Page 1 17.1 SOLUTIONS TO CONCEPTS CHAPTER 17 1. Given that, 400 m < ? < 700 nm. 1 1 1 700nm 400nm ? ? ? ? 8 8 7 7 7 7 1 1 1 3 10 c 3 10 7 10 4 10 7 10 4 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (Where, c = speed of light = 3 ? 10 8 m/s) ? 4.3 ? 10 14 < c/ ? < 7.5 ? 10 14 ? 4.3 ? 10 14 Hz < f < 7.5 ? 10 14 Hz. ? 2. Given that, for sodium light, ? = 589 nm = 589 ? 10 –9 m a) f a = 8 9 3 10 589 10 ? ? ? = 5.09 ? 10 14 1 c sec f ? ? ? ? ? ? ? ? ? ? b) a w w w 9 w a 1 1.33 589 10 ? ? ? ? ? ? ? ? ? ? ? ? = 443 nm c) f w = f a = 5.09 ? 10 14 sec –1 [Frequency does not change] d) 8 a a a w w w a w v v 3 10 v v 1.33 ? ? ? ? ? ? ? ? ? = 2.25 ? 10 8 m/sec. 3. We know that, 2 1 1 2 v v ? ? ? So, 8 8 400 400 1472 3 10 v 2.04 10 m/ sec. 1 v ? ? ? ? ? [because, for air, ? = 1 and v = 3 ? 10 8 m/s] Again, 8 8 760 760 1452 3 10 v 2.07 10 m/ sec. 1 v ? ? ? ? ? ? 4. 8 t 8 1 3 10 1.25 (2.4) 10 ? ? ? ? ? ? velocity of light in vaccum since, = velocity of light in the given medium ? ? ? ? ? ? ? 5. Given that, d = 1 cm = 10 –2 m, ? = 5 ? 10 –7 m and D = 1 m a) Separation between two consecutive maxima is equal to fringe width. So, ? = 7 2 D 5 10 1 d 10 ? ? ? ? ? ? m = 5 ? 10 –5 m = 0.05 mm. b) When, ? = 1 mm = 10 –3 m 10 –3 m = 7 5 10 1 D ? ? ? ? D = 5 ? 10 –4 m = 0.50 mm. ? 6. Given that, ? = 1 mm = 10 –3 m, D = 2.t m and d = 1 mm = 10 –3 m So, 10 –3 m = 3 25 10 ? ? ? ? ? = 4 ? 10 –7 m = 400 nm. ? 7. Given that, d = 1 mm = 10 –3 m, D = 1 m. So, fringe with = D d ? = 0.5 mm. a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm b) No. of fringes = 10 / 0.5 = 20. 8. Given that, d = 0.8 mm = 0.8 ? 10 –3 m, ? = 589 nm = 589 ? 10 –9 m and D = 2 m. So, ? = D d ? = 9 3 589 10 2 0.8 10 ? ? ? ? ? = 1.47 ? 10 –3 m = 147 mm. ? Page 2 17.1 SOLUTIONS TO CONCEPTS CHAPTER 17 1. Given that, 400 m < ? < 700 nm. 1 1 1 700nm 400nm ? ? ? ? 8 8 7 7 7 7 1 1 1 3 10 c 3 10 7 10 4 10 7 10 4 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (Where, c = speed of light = 3 ? 10 8 m/s) ? 4.3 ? 10 14 < c/ ? < 7.5 ? 10 14 ? 4.3 ? 10 14 Hz < f < 7.5 ? 10 14 Hz. ? 2. Given that, for sodium light, ? = 589 nm = 589 ? 10 –9 m a) f a = 8 9 3 10 589 10 ? ? ? = 5.09 ? 10 14 1 c sec f ? ? ? ? ? ? ? ? ? ? b) a w w w 9 w a 1 1.33 589 10 ? ? ? ? ? ? ? ? ? ? ? ? = 443 nm c) f w = f a = 5.09 ? 10 14 sec –1 [Frequency does not change] d) 8 a a a w w w a w v v 3 10 v v 1.33 ? ? ? ? ? ? ? ? ? = 2.25 ? 10 8 m/sec. 3. We know that, 2 1 1 2 v v ? ? ? So, 8 8 400 400 1472 3 10 v 2.04 10 m/ sec. 1 v ? ? ? ? ? [because, for air, ? = 1 and v = 3 ? 10 8 m/s] Again, 8 8 760 760 1452 3 10 v 2.07 10 m/ sec. 1 v ? ? ? ? ? ? 4. 8 t 8 1 3 10 1.25 (2.4) 10 ? ? ? ? ? ? velocity of light in vaccum since, = velocity of light in the given medium ? ? ? ? ? ? ? 5. Given that, d = 1 cm = 10 –2 m, ? = 5 ? 10 –7 m and D = 1 m a) Separation between two consecutive maxima is equal to fringe width. So, ? = 7 2 D 5 10 1 d 10 ? ? ? ? ? ? m = 5 ? 10 –5 m = 0.05 mm. b) When, ? = 1 mm = 10 –3 m 10 –3 m = 7 5 10 1 D ? ? ? ? D = 5 ? 10 –4 m = 0.50 mm. ? 6. Given that, ? = 1 mm = 10 –3 m, D = 2.t m and d = 1 mm = 10 –3 m So, 10 –3 m = 3 25 10 ? ? ? ? ? = 4 ? 10 –7 m = 400 nm. ? 7. Given that, d = 1 mm = 10 –3 m, D = 1 m. So, fringe with = D d ? = 0.5 mm. a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm b) No. of fringes = 10 / 0.5 = 20. 8. Given that, d = 0.8 mm = 0.8 ? 10 –3 m, ? = 589 nm = 589 ? 10 –9 m and D = 2 m. So, ? = D d ? = 9 3 589 10 2 0.8 10 ? ? ? ? ? = 1.47 ? 10 –3 m = 147 mm. ? Chapter 17 17.2 9. Given that, ? = 500 nm = 500 ? 10 –9 m and d = 2 ? 10 –3 m As shown in the figure, angular separation ? = D D dD d ? ? ? ? ? So, ? = 9 3 500 10 D d 2 10 ? ? ? ? ? ? ? ? = 250 ? 10 –6 = 25 ? 10 –5 radian = 0.014 degree. ? 10. We know that, the first maximum (next to central maximum) occurs at y = D d ? Given that, ? 