 2° Ans. h - 10 = 0.62 and m = 1.62 h 16. Using lens formula for L2, Solving this, we get P h = 26.65 cm Ans. I2N (say) 18. BO = OC \\ ÐOBC = ÐBCO = r L1 I1N y A iB f/2 L2 x I2 r O 60° I1 ON ir O C D f f PN 1 - - 1 = 1 Let angle of incidence be i, v f /2 f or v = - f i = r + r = 2r (external angle) This f length will be along PP¢ from point O ¢ (towards P). \\ m = sin i = sin 2r » 2r = 2 Ans. sin r sin r r \\ O¢I2¢ = f On x-axis this distance will be f sec 60° = 2 f . 19. dTotal = dRefraction + 2dReflection + dRefraction Since, OO ¢ = 2 f , therefore image will be formed at or d = (i – r) + 2(180° - 2r) + (i - r) = 360° + 2i - 6r origin.

Chapter 31 Refraction of Light — 469 = 360° + 2i - 6 sin -1 æç sin i ÷ö So, length of final image è m ø A3 B3 = 1 A2 B2 = 2 mm For deviation to be minimum, dd = 0 3 di Point B2 is 2 mm below the optic axis of second half By putting first derivative of d (w.r.t. i) equal to lens. Hence, its image B3 is formed 2/3 mm above zero, we get the desired result. the principal axis. Similarly, point A2 is 8 mm below the principal axis. Hence, its image is 8/3 mm 20. (a) For refraction at first half lens ççæè 1 - 1 = 1 ÷÷öø above it. Therefore, image is at a distance of 20 cm v u f behind the second half lens and at a distance of 2/3 mm above the principal axis. The size of image is 2 1 - 1 = 1 mm and is inverted as compared to the given object. v -20 15 Image formed by second half lens is shown in figure (c). \\ v = 60 cm Magnification, m = v = 60 = -3 A3 u -20 B3 The image formed by first half lens is shown in O¢ O ¢¢ B2 figure (a). AB = 2 mm, A1B1 = 6 mm, AO1 = 20 cm, O1F = 15 cm and O1A1 = 60 cm. B A1 A2 (c) F C1 (b) Ray diagram for final image is shown in A O1 B1 figure (d). (a) 4 mm B A3 2 mm A1 B3 Point B1 is 6 mm below the principal axis of the A B2 lenses. Plane mirror is 4 mm below it. B1 B2 A2 2 mm (d) C2 21. (i) PO = OQ 4 mm \\ ÐOPQ = ÐOQP = r (say) A2 iP (b) h r Radius = 0.1 m Hence, 4 mm length of A1B1 (i.e. A1C1) acts as i rQ real object for mirror. Mirror forms its virtual RO i image A2C2. 2 mm length of A1B1 (i.e. C1B1) acts as virtual object for mirror. Real image C2B2 is formed of this part. Image formed by plane mirror is shown in figure (b). Also, i = r + r = 2r In DPOR, h = OP sin i = 0.1 sin i For the second half lens, 1 - 1 = 1 v -60 15 or = 0.1 sin 2r Also, h = 0.2 sin r cos r \\ v = + 20 3 = sin i = 2 sin r cos r …(i) m = v = 20 = - 1 sin r sin r u -60 3

470 — Optics and Modern Physics = 2 cos r Further, 1 - m = 1-m BI 2 (AI1 - -R \\ r = 30° R) Substituting in Eq. (i), we get Solving this equation, we get h = 0.2 ´ 1 ´ 3 BI 2 = - 2R(4m - 1) 2 2 3m - 1 = 0.086 m + ve Hence, height from the mirror is 0.1 + 0.086 = 0.186 m (ii) Use the principle of reversibility. QS C O AB Eye Oi R 2R M i = 2r = 60° \\ Distance between the final image and object is QS = cot i = cot 60° = 1 Now, MS 3 d = 3R - 2R(4m - 1) 3m - 1 \\ QS = MS = 0.1 = (m - 1)R Hence proved. 33 (3m - 1) \\ The desired distance, 24. dTotal = dP + dQ OC = 2 ´ 0.1 + 2 ´ 0.1 3 = 0.315 Ans. a 22. Vi = rS = r = 1 P Q V rL 2r 2 ir ri b i.e. half the sphere is inside the liquid. For the O image to coincide with the object light should fall normally on the sphere. Using m 2 - m 1 = m 2 - m 1 twice, we have vu R 3/2 - 1 = 3/2 - 1 \\ a = (i - r) + (i - r) v1 -8 +2 or i - r = a …(i) \\ v1 = 12 cm 2 Further, 4/3 - 3/2 = 4/3 - 3/2 Further, in DOPQ, r + r + b = 180° h - 10 8 -2 \\ r = 90° - b …(ii) 2 Solving this equation, we get h = 15 cm Ans. From Eq. (i), 23. We have to see the image of O from the other side. i = r + a = 90° + æç a - b ö÷ …(iii) 2 è 2ø Applying, m 2 - m 1 = m 2 - m 1 twice, we have vu R sin é 90° + çèæ a - b ÷øö ù êë 2 ûú m - 1 =m -1 m = sin i = AI1 -2R -R sin r æç 90° b ö÷ sin è - 2 ø \\ AI 1 = 2 mR 1- 2m

Chapter 31 Refraction of Light — 471 cos çèæ b - a ÷öø Further, 1- m = 1-m 2 \\ = v2 -(3R - v1) R/2 cos æç b ö÷ è 2ø v2 = R(9 - 4m ) Ans. (10 m - 9)(m - 2) or cos çæ b - a ö÷ = m cos b è 2ø 2 Hence proved. Final image is real if, v2 > 0. 25. q = 90° - i As 10m - 9 is always positive (m > 1). Therefore, for v2 > 0, either (9 - 4m ) and (m - 2) both should y be greater than zero or both should be less than P (x, y) zero. For the first condition (when both > 0) i 2 < m < 2.25 and for the second condition (when both < 0), m < 2 and m > 2.25 which is not possible. Hence, m should lie between 2 and 2.25. x 27. For the lens, u = - 2.0 m, f = + 1.5 m \\ 1 - 1 = 1 v -2.0 1.5 tan q = cot i or dy = cot i or v = 6.0 m dx …(i) m = 6.0 = - 3.0 m 0 sin i0 = m P sin iP –2.0 (i) sin 90° = ( 1 + ay) sin i I2 90°– q \\ sin i = 1 0.3 m q q M 1 + ay O N Pq \\ cot i = ay = dy I1 dx d y dy x y 6.0 m ò ò\\ = dx or x=2 0 ay 0 a Therefore, y-coordinate of image formed by lens is Substituting y = 2 m, m(0.1) = - 0.3 m. and a = 2.0 ´ 10-6 m-1 0.3 = tan q = 0.3 We get xmax = 2000 m = 2 km Ans. NP 26. Applying, m 2 - m 1 = m 2 - m 1 twice, we have \\ NP = MP = 1.0 m vu R or d = 6.0 – 1.0 m 1 m -1 = 5.0 m Ans. v1 -2R R - = and x-coordinate of final image I2 is x = d - 1.0 = 4.0 m 2 mR Ans. 2m - 3 \\ v1 =

32. Interference and Diffraction of Light INTRODUCTORY EXERCISE 32.1 4. A1 = 3A2 1. I1 = 9, I max = çèæç I1/I2 + 11öø÷÷ 2 So, I1 = 9I2 I2 16 I min I1/I2 - Let A2 = A0 and I2 = I0 Then, A1 = 3A0 and I1 = 9I0 2. (a) Amax = A1 + A2 Amax = (A1 + A2) = 4 A0 Amin A1 - A2 and Imax = 16I0 Now, I max çæçè Amax øö÷÷ 2 I = I1 + I2 + 2 I1I2 cos f I min Amin (b) = 3. l path difference is equivalent to 90° phase = 9I0 + I0 + 2 9I0 ´ I0 cos f 4 = 10I0 + 6I0 cos f difference = 10 ´ I max + 6 çèæ I max öø÷ cos f 16 16 Anet = 5A0 Þ Imax = 25 I0 = 5 + 3 I max cos f 8 I max 8 Ö2A0 3A0 = 5 + 3 I æç 2 cos2 f - 1ö÷ 8 I max 8 è 2 ø max 45° Ö2A0 = 1 + 3 I max × cos2 f 4 I max 4 2 INTRODUCTORY EXERCISE 32.2 = Imax çæ1 + 3cos2 fö÷ 4 è 2ø cos2 f 5. (a) I = I max cos2 f 2 2 3. I = I max = I0 cos2 30° (as f = 60° ) 3 I max = I max cos2 f or cos f = 3 = 3 I0 4 2 22 4 \\ f=p (b) 60° phase difference is equivalent to l path 26 6 p 2p 2p \\ f= 3 = çæè l ÷øö (Dx) = l æçè yd öø÷ difference. D 6. = 2a cos f lD (600 ´ 10-19 ) (1.2) aR 2 \\ y = 6d = (6) (0.25 ´ 10-2) (i) For aR = 2a, f = 0° = 48 ´ 10-6 m = 48 mm (ii) For aR = 2a, f = 90° etc Exercises LEVEL 1 Substituting in Eq. (i), we get Assertion and Reason I = I0 Þ Dx = çèæ l ÷øö f 2p f 1. I = 4I0 cos2 2 …(i) For f= 2p , Dx = l 3 3 f = 2p or 120° Þ f = 60° 32 2. The whole fringe pattern will shift upwards.

Chapter 32 Interference and Diffraction of Light — 473 3. No reflected ray reaches below O. Objective Questions 4. Imax = ( I1 + I2 )2 and 1. A = (8)2 + (6)2 = 10 mm Imin = ( I1 - I2 )2 6 mm 6 mm A Þ When I1 = I2 = I0, then Imax = 4I0 and Imin = 0 When slit of one width is slightly increased, then 4 mm 12 mm 8 mm intensity due to that slit becomes greater than I0 . 2. I1 = b2 In that case, we can see that I2 Imax > 4I0 and Imin > 0 So let, I2 = 1 unit, then I1 = b 6. Locus of points of equal path difference in the Imax = ( I1 + I2 )2 = (1 + b)2 shown case is circle. Imin = ( I1 - I2 )2 = (1- b)2 7. At centre, path difference is maximum and this is = Imax - Imin = 4b equal to S1S2. Then, path difference decreases as Imax + Imin = 2(1 + b2) we move away on the screen. So, order of fringe also decreases. Hence, 11th order maxima occurs before 10th order maxima. 8. At points P and R \\ The asked ratio is 2b . 1+ b2 | Dx | = S1S2 = 4 l, therefore maxima. 3. d sin q= l Q 2 RP \\ q = sin -1 çæè l öø÷ = sin -1 çæçè 5460 ´ 10-10 ø÷÷ö S1 S2 2d 2 ´ 0.1 ´ 10-3 = 0.16° S 4. 6th dark fringe distance in vacuum = 10th bright At Q and S fringe distance in liquid. Dx = 0, therefore again maxima. Then, three \\ 5.5 w = 10 w¢ maxima in each quadrant (between P and Q or between Q and R etc.) corresponding to or w =m = 10 = 1.81 | Dx | = l, 2l and 3l. Therefore, there are total w¢ 5.5 16 maxima. 5. Shift = (m - 1)tD d In each quadrant there are four minima = (1.5 - 1) (10 ´ 10-6) (1.0) corresponding to | Dx | = 0.5l, 1.5l, 2.5l and 2.5 ´ 10-3 3.5l. Hence, there are total 16 minima. = 2 ´ 10- 3 m 9. d sin q = 2l (for second order maxima) = 2 mm sin q = 2 l = 2 ´ 1 = 1 d 42 6. Distance = w = (lD /d) = lD \\ q = 30° 2 2 2d For maxima, d sin q = nl 7. Path difference at centre is always zero. Hence, all \\ nmax = d = 4 (for q = 90°) wavelengths under all conditions always interfere l constructively. So, there are total 7 maxima corresponding to 8. w = lD or w µ l n = 0, ± 1, ± 2 and ± 3. d We cannot take n = 4, as it is for q = 90°, which is out of screen. w will increase 6000 4000 10. Shift = (m - 1) tD is independent of l. or 1.5 times. d Hence, number of fringes in same distance will decrease 1.5 times.