1 = 480 nm, ? 2 = 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25 ? 10 –3 m So, y 1 = 9 1 3 D 1.5 480 10 d 0.25 10 ? ? ? ? ? ? ? = 2.88 mm y 2 = 9 3 1.5 600 10 0.25 10 ? ? ? ? ? = 3.6 mm. So, the separation between these two bright fringes is given by, ? separation = y 2 – y 1 = 3.60 – 2.88 = 0.72 mm. 11. Let m th bright fringe of violet light overlaps with n th bright fringe of red light. ? m 400nm D n 700nm D m 7 d d n 4 ? ? ? ? ? ? ? ? 7 th bright fringe of violet light overlaps with 4 th bright fringe of red light (minimum). Also, it can be seen that 14 th violet fringe will overlap 8 th red fringe. Because, m/n = 7/4 = 14/8. 12. Let, t = thickness of the plate Given, optical path difference = ( ? – 1)t = ?/2 ? t = 2( 1) ? ? ? ? 13. a) Change in the optical path = ?t – t = ( ? – 1)t b) To have a dark fringe at the centre the pattern should shift by one half of a fringe. ? ( ? – 1)t = t 2 2( 1) ? ? ? ? ? ? . ? 14. Given that, ? = 1.45, t = 0.02 mm = 0.02 ? 10 –3 m and ? = 620 nm = 620 ? 10 –9 m We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( ? – 1)t. Again, for shift of one fringe, the optical path should be changed by ?. So, no. of fringes crossing through the centre is given by, n = 3 9 ( 1)t 0.45 0.02 10 620 10 ? ? ? ? ? ? ? ? ? = 14.5 ? 15. In the given Young’s double slit experiment, ? = 1.6, t = 1.964 micron = 1.964 ? 10 –6 m We know, number of fringes shifted = ( 1)t ? ? ? So, the corresponding shift = No.of fringes shifted ? fringe width = ( 1)t D ( 1)tD d d ? ? ? ? ? ? ? ? … (1) Again, when the distance between the screen and the slits is doubled, Fringe width = (2D) d ? …(2) From (1) and (2), ( 1)tD d ? ? = (2D) d ? ? ? = ( 1)t ? ? ? = 6 (1.6 1) (1.964) 10 2 ? ? ? ? = 589.2 ? 10 –9 = 589.2 nm. ? B S 1 S 2 ? ? D Page 3 17.1 SOLUTIONS TO CONCEPTS CHAPTER 17 1. Given that, 400 m < ? < 700 nm. 1 1 1 700nm 400nm ? ? ? ? 8 8 7 7 7 7 1 1 1 3 10 c 3 10 7 10 4 10 7 10 4 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (Where, c = speed of light = 3 ? 10 8 m/s) ? 4.3 ? 10 14 < c/ ? < 7.5 ? 10 14 ? 4.3 ? 10 14 Hz < f < 7.5 ? 10 14 Hz. ? 2. Given that, for sodium light, ? = 589 nm = 589 ? 10 –9 m a) f a = 8 9 3 10 589 10 ? ? ? = 5.09 ? 10 14 1 c sec f ? ? ? ? ? ? ? ? ? ? b) a w w w 9 w a 1 1.33 589 10 ? ? ? ? ? ? ? ? ? ? ? ? = 443 nm c) f w = f a = 5.09 ? 10 14 sec –1 [Frequency does not change] d) 8 a a a w w w a w v v 3 10 v v 1.33 ? ? ? ? ? ? ? ? ? = 2.25 ? 10 8 m/sec. 3. We know that, 2 1 1 2 v v ? ? ? So, 8 8 400 400 1472 3 10 v 2.04 10 m/ sec. 1 v ? ? ? ? ? [because, for air, ? = 1 and v = 3 ? 10 8 m/s] Again, 8 8 760 760 1452 3 10 v 2.07 10 m/ sec. 1 v ? ? ? ? ? ? 4. 8 t 8 1 3 10 1.25 (2.4) 10 ? ? ? ? ? ? velocity of light in vaccum since, = velocity of light in the given medium ? ? ? ? ? ? ? 5. Given that, d = 1 cm = 10 –2 m, ? = 5 ? 10 –7 m and D = 1 m a) Separation between two consecutive maxima is equal to fringe width. So, ? = 7 2 D 5 10 1 d 10 ? ? ? ? ? ? m = 5 ? 10 –5 m = 0.05 mm. b) When, ? = 1 mm = 10 –3 m 10 –3 m = 7 5 10 1 D ? ? ? ? D = 5 ? 10 –4 m = 0.50 mm. ? 6. Given that, ? = 1 mm = 10 –3 m, D = 2.t m and d = 1 mm = 10 –3 m So, 10 –3 m = 3 25 10 ? ? ? ? ? = 4 ? 10 –7 m = 400 nm. ? 7. Given that, d = 1 mm = 10 –3 m, D = 1 m. So, fringe with = D d ? = 0.5 mm. a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm b) No. of fringes = 10 / 0.5 = 20. 8. Given that, d = 0.8 mm = 0.8 ? 10 –3 m, ? = 589 nm = 589 ? 10 –9 m and D = 2 m. So, ? = D d ? = 9 3 589 10 2 0.8 10 ? ? ? ? ? = 1.47 ? 10 –3 m = 147 mm. ? Chapter 17 17.2 9. Given that, ? = 500 nm = 500 ? 10 –9 m and d = 2 ? 10 –3 m As shown in the figure, angular separation ? = D D dD d ? ? ? ? ? So, ? = 9 3 500 10 D d 2 10 ? ? ? ? ? ? ? ? = 250 ? 10 –6 = 25 ? 10 –5 radian = 0.014 degree. ? 10. We know that, the first maximum (next to central maximum) occurs at y = D d ? Given that, ? 