474 — Optics and Modern Physics 9. w = lD 1m 4m d A P1 B Þ Dw = l(DD ) 4m 1m d A P2 B \\ l = (d) (Dw) = (10-3) (3 ´ 10-5) At P1 BP1 - AP1 = 3m = l DD 5 ´ 10-2 2 = 0.6 ´ 10-6 m = 6000 Å At P2 AP2 - BP2 = 3m = l 2 10. Imax = ( I1/I2 + 1)2 = 49 I min ( I1/I2 - 1)2 9 3 ´ 108 3. Q l= c = 120 ´ 106 = 2.5 m Solving, we get I1 = 25 f I2 4 x (9 – x) 11. y = 3lD AP B d Dx = (BP - AP) = (9 - 2x) = nl \\ l = yd = (7.5 ´ 10-3) (0.2 ´ 10-3) \\ x = 9 - nl = 9 - 2.5n 3D (3) (1) 22 = 500 ´ 10- 9 m = 500 nm Now, substituting n = 1, 2,L etc. 12. n1l1D = n2l2D We can find different values of x. dd x1 = 3.25 m for n = 1 x2 = 2.0 m for n = 2 or n1 = l1 = 5200 = 4 n2 l2 6500 5 and x3 = 0.75 m for n = 3 \\ 4th maxima of l1 coincides with 5th maxima Similarly, we will get three points at same distance of l2. from other point B. ymin = 4 l1D 4. Q w = lD d d = 4 ´ 6500 ´ 10-10 ´ 1.2 \\ l = wd = (2.82 ´ 10-3) (0.46 ´ 10-3) 2 ´ 10-3 D 2.2 = 1.56 ´ 10- 3 m = 0.156 cm = 0.589 ´ 106 m 13. Only fringe pattern will shift. Number of fringes » 0.590 nm on screen will remain unchanged. 5. q = l = 500 ´ 10-9 radian d 2.0 ´ 10-3 Subjective Questions » 0.014° 1. Amax = 5 + 3 = 8 units 6. Wavelength in water, l¢ = l m Amin = 5 - 3 = 2 units Amax = 4 Fringe width, w= l¢D = lD Amin d md \\ Imax = (4)2 = 16 = (700 ´10- 9 )(0.48) I min (4 /3) (0.25 ´ 10-3) 2. (a) At centre path difference is zero. Therefore, = 10-3m construction interference will be obtained. = 1 mm (b) l = 3 m. At a distance, where path difference 7. Distance = 3l2D - 3l1D 2 is l or 3 m destructive interference will be dd = 3(l2 - l1)D 2 obtained. d

Chapter 32 Interference and Diffraction of Light — 475 = 3 ´(600 - 480) ´ 10-9 ´ 1.0 12. (a) Dx = yd 5.0 ´ 10-3 D = 7.2 ´ 10-5 m and f = 2p × Dx l = 0.072 mm (2p) yd 8. The required distance = one fringe width = lD = w = lD = (2 ´ 180) yd degree d lD = (500 ´ 10-9 ) (0.75) = 360 ´ 0.3 ´ 10-3 ´ 10 ´ 10-3 (0.45 ´ 10-3) 546 ´ 10-9 ´ 1.0 = 8.33 ´ 10-4 m = 1978° = 0.83 mm cos2 f 2 9. y = w1 = w2 Now, I = I0 2 = 2.97 ´ 10- 4I0 \\ l1D = l2D d 2d » 3.0 ´ 10- 4I0 (b) w = lD Þ l2 = 2l1 = 1200 nm l d 10. d sin q1 = 2 l 550 ´ 10-9 \\ Number of fringes between central fringe and P 2d 2 ´ 1.8 ´ 10-6 sin q1 = = N = y = yd w lD \\ q1 = 8.78° = (10 ´ 10-3) (0.3 ´ 10-3) (546 ´ 10-9 ) (1.0) y1 = tan q1 D = 5.49 \\ y1 = D tan q1 = 35tan 8.78° So, bright fringes are five. = 5.41 cm 13. Shift = (m - 1) tD d sin q2 = 3l d 2 S1 = (1.6 - 1) (10 ´ 10- 6) (1.5) 1.5 ´ 10-3 \\ sin q2 = 3l = 3 ´ 550 ´ 10-9 2d 2´ 1.8 ´ 10-6 = 0.6 ´ 10-2 m \\ q2 = 27.27° = 0.6 m \\ y2 = D tan q2 S2 = (1.2 - 1) (15 ´ 10-6) (1.5) = 35tan 27.27° 1.5 ´ 10-3 = 18 cm = 0.3 ´ 10-2 m Dy = y2 - y1 = 12.6 cm = 0.3 cm 11. l(in Å) = 150 = DS = S1 - S2 = 0.3 cm = 3 mm E(in eV) 14. l(2D ) = (m - 1)tD (fringe width = shift) (de-Broglie wavelength of electron) dd = 150 = 1.22 Å \\ l = (m - 1)t 100 2 w = lD = (1.22 Å) (3m) = (1.6 - 1) (1.964 ´ 10-6) m d (10 Å) 2 = 0.366 m = 0.5892 ´ 10-6 m = 589 nm = 36.6 cm

476 — Optics and Modern Physics 15. Q d = 2 cm For n = 4, l = 424 nm For n = 5, l = 329 nm w = lD = (500 ´ 10-9 ) (100) d 2 ´ 10-2 Therefore, two wavelengths lying in the given range are 424 nm and 594 nm. = 2.5 ´ 10-3 m = 2.5 mm (b) Ray-1 and ray-2 are in same phase. S nt dP 12 S¢ Hence, In this figure, P is in a dark fringe, as conditions of maxima and minima are interchanged. Dx = 2nt = ml (for maximum intensity) Hence, next dark fringe will be obtained at a distance w or 2.5 mm from P. \\ l = 2nt = (2) (1.53) (485 nm) mm 16. 1 and 2 both are reflected from denser medium. 12 n1 t = çæè 1484 öø÷ nm n2 Glass m Hence, 2n1t = l for first order minima Substituting m = 1, 2, 3L etc, or 2 l1 = 1484 nm, l2 = 742 nm, l3 = 495 nm, l4 = 371nm etc. Therefore, only wavelength lying in the given range is 495 nm. tmin = l = 4 650 18. Ray-1 and ray-2 both are reflected from denser 4 n1 ´ 1.42 medium. Hence, they are in phase. = 114 nm 12 17. (a) Ray-1 is reflected from a denser medium and 1.3 t ray-2 by a rarer medium. 12 1.5 tn \\ 2 mt = l for minima 2 or t = l = 600 4m 4 ´ 1.3 l Dx = 2nt = (2m - 1) 2 for maximum intensity = 1154 Å \\ l = 4 nt 19. Ray-1 is reflected from denser medium and ray-2 (2m - 1) from denser medium. = (4 ) (1.53) (485 nm) 12 2m -1 = æèçç 22m96-81øö÷÷ nm Oil m t Water For m = 1, l = 2968 nm \\ Dx = 2 mt = l = 800 nm for destruction For m = 2, l = 989 nm interference. For m = 3, l = 594 nm

Chapter 32 Interference and Diffraction of Light — 477 \\ mt = 400 nm \\ y1 = lD = b Further, 4d 4 For constructive interference, Dx = 2 mt = (2n - 1) l I0 = 4I0 cos2 èçæ f2 öø÷ 2 2 \\ l = 4mt = 1600 \\ f2 = 2p = æçè 2p ö÷ø (D x2 ) = èæç 2p ø÷ö èçæ y2d öø÷ 2n - 1 2n - 1 3 l l D For n = 1, l = 1600 nm \\ y2 = lD = b For n = 2, l = 533 nm 3d 3 For n = 3, l = 320 nm Dy = y2 - y1 = b The only wavelength lying in the given range is 12 533 nm. 2. At path difference l, we get maximum intensity. 20. 2mt = l/2 \\ Imax = I This is the condition for destructive interference. cos2æç fö÷ IR = I max è 2ø \\ t= l = 3.0 = 4m 4 ´ 1.5 0.5 cm 21. Path difference produced by slab, \\ I = I cos2æç fö÷ 4 è 2ø l Dx = (m - 1)t = 2 or cos æèç f2øö÷ = ± 1 2 l path difference is equivalent to 180° phase 2 \\ f = 60° or 120° 2 difference. Hence, maxima and minima \\ f = 120° or 240° or 2p and 4p interchange their positions. 33 22. Dx = d sin q = nl æèç l ø÷ö 2p n = d sin q From the relation, Dx = × f l We see that, nmax = d = 4.0 ´ 10-6 Dx = l and 2l l 600 ´ 10-9 33 = 6.66 3. Ray-1 is reflected from a denser medium (D f = p) Highest integer is 6. while ray-2 comes after reflecting from a rarer 23. medium (D f = 0° ). \\ Dx = 2mt = (2n -1) l for maximum intensity. 2 45º A (each) or l = 4mt (2n - 1) =4 ´1.5 ´500 = èççæ 23n00-01÷öø÷ nm 2n - 1 Anet = 0 12 LEVEL 2 Single Correct Option mt 1. 2 I0 = 4 I0 cos2 èçæ q1 ÷öø 2 Substituting n = 1, 2, 3××× etc, we get l = 3000 nm, \\ f1 = p = èçæ 2p öø÷ (D x1 ) = çèæ 2p öø÷ èçæ y1d ö÷ 1000 nm, 600 nm etc. 2 l l Dø \\ Answer is 600 nm.