1 = 480 nm, ? 2 = 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25 ? 10 –3 m So, y 1 = 9 1 3 D 1.5 480 10 d 0.25 10 ? ? ? ? ? ? ? = 2.88 mm y 2 = 9 3 1.5 600 10 0.25 10 ? ? ? ? ? = 3.6 mm. So, the separation between these two bright fringes is given by, ? separation = y 2 – y 1 = 3.60 – 2.88 = 0.72 mm. 11. Let m th bright fringe of violet light overlaps with n th bright fringe of red light. ? m 400nm D n 700nm D m 7 d d n 4 ? ? ? ? ? ? ? ? 7 th bright fringe of violet light overlaps with 4 th bright fringe of red light (minimum). Also, it can be seen that 14 th violet fringe will overlap 8 th red fringe. Because, m/n = 7/4 = 14/8. 12. Let, t = thickness of the plate Given, optical path difference = ( ? – 1)t = ?/2 ? t = 2( 1) ? ? ? ? 13. a) Change in the optical path = ?t – t = ( ? – 1)t b) To have a dark fringe at the centre the pattern should shift by one half of a fringe. ? ( ? – 1)t = t 2 2( 1) ? ? ? ? ? ? . ? 14. Given that, ? = 1.45, t = 0.02 mm = 0.02 ? 10 –3 m and ? = 620 nm = 620 ? 10 –9 m We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( ? – 1)t. Again, for shift of one fringe, the optical path should be changed by ?. So, no. of fringes crossing through the centre is given by, n = 3 9 ( 1)t 0.45 0.02 10 620 10 ? ? ? ? ? ? ? ? ? = 14.5 ? 15. In the given Young’s double slit experiment, ? = 1.6, t = 1.964 micron = 1.964 ? 10 –6 m We know, number of fringes shifted = ( 1)t ? ? ? So, the corresponding shift = No.of fringes shifted ? fringe width = ( 1)t D ( 1)tD d d ? ? ? ? ? ? ? ? … (1) Again, when the distance between the screen and the slits is doubled, Fringe width = (2D) d ? …(2) From (1) and (2), ( 1)tD d ? ? = (2D) d ? ? ? = ( 1)t ? ? ? = 6 (1.6 1) (1.964) 10 2 ? ? ? ? = 589.2 ? 10 –9 = 589.2 nm. ? B S 1 S 2 ? ? D Chapter 17 17.3 16. Given that, t 1 = t 2 = 0.5 mm = 0.5 ? 10 –3 m, ? m = 1.58 and ? p = 1.55, ? = 590 nm = 590 ? 10 –9 m, d = 0.12 cm = 12 ? 10 –4 m, D = 1 m a) Fringe width = 9 4 D 1 590 10 d 12 10 ? ? ? ? ? ? ? = 4.91 ? 10 –4 m. b) When both the strips are fitted, the optical path changes by ?x = ( ? m – 1)t 1 – ( ? p – 1)t 2 = ( ? m – ? p )t = (1.58 – 1.55) ? (0.5)(10 –3 ) = 0.015 ? 10 –13 m. So, No. of fringes shifted = 3 3 0.015 10 590 10 ? ? ? ? = 25.43. ? There are 25 fringes and 0.43 th of a fringe. ? There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe. So, position of first maximum on both sides will be given by ? x = 0.43 ? 4.91 ? 10 –4 = 0.021 cm x ? = (1 – 0.43) ? 4.91 ? 10 –4 = 0.028 cm (since, fringe width = 4.91 ? 10 –4 m) 17. The change in path difference due to the two slabs is ( ? 1 – ? 2 )t (as in problem no. 16). For having a minimum at P 0 , the path difference should change by ?/2. So, ? ?/2 = ( ? 1 – ?? 2 )t ? t = 1 2 2( ) ? ? ? ? . ? 18. Given that, t = 0.02 mm = 0.02 ? 10 –3 m, ? 1 = 1.45, ? = 600 nm = 600 ? 10 –9 m a) Let, I 1 = Intensity of source without paper = I b) Then I 2 = Intensity of source with paper = (4/9)I ? 1 1 2 2 I r 9 3 I 4 r 2 ? ? ? [because I ? r 2 ] where, r 1 and r 2 are corresponding amplitudes. So, 2 max 1 2 2 min 1 2 I (r r ) I (r r ) ? ? ? = 25 : 1 b) No. of fringes that will cross the origin is given by, n = ( 1)t ? ? ? = 3 9 (1.45 1) 0.02 10 600 10 ? ? ? ? ? ? = 15. 19. Given that, d = 0.28 mm = 0.28 ? 10 –3 m, D = 48 cm = 0.48 m, ? a = 700 nm in vacuum Let, ? w = wavelength of red light in water Since, the fringe width of the pattern is given by, ??= 9 w 3 D 525 10 0.48 d 0.28 10 ? ? ? ? ? ? ? = 9 ? 10 –4 m = 0.90 mm. ? 20. It can be seen from the figure that the wavefronts reaching O from S 1 and S 2 will have a path difference of S 2 X. In the ? S 1 S 2 X, sin ???= 2 1 2 S X S S So, path difference = S 2 X = S 1 S 2 sin ? = d sin ? = d ? ?/2d = ?/2 As the path difference is an odd multiple of ?