478 — Optics and Modern Physics 4. D x = yd 7. Q q= d D D P S1 q y d S2 D For destructive interference at P. Third minima, Dx = yd = (2n - 1) l y = ± 2.5 w = ± 2.5 èæç lD ø÷ö = ± 2.5l D 2 d q 2 yd \\ l = (2n - 1) D 8. At points P and Q, Substituting n = 1, 3, 5×××etc we get R l = 2 yd , 2 æçè yd ø÷ö , 2 æçè yd øö÷ etc …(i) D 3 D 5 D Here, yd = (2 ´ 10-3) (0.1´ 10-3) QP D 1.0 S1 O S2 3a = 15l = 2 ´ 10-7 m = 2000 Å Substituting in Eq. (i), we get l = 4000 Å, 2680 Å, S 1600Å etc. | Dx | = 15, therefore maxima So that answer is 4000 Å. At points R and S Dx = 0, therefore maxima. 5. 3 I max = I max cos2çæ fö÷ 4 è 2ø Between P and R (and similarly in other three \\ f= p 5p 2 12 and 6 quadrants), we will get 14 maxima corresponding \\ f= p to, Dx = l, 2l ××× 14l. 6 Therefore, total maximas are 60. and 5p = èæç 2p ö÷ø (Dx) = èæç 2p ø÷ö æèç yd öø÷ 9. Dx = yd = (w/4) d = l æç w = lD ö÷ 3 l l D D D 4è d ø \\ y1 = lD = (6000 ´ 10-10) (1) f = çèæ 2p ø÷ö Dx = p or 90° 12d (12) (10-3) l 2 = 0.05 ´ 10-3m = 0.05 mm I = I max cos2 q 2 æç lD ÷ö y2 = 5 è 12d ø = 5 ´ 0.05 mm \\ I = cos2 45° = 1 I max 2 = 0.25 mm 10. Shift = (m - 1)tD D y = y2 - y1 = 0.2 mm d 6. Shift = (m - 1) tD = 3.5, w = 3.5 lD At m = 1,shift = 0 dd t = 3.5l P m -1 I0 = (3.5) (6000´10-10) Q Zero 1.5 - 1 R I0 = 4.2 ´ 10-6m Therefore, intensity at centre is maximum or I0. = 4.2 mm

Chapter 32 Interference and Diffraction of Light — 479 As m increases fringes shift upwards as shown in 5. Fringe pattern shifts in the direction of slab. But, figure. Shift = (m - 1) tD So, intensity at P first decreases to zero (as Q d reaches at P), then it further increases to I0 (as So, actual shift will depend on the values of m ,t,D point R reaches to P). and d. 11. 3 I max = I max cos2 f 6. For overlapping of maxima 4 2 n1l1D = n2l2D \\ f=p or f= p = èæç 2p ø÷ö (Dx) = çæè 2p ø÷ö (m - 1) t dd 26 3 l l n1 = l2 = 7, 14 ××× l 6000´ 10-10 or n2 l1 5 10 (m - 6 (1.5 - 1) \\ t = 6 1) = = 0.2 ´ 10-6m = 0.2 mm Þ 14th order maxima of l1 will coincide with 10th order maxima of l2. 12. Dx1 = (m 1 - 1), Dx2 = (m 2 - 1)t For overlapping of minima \\ Dx = (m 1 -m 2) t (2n1 - 1) l1D = (2n2 - 1) l2D = (1.52 - 1.40) (10400 nm ) 2d 2d = 1248 nm \\ 2n1 - 1= l2 = 7 2n2 - 1 l1 5 For maximum intensity, (c) Option with n1 = 11 Dx = 1248 = nl and n2 = 8 gives this ratio. \\ l = 1248 (n = 1, 2, 3×××) n Comprehension Based Questions For n = 2, l = 624 nm 1. l = l0 , w = lD = l0D and for n = 3, l = 416 nm m d md More than One Correct Options = (6300 ´ 10-10) (1.33) 1.33 ´ 10-3 1. lv is least. Therefore wv is minimum (as w µ l). = 0.63 ´10-3 m Hence, the fringe next to centre will be violet. At centre, Dx = 0 for all wavelengths. Hence, all = 0.63 mm wavelengths interfere constructively at centre. So, 2. Dy = 7w - 3w = 4w it is white. = 4 ´ 0.63 mm 2. (c) Imax = ( I1 + I2 )2 = 2.52 mm Imin = ( I1 - I2 )2 çèçæ m 1öø÷÷ l m 2 When I1 =I2 = I0 3. Dx = 2 - t = When 1 Imax = 4 I0 and Imin = 0 l = I0 , \\ t = I2 2 èæçç m 1÷øö÷ 2 m 2 - then Imax < 4I0 and Imin > 0. 1 3. Dx0 = d sin q = (10-3) çèæ 1 ÷øö = (6300 ´ 10-10 / 1.33) 2 2 èçæ 1.53 - 1÷øö = 5 ´ 10-4m = (103) l 1.33 Since, Dx0 is integer multiple of l, it will produce = 1.575 ´ 10-3m maximum intensity or 4I0 at O. = 1.575 mm ´ 10-7 w = lD = (5 (10-3 ) (2) 4. Fringe width remains unchanged by the d ) introduction of glass sheet. = 10-3m = 1 mm At 4 mm, we will get 4th order maxima.

480 — Optics and Modern Physics Match the Columns \\ I0 = 4IS3 or IS4 = 3I 1. I = 4I0 cos2 æèç f2øö÷ Same explanations can be given for (c) and (d) options. 2. f= 2p Dx and then apply, l 6. (a) Fringe pattern will shift in the direction of slab. I = 4I0 cos2 æç f÷ö (b) No interference pattern will be obtained due to è 2ø single slit. \\ I = IS2 = I0 » uniform 3. y6 Y6 Y ® maxima (c) Fringe pattern will shift in the direction of y5 Y5 y ® minima slab. y4 Y4 y3 Y3 (d) No interference pattern will be obtained due to y2 Y2 two real incoherent sources. y1 Y1 w I = IS1 + IS2 » uniform 4. (a) Inclined rays and slab both will shift the fringe Subjective Questions pattern upwards. So, zero order maxima will 1. I1 = 0.1I0 , I2 = 0.081I0 definitely lie about O. 2 0.081I0 (b) Inclined rays will shift the fringe pattern upwards, but slab will shift the fringe pattern 1 0.1I0 0.09I0 downwards. Hence, zero order maxima may 0.9I0 lie above O, below O or at O. I (c) Same explanations can be given for this option. (d) S1 \\ I1 = 10 S2 \\ I2 9 P 2. Q I max = èççæ I1 /I2 + 11øö÷÷ 2 O I min I1 / I2 – \\ DX p = 0, where S1 P = S2 P = (19)2 \\ I \\ = 361 Ans. 4 5. I = Imax = 4I0 Þ I0 = m = sin i sin r (a) For y= lD , Dx yd =l 2d D 2 4 = sin 53° = 4 /5 3 sin r sin r \\ IS3 = IS4 = 0 for Dx = l/2 \\ I0 = 0 (b) For y = lD , Dx = yd =l 37° 6d D 6 54 f= æçè 2p ø÷ö (Dx) = 2p = 60° l 6 53° Now, I S3 = IS4 = Imax cos2 q 3 2 sin r = 3 = I cos2 30° = 3 I 5 4 r = 37°