/2, there will be a dark fringe at point P 0 . 21. a) Since, there is a phase difference of ? between direct light and reflecting light, the intensity just above the mirror will be zero. b) Here, 2d = equivalent slit separation D = Distance between slit and screen. We know for bright fringe, ?x = y 2d D ? = n ? But as there is a phase reversal of ?/2. ? y 2d D ? + 2 ? = n ? ? y 2d D ? = n ? – 2 ? ? y = D 4d ? ? mica Screen polysterene S 2 S 1 ? ? P 0 ? ? x S 2 S 1 Dark fringe (1 – 0.43) ?? 0.43 ?? S 2 S 1 2d D Screen Page 4 17.1 SOLUTIONS TO CONCEPTS CHAPTER 17 1. Given that, 400 m < ? < 700 nm. 1 1 1 700nm 400nm ? ? ? ? 8 8 7 7 7 7 1 1 1 3 10 c 3 10 7 10 4 10 7 10 4 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (Where, c = speed of light = 3 ? 10 8 m/s) ? 4.3 ? 10 14 < c/ ? < 7.5 ? 10 14 ? 4.3 ? 10 14 Hz < f < 7.5 ? 10 14 Hz. ? 2. Given that, for sodium light, ? = 589 nm = 589 ? 10 –9 m a) f a = 8 9 3 10 589 10 ? ? ? = 5.09 ? 10 14 1 c sec f ? ? ? ? ? ? ? ? ? ? b) a w w w 9 w a 1 1.33 589 10 ? ? ? ? ? ? ? ? ? ? ? ? = 443 nm c) f w = f a = 5.09 ? 10 14 sec –1 [Frequency does not change] d) 8 a a a w w w a w v v 3 10 v v 1.33 ? ? ? ? ? ? ? ? ? = 2.25 ? 10 8 m/sec. 3. We know that, 2 1 1 2 v v ? ? ? So, 8 8 400 400 1472 3 10 v 2.04 10 m/ sec. 1 v ? ? ? ? ? [because, for air, ? = 1 and v = 3 ? 10 8 m/s] Again, 8 8 760 760 1452 3 10 v 2.07 10 m/ sec. 1 v ? ? ? ? ? ? 4. 8 t 8 1 3 10 1.25 (2.4) 10 ? ? ? ? ? ? velocity of light in vaccum since, = velocity of light in the given medium ? ? ? ? ? ? ? 5. Given that, d = 1 cm = 10 –2 m, ? = 5 ? 10 –7 m and D = 1 m a) Separation between two consecutive maxima is equal to fringe width. So, ? = 7 2 D 5 10 1 d 10 ? ? ? ? ? ? m = 5 ? 10 –5 m = 0.05 mm. b) When, ? = 1 mm = 10 –3 m 10 –3 m = 7 5 10 1 D ? ? ? ? D = 5 ? 10 –4 m = 0.50 mm. ? 6. Given that, ? = 1 mm = 10 –3 m, D = 2.t m and d = 1 mm = 10 –3 m So, 10 –3 m = 3 25 10 ? ? ? ? ? = 4 ? 10 –7 m = 400 nm. ? 7. Given that, d = 1 mm = 10 –3 m, D = 1 m. So, fringe with = D d ? = 0.5 mm. a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm b) No. of fringes = 10 / 0.5 = 20. 8. Given that, d = 0.8 mm = 0.8 ? 10 –3 m, ? = 589 nm = 589 ? 10 –9 m and D = 2 m. So, ? = D d ? = 9 3 589 10 2 0.8 10 ? ? ? ? ? = 1.47 ? 10 –3 m = 147 mm. ? Chapter 17 17.2 9. Given that, ? = 500 nm = 500 ? 10 –9 m and d = 2 ? 10 –3 m As shown in the figure, angular separation ? = D D dD d ? ? ? ? ? So, ? = 9 3 500 10 D d 2 10 ? ? ? ? ? ? ? ? = 250 ? 10 –6 = 25 ? 10 –5 radian = 0.014 degree. ? 10. We know that, the first maximum (next to central maximum) occurs at y = D d ? Given that, ? 1 = 480 nm, ? 2 = 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25 ? 10 –3 m So, y 1 = 9 1 3 D 1.5 480 10 d 0.25 10 ? ? ? ? ? ? ? = 2.88 mm y 2 = 9 3 1.5 600 10 0.25 10 ? ? ? ? ? = 3.6 mm. So, the separation between these two bright fringes is given by, ? separation = y 2 – y 1 = 3.60 – 2.88 = 0.72 mm. 11. Let m th bright fringe of violet light overlaps with n th bright fringe of red light. ? m 400nm D n 700nm D m 7 d d n 4 ? ? ? ? ? ? ? ? 7 th bright fringe of violet light overlaps with 4 th bright fringe of red light (minimum). Also, it can be seen that 14 th violet fringe will overlap 8 th red fringe. Because, m/n = 7/4 = 14/8. 12. Let, t = thickness of the plate Given, optical path difference = ( ? – 1)t = ?/2 ? t = 2( 1) ? ? ? ? 13. a) Change in the optical path = ?t – t = ( ? – 1)t b) To have a dark fringe at the centre the pattern should shift by one half of a fringe. ? ( ? – 1)t = t 2 2( 1) ? ? ? ? ? ? . ? 14. Given that, ? = 1.45, t = 0.02 mm = 0.02 ? 10 –3 m and ? = 620 nm = 620 ? 10 –9 m We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( ? – 1)t. Again, for shift of one fringe, the optical path should be changed by ?. So, no. of fringes crossing through the centre is given by, n = 3 9 ( 1)t 0.45 0.02 10 620 10 ? ? ? ? ? ? ? ? ? = 14.5 ? 15. In the given Young’s double slit experiment, ? = 1.6, t = 1.964 micron = 1.964 ? 10 –6 m We know, number of fringes shifted = ( 1)t ? ? ? So, the corresponding shift = No.of fringes shifted ? fringe width = ( 1)t D ( 1)tD d d ? ? ? ? ? ? ? ? … (1) Again, when the distance between the screen and the slits is doubled, Fringe width = (2D) d ? …(2) From (1) and (2), ( 1)tD d ? ? = (2D) d ? ? ? = ( 1)t ? ? ? = 6 (1.6 1) (1.964) 10 2 ? ? ? ? = 589.2 ? 10 –9 = 589.2 nm. ? B S 1 S 2 ? ? D Chapter 17 17.3 16. Given that, t 1 = t 2 = 0.5 mm = 0.5 ? 10 –3 m, ? m = 1.58 and ? p = 1.55, ? = 590 nm = 590 ? 10 –9 m, d = 0.12 cm = 12 ? 10 –4 m, D = 1 m a) Fringe width = 9 4 D 1 590 10 d 12 10 ? ? ? ? ? ? ? = 4.91 ? 10 –4 m. b) When both the strips are fitted, the optical path changes by ?x = ( ? m – 1)t 1 – ( ? p – 1)t 2 = ( ? m – ? p )t = (1.58 – 1.55) ? (0.5)(10 –3 ) = 0.015 ? 10 –13 m. So, No. of fringes shifted = 3 3 0.015 10 590 10 ? ? ? ? = 25.43. ? There are 25 fringes and 0.43 th of a fringe. ? There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe. So, position of first maximum on both sides will be given by ? x = 0.43 ? 4.91 ? 10 –4 = 0.021 cm x ? = (1 – 0.43) ? 4.91 ? 10 –4 = 0.028 cm (since, fringe width = 4.91 ? 10 –4 m) 17. The change in path difference due to the two slabs is ( ? 1 – ? 2 )t (as in problem no. 16). For having a minimum at P 0 , the path difference should change by ?/2. So, ? ?/2 = ( ? 1 – ?? 2 )t ? t = 1 2 2( ) ? ? ? ? . ? 18. Given that, t = 0.02 mm = 0.02 ? 10 –3 m, ? 1 = 1.45, ? = 600 nm = 600 ? 10 –9 m a) Let, I 1 = Intensity of source without paper = I b) Then I 2 = Intensity of source with paper = (4/9)I ? 1 1 2 2 I r 9 3 I 4 r 2 ? ? ? [because I ? r 2 ] where, r 1 and r 2 are corresponding amplitudes. So, 2 max 1 2 2 min 1 2 I (r r ) I (r r ) ? ? ? = 25 : 1 b) No. of fringes that will cross the origin is given by, n = ( 1)t ? ? ? = 3 9 (1.45 1) 0.02 10 600 10 ? ? ? ? ? ? = 15. 19. Given that, d = 0.28 mm = 0.28 ? 10 –3 m, D = 48 cm = 0.48 m, ? a = 700 nm in vacuum Let, ? w = wavelength of red light in water Since, the fringe width of the pattern is given by, ??= 9 w 3 D 525 10 0.48 d 0.28 10 ? ? ? ? ? ? ? = 9 ? 10 –4 m = 0.90 mm. ? 20. It can be seen from the figure that the wavefronts reaching O from S 1 and S 2 will have a path difference of S 2 X. In the ? S 1 S 2 X, sin ???= 2 1 2 S X S S So, path difference = S 2 X = S 1 S 2 sin ? = d sin ? = d ? ?/2d = ?/2 As the path difference is an odd multiple of ?/2, there will be a dark fringe at point P 0 . 21. a) Since, there is a phase difference of ? between direct light and reflecting light, the intensity just above the mirror will be zero. b) Here, 2d = equivalent slit separation D = Distance between slit and screen. We know for bright fringe, ?x = y 2d D ? = n ? But as there is a phase reversal of ?/2. ? y 2d D ? + 2 ? = n ? ? y 2d D ? = n ? – 2 ? ? y = D 4d ? ? mica Screen polysterene S 2 S 1 ? ? P 0 ? ? x S 2 S 1 Dark fringe (1 – 0.43) ?? 0.43 ?? S 2 S 1 2d D Screen Chapter 17 17.4 22. Given that, D = 1 m, ? = 700 nm = 700 ? 10 –9 m Since, a = 2 mm, d = 2a = 2mm = 2 ? 10 –3 m (L loyd’s mirror experiment) Fringe width = 9 3 D 700 10 m 1 m d 2 10 m ? ? ? ? ? ? ? = 0.35 mm. 23. Given that, the mirror reflects 64% of energy (intensity) of the light. So, 1 1 2 2 I r 16 4 0.64 I 25 r 5 ? ? ? ? So, 2 max 1 2 2 min 1 2 I (r r ) I (r r ) ? ? ? = 81 : 1. 24. It can be seen from the figure that, the apparent distance of the screen from the slits is, D = 2D 1 + D 2 So, Fringe width = 1 2 (2D D ) D d d ? ? ? ? 25. Given that, ? = (400 nm to 700 nm), d = 0.5 mm = 0.5 ? 10 –3 m, D = 50 cm = 0.5 m and on the screen y n = 1 mm = 1 ? 