Chapter 32 Interference and Diffraction of Light — 481 Refer figure (a) 1 2 1 S1 0.5 mm B i ii 0.25 mm A rr C r S2 0.25 mm t 0.5 mm S2 D 15 cm 30 cm D (a) 60 cm Dx1 = between 2 and 1 = 2 (AD ) Distance between two slits, = 2BD sec r d = 1.5 mm, D = 30 cm = 2t sec r Fringe width, w = lD Their optical path Dx1 = 2mt secr. Refer figure (b) d 1 = (5.0 ´ 10–7 ) (0.3) = 10–4 m E (1.5 ´ 10–3) 2 = 0.1 mm Ans. i 4. l = 0.25 m, d = 2 m = 8l AB C B r D CA (b) S1 S2 d D x2 = AC sin i = (2t tan r) sin i \\ (Dx)net = Dx1 – Dx2 D = 2 mt sec r – 2t (tan r) (sin i) At A and C , Dx = d = 8l, i.e. maximum intensity is obtained. = 2 ´ 4 ´t ´ 5 – 2 ´t ´ 3 ´ 4 = 32 t 3 4 4 5 15 At B and D, Dx = 0, i.e. again maximum intensity will be obtained. Phase difference between 1 and 2 is p. Between A and B seven maximas corresponding to \\ For constructive interference, Dx = 7l, 6l, 5l, 4l, 3l, 2l and l will be obtained. Similarly, between B and C ,C and D , and D and 32 = l or t = 15l = 15 ´ 0.6 A. 15 t 2 64 64 \\ Total number of maximas = 0.14 mm Ans. =4 ´7+ 4 = 32 Ans. 3. Applying lens formula, 1 – 1 = 1 vu f 1+ 1 = 1 5. (a) Q Dx = d cos q v 15 10 \\ cos q = 1 – q2 \\ v = 30 cm 2 (When q is small) m = v = 30 = – 2 Dx = d ççæè1 – q2 ÷÷øö u –15 2

482 — Optics and Modern Physics = d ççèæ1 – y2 ö÷÷ø 7. (a) d sin f = Dx1 2D 2 = (50 ´ 10-4) sin 30° P f q = 30° y q q C O S1 S2 d Dxnet = 0 For nth maxima Dx = nl = 2.5 ´ 10-3 cm \\ y = radius of nth bright ring Dx2 = (m - 1)t =D 2 èçæ1 – nl ÷øö Ans. = èæç 3 - 1ö÷ø (0.01) d 2 (b) d = 1000l = 5.0 ´ 10-3 cm At O , D x = d = 1000l Dx2 - Dx1 = 2.5 ´ 10-3 cm = Dx1 (also) i.e. at O, 1000th order maxima is obtained. \\ Central maxima will be obtained at q = 30° Substituting n = 998 in, below C. y=D 2èçæ1 – nl ö÷ø (b) At C DX net = 2.5 ´ 10-3 cm = nl d \\ n = 2.5 ´ 10-3 l We get the radius of second closest ring 2.5 ´ 10-3 r = 6.32 cm Ans. = 500 ´ 10-7 (c) n = 998 Ans. = 50 Ans. 6. (a) The optical path difference between the two (c) Number of fringes that will pass if we remove the slab waves arriving at P is Dx = y1d + y2d = Path difference due to slab D1 D2 l = (1) (10) + (5) (10) = 5 ´ 10-3 103 2 ´ 103 500 ´ 10-7 = 3.5 ´ 10–2 mm = 0.035 mm = 100 As, Dx = 70l Ans. \\ 70th order maxima is obtained at P. 8. (a) (Dx)net = 0 (b) At O , Dx = y1d = 10–2 mm \\ y1d = y2d D1 D1 D2 = 0.01 mm \\ d/2 = y 1.5 2.0 As Dx = 20l \\ 20th order maxima is obtained at O. or y = d (c) (m – 1) t = 0.01 mm 1.5 \\ t = 0.01 = 0.02 mm = 20 mm Ans. =6 1.5 – 1 1.5 Since, the pattern has to be shifted upwards, = 4 mm Ans. therefore, the film must be placed in front of S1.

Chapter 32 Interference and Diffraction of Light — 483 (b) At O , net path difference, f= p= çèæ 2p ö÷ø (m – 1) t 3 l Dx = y1d D1 \\ t = 6 l 1) (m – = (d /2) (d) D1 = 6000 6 (1.5 – 1) = (6 ´ 10–3)2 2 ´ 1.5 = 2000 Å Ans. = 12 ´ 10–6 m 9. (a) æç1 – 1 ö÷ t = 3l Ans. è m ø m Ans. = 120 ´ 10–7 m Ans. 3l l = 6000 Å \\ t = (m – 1) = 6 ´ 10–7 m = 3 ´ 0.78 1.3 – 1 As, Dx = 20l, therefore at O bright fringe of = 7.8 mm order 20 will be obtained. (c) I = Imax cos2 çèæ 2f÷øö (b) Upwards yd – æç1 – 1 ö÷ t = 4l D è m ø m 3 I max = I max cos2 èçæ 2fø÷ö Solving, we get y = 4.2 mm 4 Downwards t æç1 – 1 ö÷ + yd = 4l \\ f=p è m ø D m 26 Solving, we get y = 0.6 mm

33. Modern Physics - I INTRODUCTORY EXERCISE 33.1 2. En = - 13.6 = - 1.51 Þ n=3 (n)2 1. E = hc , P = h Now, Ln = n èæç h ÷øö l l 2p 2. Number of photons emitted per second, 3. The expressions of kinetic energy, potential energy N 1 = Power of source = P = Pl and total energy are Energy of one photon (hc/ l) hc me4 1 At a distance r, these photons are falling on an area Kn = 8e20n2h2 Þ Kn µ n2 4 pr2. - me4 1 4 e20n2h2 n2 \\ Number of photons incident per unit area per Un = Þ Un µ - and unit time, Pl N2 = N1 = (4 pr2 ) hc En = - me4 Þ En µ - 1 4 pr2 8e20n2h2 n2 3. l = h Þ l µ 1 In the transition from some excited state to ground 2qVm qm state, the value of n decreases, therefore kinetic lP = (qm)d = 1´2 = 2 energy increases, but potential and total energy ld (qm)P 1´1 hÞ decrease. 2Km 4. l = lµ 1 4. n = 4 n=4 m n=3 n=3 ld = ma = 4 = 2 n=2 n=2 la md 2 5. Kinetic energy in magnetic field remains n=1 n=1 First line of Second line of unchanged while in electric field it will increase. Balmer series Balmer series Further, For hydrogen or hydrogen type atoms, lµ 1 K 1 = RZ 2 æç 1 - 1 ö÷ l çè n2f ni2 ÷ø 6. l = h mv In the transition from ni ¾® nf , 6.63 ´ 10-34 (a) l= (46 ´ 10-3) (30) \\ l µ 1 = 4.8 ´ 10-34 m Z2 æç 1 - 1 ÷ö çè n2f ni2 ÷ø 6.63 ´ 10-34 (b) l = 9.31 ´ 10-31 ´ 107 Z12 çæ 1 1 ÷ö èç n2f ni2 ø÷ 1 = 7.12 ´ 1011 m \\ l2 = - 1 ÷ö l1 - ni2 ø÷ 2 Z22 çæ 1 INTRODUCTORY EXERCISE 33.2 èç n2f 1. Z = 3 for doubly ionized atom E µ Z 2 l1Z12 æç 1 - 1 ÷ö çè n2f ni2 ø÷ 1 Ionization energy of hydrogen atom is 13.6 eV. l2 = \\ Ionisation energy of this atom æç 1 ö÷ Z22 çè 1 - ni2 ÷ø 2 = (3)2(13.6) n2f = 122.4 eV

Chapter 33 Modern Physics-I — 485 Substituting the values, we have E1 = (-13.6) (3)2 (1)2 (6561 Å) (1)2 æçè 1 - 1 ö÷ø 22 32 = - 122.4 eV = = 1215 Å æçè 1 1 ø÷ö \\ Ionization energy of an electron in ground state (2)2 22 - 42 of doubly ionized lithium atom will be 122.4 eV. \\ Correct option is (a). 10. Shortest wavelength will correspond to maximum 5. The series in U-V region is Lyman series. Largest energy. As value of atomic number (Z) increases, wavelength corresponds to minimum energy which the magnitude of energy in different energy states occurs in transition from n = 2 to n = 1. gets increased. Value of Z is maximum for doubly 1 ionised lithium atom (Z = 3) among the given elements. Hence, wavelength corresponding to this \\ 122 = èæç 1 R 1 ø÷ö …(i) 12 - 22 will be least. \\ Correct option is (d). The smallest wavelength in the infrared region 11. 5E - E = hf Þ E = hf …(i) corresponds to maximum energy of Paschen series. 4 1 Between 5E and 4E \\ l= R 5E - 4E = hf1 æèç 1 1 öø÷ …(ii) \\ f1 = E = f [from Eq. (i)] 32 ¥ h 4 - Solving Eqs. (i) and (ii), we get Between 4E and E l = 823.5 nm 4E - E = hf2 \\ Correct option is (b). \\` f2 = 3E = 3 æçè f öø÷ = 3 f h 4 4 6. The first photon will excite the hydrogen atom (in 12. Longest wavelength means minimum energy. ground state) to first excited state (DE)min = E3 - E2 (as E2 - E1 = 10.2 eV). Hence, during = - 13.6 + 13.6 = 1.9 eV de-excitation a photon of 10.2 eV will be released. 94 The second photon of energy 15 eV can ionise the atom. Hence, the balance energy, i.e. l (in Å) = 12375 = 6513 (15 - 13.6) eV = 1.4 eV is retained by the electron. 1.9 Therefore, by the second photon an electron of energy 1.4 eV will be released. or l » 651 nm \\ Correct answer is (c). INTRODUCTORY EXERCISE 33.3 7. In second excited state n = 3, So, lH = lLi = 3 çæè h ÷øö 1. Ka transition takes place from n1 = 2 to n2 = 1 2p \\ 1 = R (Z - b)2 é1 - 1ù while E µ Z 2 and ZH = 1, ZLi = 3 l ëê (1)2 (2)2 ûú So, | ELi | = 9 | EH | For K-series, b = 1 or | EH | < | ELi | \\ 1 µ (Z - 1)2 l 8. Energy of infrared radiation is less than the energy - 1)2 - 1)2 of ultraviolet radiation. In options (a), (b) and (c), Þ lCu = (ZMo - 1)2 = (42 - 1)2 lMo (ZCu (29 energy released will be more, while in option (d) only, energy released will be less. 41 ´ 41 1681 28 ´ 28 784 9. For hydrogen and hydrogen like atoms, = = = 2.144 En = - 13.6 (Z 2) eV 2. Cut off wavelength depends on the applied voltage (n2) not on the atomic number of the target. Therefore, ground state energy of doubly ionized Characteristic wavelengths depend on the atomic lithium atom (Z = 3, n = 1) will be number of target.