10 –3 m a) We know that for zero intensity (dark fringe) y n = n D 2n 1 2 d ? ? ? ? ? ? ? ? where n = 0, 1, 2, ……. ? ? n = 3 3 6 3 n d 2 2 10 0.5 10 2 2 10 m 10 nm (2n 1) D 2n 1 0.5 (2n 1) (2n 1) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? If n = 1, ? 1 = (2/3) ? 1000 = 667 nm If n = 1, ? 2 = (2/5) ? 1000 = 400 nm So, the light waves of wavelengths 400 nm and 667 nm will be absent from the out coming light. b) For strong intensity (bright fringes) at the hole ? y n = n n n n D y d d nD ? ? ? ? When, n = 1, ? 1 = n y d D = 3 3 6 10 0.5 10 10 m 1000nm 0.5 ? ? ? ? ? ? ? . 1000 nm is not present in the range 400 nm – 700 nm Again, where n = 2, ? 2 = n y d 2D = 500 nm So, the only wavelength which will have strong intensity is 500 nm. ? 26. From the diagram, it can be seen that at point O. Path difference = (AB + BO) – (AC + CO) = 2(AB – AC) [Since, AB = BO and AC = CO] = 2 2 2( d D D) ? ? For dark fringe, path difference should be odd multiple of ?/2. So, 2 2 2( d D D) ? ? = (2n + 1)( ?/2) ? 2 2 d D ? = D + (2n + 1) ?/4 ? D 2 + d 2 = D 2 + (2n+1) 2 ? 2 /16 + (2n + 1) ?D/2 Neglecting, (2n+1) 2 ? 2 /16, as it is very small We get, d = D (2n 1) 2 ? ? For minimum ‘d’, putting n = 0 ? d min = D 2 ? . ? C O P B d A x D D D y n d=0.5mm 50cm 1 mm Page 5 17.1 SOLUTIONS TO CONCEPTS CHAPTER 17 1. Given that, 400 m < ? < 700 nm. 1 1 1 700nm 400nm ? ? ? ? 8 8 7 7 7 7 1 1 1 3 10 c 3 10 7 10 4 10 7 10 4 10 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? (Where, c = speed of light = 3 ? 10 8 m/s) ? 4.3 ? 10 14 < c/ ? < 7.5 ? 10 14 ? 4.3 ? 10 14 Hz < f < 7.5 ? 10 14 Hz. ? 2. Given that, for sodium light, ? = 589 nm = 589 ? 10 –9 m a) f a = 8 9 3 10 589 10 ? ? ? = 5.09 ? 10 14 1 c sec f ? ? ? ? ? ? ? ? ? ? b) a w w w 9 w a 1 1.33 589 10 ? ? ? ? ? ? ? ? ? ? ? ? = 443 nm c) f w = f a = 5.09 ? 10 14 sec –1 [Frequency does not change] d) 8 a a a w w w a w v v 3 10 v v 1.33 ? ? ? ? ? ? ? ? ? = 2.25 ? 10 8 m/sec. 3. We know that, 2 1 1 2 v v ? ? ? So, 8 8 400 400 1472 3 10 v 2.04 10 m/ sec. 1 v ? ? ? ? ? [because, for air, ? = 1 and v = 3 ? 10 8 m/s] Again, 8 8 760 760 1452 3 10 v 2.07 10 m/ sec. 1 v ? ? ? ? ? ? 4. 8 t 8 1 3 10 1.25 (2.4) 10 ? ? ? ? ? ? velocity of light in vaccum since, = velocity of light in the given medium ? ? ? ? ? ? ? 5. Given that, d = 1 cm = 10 –2 m, ? = 5 ? 10 –7 m and D = 1 m a) Separation between two consecutive maxima is equal to fringe width. So, ? = 7 2 D 5 10 1 d 10 ? ? ? ? ? ? m = 5 ? 10 –5 m = 0.05 mm. b) When, ? = 1 mm = 10 –3 m 10 –3 m = 7 5 10 1 D ? ? ? ? D = 5 ? 10 –4 m = 0.50 mm. ? 6. Given that, ? = 1 mm = 10 –3 m, D = 2.t m and d = 1 mm = 10 –3 m So, 10 –3 m = 3 25 10 ? ? ? ? ? = 4 ? 10 –7 m = 400 nm. ? 7. Given that, d = 1 mm = 10 –3 m, D = 1 m. So, fringe with = D d ? = 0.5 mm. a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm b) No. of fringes = 10 / 0.5 = 20. 8. Given that, d = 0.8 mm = 0.8 ? 10 –3 m, ? = 589 nm = 589 ? 10 –9 m and D = 2 m. So, ? = D d ? = 9 3 589 10 2 0.8 10 ? ? ? ? ? = 1.47 ? 10 –3 m = 147 mm. ? Chapter 17 17.2 9. Given that, ? = 500 nm = 500 ? 10 –9 m and d = 2 ? 10 –3 m As shown in the figure, angular separation ? = D D dD d ? ? ? ? ? So, ? = 9 3 500 10 D d 2 10 ? ? ? ? ? ? ? ? = 250 ? 10 –6 = 25 ? 10 –5 radian = 0.014 degree. ? 10. We know that, the first maximum (next to central maximum) occurs at y = D d ? Given that, ? 1 = 480 nm, ? 2 = 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25 ? 10 –3 m So, y 1 = 9 1 3 D 1.5 480 10 d 0.25 10 ? ? ? ? ? ? ? = 2.88 mm y 2 = 9 3 1.5 600 10 0.25 10 ? ? ? ? ? = 3.6 mm. So, the separation between these two bright fringes is given by, ? separation = y 2 – y 1 = 3.60 – 2.88 = 0.72 mm. 11. Let m th bright fringe of violet light overlaps with n th bright fringe of red light. ? m 400nm D n 700nm D m 7 d d n 4 ? ? ? ? ? ? ? ? 