486 — Optics and Modern Physics 3. 1 µ (Z - 1)2 2. Kmax = E - W = hf - hf0 l = h ( f - f0) l1 æèçç Z2 - 11÷÷øö 2 1 çèæç Z2 - 11÷÷öø 2 \\ Kmax µ ( f - f0) l2 Z1 - 4 11 - \\ = or = 3. Kmax = E - W Solving this, we get Z2 = 6 1.2 = E - W …(i) \\ Correct answer is (a). 4.2 = 1.5 E - W …(ii) 4. Wavelength lk is independent of the accelerating Solving these equations, we get voltage (V), while the minimum wavelength lc is W = 4.8 eV = hf0 inversely proportional to V. Therefore, as V is 4.8 ´ 1.6 ´ 10-19 \\ f0 = 6.63 ´ 10-34 increased, olrklrkem-alincswuinllcihnacnrgeaesdew. hereas lc decreases = 1.16 ´ 1015 Hz 5. The continuous X-ray spectrum is shown in figure. El 4. Energy corresponding to 248 nm wavelength = 1240 eV = 5 eV 248 Energy corresponding to 310 nm wavelength l = 1240 eV = 4 eV 310 lmin KE1 = u12 = 4 KE2 u22 1 All wavelengths >lmin are found, where 12375 Å = 5 eV - W lmin = (in volt) 4 eV - W V Þ 16 - 4W = 5 - W Here, V is the applied voltage. Þ 11 = 3W DE = hn = (Z - b)2 æèçç 1 - 1 ÷÷øö 6. Rhc n12 n22 = 11 = 3.67 eV 3 Þ W For K-series, b = 1 =~ 3.7 eV \\ n = Rc (Z - 1)2 çèæç 1 - 1 ø÷ö÷ 5. Saturation current is proportional to intensity while n12 n22 stopping potential increases with increase in Substituting the values, frequency. 4.2 ´ 1018 = (1.1 ´ 107 ) (3 ´ 108 ) (Z - 1)2 èçæ 1 - 1 øö÷ Hence, fa = fb while Ia < Ib 1 4 Therefore, the correct option is (a). \\ (Z - 1)2 = 1697 6. l (in Å) = 12375 or Z - 1 » 41 W (eV) or Z = 42 = 12375 Å » 3093 Å INTRODUCTORY EXERCISE 33.4 4.0 or l » 309.3 nm or » 310 nm 1. Kmin = 0 Note l(in Å) = 12375 comes from W = hc W (eV) l and Kmax = E - W = 12375 - 3.0 2000 7. Stopping potential is the negative potential applied » 3.19 eV to stop the electrons having maximum kinetic energy. Therefore, stopping potential will be 4 V.

Exercises LEVEL 1 10. In X-ray spectrum, all wavelengths greater than Assertion and Reason lmin are obtained. 2. E = hc and P = h Objective Questions l l 1. Kmax = hf - W 1 \\ E and P µ l Kmax versus f graph is a straight line of slope h (a universal constant) Speed of all wavelengths (in vacuum) is c. c 3. Intensity = energy incident per unit area per unit 2. v1H = 2.19 ´ 106 m/s » 137 time or I = n(hf ) . 3. Let a-particles are n and b-particles are m. Then, Here, n = number of photons incident per unit area 86 - 2n + m = 84 …(i) per unit time. f = frequency of incident photons. Hence, I µ nf . 222 - 4n = 210 …(ii) Hence, intensity can be increased either by Solving these two equations, we get n = 3 and increasing n or f . But saturation current only depends on n (Is µ m). b=4 By increasing n and decreasing f , we can increase 4. lmin = 12375 in Å the saturation current even without increasing the V (in volts) intensity. = 12375 = 0.62 Å 5. Let us take n = 3 as N = 1. Then, n = 6 means 20 ´ 103 N = 4. 5. K max = 18 ´ 103 eV = 1 mvm2 ax 2 So, total number of emission lines between N = 1 and N = 4 are \\ vmax = 2 ´ 18 ´ 103 ´ 1.6 ´ 10-19 9.1 ´ 10-31 N (N - 1) = 4 ´ 3 = 6 22 = 8 ´ 107 m/s 6. eV0 = hn - hn0 6. (2pr2) = 2l2 or V0 = h (n - n0) (2pr3) = 3l3 e \\ V0¢ = h (2n - n0) + hn0 Now, l2 = 3r2 …(i) e e l3 2r3 = 2V0 = hn0 > 2V0 r µ n2 e \\ r2 = çæ 2÷ö 2 7. Energy of X-ray > 13.6 eV r3 è 3ø 8. Dl = lKa - lmin Substituting in Eq. (i), we get Here, lmin (in Å) = V 12375 l2 = 2 (in volts) l3 3 If V is increased, lmin decreases. Therefore, Dl 7. E µ Z 2 increases. \\ (- 13.6) (Z 2) = - 122.4 E2 = - 3.4 eVü 9. U2 = - 6.8 eVýþ \\ Z =3 E1 = - 13.6 eVü 8. lmin µ1 µ V -1 U1 = - 27.2 eVýþ V E2 > E1 , similarly U 2 > U 1 % change in lmin = (- 1) (% change in V ) for small % changes