7 th bright fringe of violet light overlaps with 4 th bright fringe of red light (minimum). Also, it can be seen that 14 th violet fringe will overlap 8 th red fringe. Because, m/n = 7/4 = 14/8. 12. Let, t = thickness of the plate Given, optical path difference = ( ? – 1)t = ?/2 ? t = 2( 1) ? ? ? ? 13. a) Change in the optical path = ?t – t = ( ? – 1)t b) To have a dark fringe at the centre the pattern should shift by one half of a fringe. ? ( ? – 1)t = t 2 2( 1) ? ? ? ? ? ? . ? 14. Given that, ? = 1.45, t = 0.02 mm = 0.02 ? 10 –3 m and ? = 620 nm = 620 ? 10 –9 m We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( ? – 1)t. Again, for shift of one fringe, the optical path should be changed by ?. So, no. of fringes crossing through the centre is given by, n = 3 9 ( 1)t 0.45 0.02 10 620 10 ? ? ? ? ? ? ? ? ? = 14.5 ? 15. In the given Young’s double slit experiment, ? = 1.6, t = 1.964 micron = 1.964 ? 10 –6 m We know, number of fringes shifted = ( 1)t ? ? ? So, the corresponding shift = No.of fringes shifted ? fringe width = ( 1)t D ( 1)tD d d ? ? ? ? ? ? ? ? … (1) Again, when the distance between the screen and the slits is doubled, Fringe width = (2D) d ? …(2) From (1) and (2), ( 1)tD d ? ? = (2D) d ? ? ? = ( 1)t ? ? ? = 6 (1.6 1) (1.964) 10 2 ? ? ? ? = 589.2 ? 10 –9 = 589.2 nm. ? B S 1 S 2 ? ? D Chapter 17 17.3 16. Given that, t 1 = t 2 = 0.5 mm = 0.5 ? 10 –3 m, ? m = 1.58 and ? p = 1.55, ? = 590 nm = 590 ? 10 –9 m, d = 0.12 cm = 12 ? 10 –4 m, D = 1 m a) Fringe width = 9 4 D 1 590 10 d 12 10 ? ? ? ? ? ? ? = 4.91 ? 10 –4 m. b) When both the strips are fitted, the optical path changes by ?x = ( ? m – 1)t 1 – ( ? p – 1)t 2 = ( ? m – ? p )t = (1.58 – 1.55) ? (0.5)(10 –3 ) = 0.015 ? 10 –13 m. So, No. of fringes shifted = 3 3 0.015 10 590 10 ? ? ? ? = 25.43. ? There are 25 fringes and 0.43 th of a fringe. ? There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe. So, position of first maximum on both sides will be given by ? x = 0.43 ? 4.91 ? 10 –4 = 0.021 cm x ? = (1 – 0.43) ? 4.91 ? 10 –4 = 0.028 cm (since, fringe width = 4.91 ? 10 –4 m) 17. The change in path difference due to the two slabs is ( ? 1 – ? 2 )t (as in problem no. 16). For having a minimum at P 0 , the path difference should change by ?/2. So, ? ?/2 = ( ? 1 – ?? 2 )t ? t = 1 2 2( ) ? ? ? ? . ? 18. Given that, t = 0.02 mm = 0.02 ? 10 –3 m, ? 1 = 1.45, ? = 600 nm = 600 ? 10 –9 m a) Let, I 1 = Intensity of source without paper = I b) Then I 2 = Intensity of source with paper = (4/9)I ? 1 1 2 2 I r 9 3 I 4 r 2 ? ? ? [because I ? r 2 ] where, r 1 and r 2 are corresponding amplitudes. So, 2 max 1 2 2 min 1 2 I (r r ) I (r r ) ? ? ? = 25 : 1 b) No. of fringes that will cross the origin is given by, n = ( 1)t ? ? ? = 3 9 (1.45 1) 0.02 10 600 10 ? ? ? ? ? ? = 15. 19. Given that, d = 0.28 mm = 0.28 ? 10 –3 m, D = 48 cm = 0.48 m, ? a = 700 nm in vacuum Let, ? w = wavelength of red light in water Since, the fringe width of the pattern is given by, ??= 9 w 3 D 525 10 0.48 d 0.28 10 ? ? ? ? ? ? ? = 9 ? 10 –4 m = 0.90 mm. ? 20. It can be seen from the figure that the wavefronts reaching O from S 1 and S 2 will have a path difference of S 2 X. In the ? S 1 S 2 X, sin ???= 2 1 2 S X S S So, path difference = S 2 X = S 1 S 2 sin ? = d sin ? = d ? ?/2d = ?/2 As the path difference is an odd multiple of ?/2, there will be a dark fringe at point P 0 . 21. a) Since, there is a phase difference of ? between direct light and reflecting light, the intensity just above the mirror will be zero. b) Here, 2d = equivalent slit separation D = Distance between slit and screen. We know for bright fringe, ?x = y 2d D ? = n ? But as there is a phase reversal of ?/2. ? y 2d D ? + 2 ? = n ? ? y 2d D ? = n ? – 2 ? ? y = D 4d ? ? mica Screen polysterene S 2 S 1 ? ? P 0 ? ? x S 2 S 1 Dark fringe (1 – 0.43) ?? 0.43 ?? S 2 S 1 2d D Screen Chapter 17 17.4 22. Given that, D = 1 m, ? = 700 nm = 700 ? 10 –9 m Since, a = 2 mm, d = 2a = 2mm = 2 ? 10 –3 m (L loyd’s mirror experiment) Fringe width = 9 3 D 700 10 m 1 m d 2 10 m ? ? ? ? ? ? ? = 0.35 mm. 23. Given that, the mirror reflects 64% of energy (intensity) of the light. So, 1 1 2 2 I r 16 4 0.64 I 25 r 5 ? ? ? ? So, 2 max 1 2 2 min 1 2 I (r r ) I (r r ) ? ? ? = 81 : 1. 24. It can be seen from the figure that, the apparent distance of the screen from the slits is, D = 2D 1 + D 2 So, Fringe width = 1 2 (2D D ) D d d ? ? ? ? 25. Given that, ? = (400 nm to 700 nm), d = 0.5 mm = 0.5 ? 10 –3 m, D = 50 cm = 0.5 m and on the screen y n = 1 mm = 1 ? 10 –3 m a) We know that for zero intensity (dark fringe) y n = n D 2n 1 2 d ? ? ? ? ? ? ? ? where n = 0, 1, 2, ……. ? ? n = 3 3 6 3 n d 2 2 10 0.5 10 2 2 10 m 10 nm (2n 1) D 2n 1 0.5 (2n 1) (2n 1) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? If n = 1, ? 1 = (2/3) ? 1000 = 667 nm If n = 1, ? 2 = (2/5) ? 1000 = 400 nm So, the light waves of wavelengths 400 nm and 667 nm will be absent from the out coming light. b) For strong intensity (bright fringes) at the hole ? y n = n n n n D y d d nD ? ? ? ? When, n = 1, ? 1 = n y d D = 3 3 6 10 0.5 10 10 m 1000nm 0.5 ? ? ? ? ? ? ? . 1000 nm is not present in the range 400 nm – 700 nm Again, where n = 2, ? 2 = n y d 2D = 500 nm So, the only wavelength which will have strong intensity is 500 nm. ? 26. From the diagram, it can be seen that at point O. Path difference = (AB + BO) – (AC + CO) = 2(AB – AC) [Since, AB = BO and AC = CO] = 2 2 2( d D D) ? ? For dark fringe, path difference should be odd multiple of ?/2. So, 2 2 2( d D D) ? ? = (2n + 1)( ?/2) ? 2 2 d D ? = D + (2n + 1) ?/4 ? D 2 + d 2 = D 2 + (2n+1) 2 ? 2 /16 + (2n + 1) ?D/2 Neglecting, (2n+1) 2 ? 2 /16, as it is very small We get, d = D (2n 1) 2 ? ? For minimum ‘d’, putting n = 0 ? d min = D 2 ? . ? C O P B d A x D D D y n d=0.5mm 50cm 1 mm Chapter 17 17.5 27. For minimum intensity ? S 1 P – S 2 P = x = (2n +1) ?/2 From the figure, we get ? 2 2 Z (2 ) Z (2n 1) 2 ? ? ? ? ? ? ? 2 2 2 2 2 Z 4 Z (2n 1) Z(2n 1) 4 ? ? ? ? ? ? ? ? ? ? Z = 2 2 2 2 2 2 4 (2n 1) ( / 4) 16 (2n 1) (2n 1) 4(2n 1) ? ? ? ? ? ? ? ? ? ? ? ? ? …(1) Putting, n = 0 ? Z = 15 ?/4 n = –1 ? Z = –15 ?/4 n = 1 ? Z = 7 ?/12 n = 2 ? Z = –9 ?/20 ? Z = 7 ?/12 is the smallest distance for which there will be minimum intensity. ? 28. Since S 1 , S 2 are in same phase, at O there will be maximum intensity. Given that, there will be a maximum intensity at P. ? path difference = ?x = n ? From the figure, (S 1 P) 2 – (S 2 P) 2 = 2 2 2 2 2 2 ( D X ) ( (D 2 ) X ) ? ? ? ? ? = 4 ?D – 4 ? 2 = 4 ?D ( ? 2 is so small and can be neglected) ? S 1 P – S 2 P = 2 2 4 D 2 x D ? ? = n ? ?? 2 2 2D x D ? ????? ? n 2 (X 2 + D 2 ) = 4D 2 = ?X = 2 D 4 n n ? when n = 1, x = 3 D (1 st order) n = 2, x = 0 (2 nd order) ? When X = 3 D, at P there will be maximum intensity. ? 29. As shown in the figure, (S 1 P) 2 = (PX) 2 + (S 1 X) 2 …(1) (S 2 P) 2 = (PX) 2 + (S 2 X) 2 …(2) From (1) and (2), (S 1 P) 2 – (S 2 P) 2 = (S 1 X) 2 – (S 2 X) 2 = (1.5 ? + R cos ?) 2 – (R cos ? – 15 ?) 2 = 6 ? R cos ? ? (S 1 P – S 2 P) = 6 Rcos 2R ? ? = 3 ? cos ?. For constructive interference, (S 1 P – S 2 P) 2 = x = 3 ? cos ? = n ? ? cos ? = n/3 ? ? = cos –1 (n/3), where n = 0, 1, 2, …. ? ? = 0°, 48.2°, 70.5°, 90° and similar points in other quadrants. ? 30. a) As shown in the figure, BP 0 – AP 0 = ?/3 ? 2 2 (D d ) D / 3 ? ? ? ? ? D 2 + d 2 = D 2 + ( ? 2 / 9) + (2 ?D)/3 ? d = (2 D)/ 3 ? (neglecting the term ? 2 /9 as it is very small) b) To find the intensity at P 0 , we have to consider the interference of light waves coming from all the three slits. Here, CP 0 – AP 0 = 2 2 D 4d D ? ? S 2 Z S 1 P 2 ?? Screen P 0 x d d D C B A Screen x 2 ? ? O S 2 S 1 P D ? ? x S 1 1.5 ?? O P S 2 RRead More

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