SOLUTIONS TO CONCEPTS CHAPTER 17
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1 1. Given that, 400 m < < 700 nm nm 400nm SOLUTIONS TO CONCETS CHATER c 3 10 (Where, c = spee of light = 3 10 m/s) < c/ < Hz < f < Hz.. Given that, for soium light, = 59 nm = m a) f a = b) w a 9 = c sec f a w 1 w w = 443 nm c) f w = f a = sec 1 [Frequency oes not change] ) v v 3 10 v 1.33 a w a a vw w a w v1 3. We know that, v 4. 1 =.5 10 m/sec So, v m / sec. 1 v 400 [because, for air, = 1 an v = 3 10 m/s] Again, v m / sec. 1 v (.4) 10 t velocity of light in vaccum since, = velocity of light in the given meium 5. Given that, = 1 cm = 10 m, = m an = 1 m a) Separation between two consecutive maima is equal to fringe with. So, = m = m = 0.05 mm. 10 b) When, = 1 mm = 10 3 m m = = m = 0.50 mm. 6. Given that, = 1 mm = 10 3 m, =.t m an = 1 mm = 10 3 m So, 10 3 m = 5 3 = m = 400 nm Given that, = 1 mm = 10 3 m, = 1 m. So, fringe with = = 0.5 mm. a) So, istance of centre of first minimum from centre of central maimum = 0.5/ mm = 0.5 mm b) No. of fringes = 10 / 0.5 = 0.. Given that, = 0. mm = m, = 59 nm = m an = m. So, = = = m = 147 mm
2 9. Given that, = 500 nm = m an = 10 3 m As shown in the figure, angular separation = So, = = S = raian = egree. 10. We know that, the first maimum (net to central maimum) occurs at y = Given that, 1 = 40 nm, = 600 nm, = 150 cm = 1.5 m an = 0.5 mm = m So, y 1 = =. mm y = = 3.6 mm So, the separation between these two bright fringes is given by, separation = y y 1 = = 0.7 mm. 11. Let m th bright fringe of violet light overlaps with n th bright fringe of re light. m 400nm n 700nm m 7 n 4 7 th bright fringe of violet light overlaps with 4 th bright fringe of re light (minimum). Also, it can be seen that 14 th violet fringe will overlap th re fringe. Because, m/n = 7/4 = 14/. 1. Let, t = thickness of the plate Given, optical path ifference = ( 1)t = / t = ( 1) 13. a) Change in the optical path = t t = ( 1)t b) To have a ark fringe at the centre the pattern shoul shift by one half of a fringe. ( 1)t = t. ( 1) 14. Given that, = 1.45, t = 0.0 mm = m an = 60 nm = m We know, when the transparent paper is paste in one of the slits, the optical path changes by ( 1)t. Again, for shift of one fringe, the optical path shoul be change by. So, no. of fringes crossing through the centre is given by, 3 ( 1)t n = = In the given Young s ouble slit eperiment, = 1.6, t = micron = m We know, number of fringes shifte = ( 1)t So, the corresponing shift = No.of fringes shifte fringe with ( 1)t ( 1)t = (1) Again, when the istance between the screen an the slits is ouble, () Fringe with = () From (1) an (), ( 1)t = = ( 1)t = () (1.6 1) (1.964) 10 6 = = 59. nm. B
3 16. Given that, t 1 = t = 0.5 mm = m, m = 1.5 an p = 1.55, = 590 nm = m, = 0.1 cm = m, = 1 m a) Fringe with = = m b) When both the strips are fitte, the optical path changes by = ( m 1)t 1 ( p 1)t = ( m p )t = ( ) (0.5)(10 3 ) = m. So, No. of fringes shifte = = There are 5 fringes an 0.43 th of a fringe. There are 13 bright fringes an 1 ark fringes an 0.43 th of a ark fringe. So, position of first maimum on both sies will be given by = = 0.01 cm = (1 0.43) = 0.0 cm (since, fringe with = m) 17. The change in path ifference ue to the two slabs is ( 1 )t (as in problem no. 16). For having a minimum at 0, the path ifference shoul change by /. So, / = ( 1 )t t = ( ) Given that, t = 0.0 mm = m, 1 = 1.45, = 600 nm = m a) Let, I 1 = Intensity of source without paper = I b) Then I = Intensity of source with paper = (4/9)I I 1 9 r 1 3 [because I r ] I 4 r where, r 1 an r are corresponing amplitues. I ma (r1 r ) So, I (r r ) min 1 = 5 : 1 b) No. of fringes that will cross the origin is given by, 3 n = ( 1)t (1.45 1) = = Given that, = 0. mm = m, = 4 cm = 0.4 m, a = 700 nm in vacuum Let, w = wavelength of re light in water Since, the fringe with of the pattern is given by, 9 w = = m = 0.90 mm It can be seen from the figure that the wavefronts reaching O from an S will have a path ifference of S X. In the S X, SX sin = S S 1 So, path ifference = S X = S sin = sin = / = / As the path ifference is an o multiple of /, there will be a ark fringe at point a) Since, there is a phase ifference of between irect light an reflecting light, the intensity just above the mirror will be zero. b) Here, = equivalent slit separation = istance between slit an screen. We know for bright fringe, = y = n But as there is a phase reversal of /. y y + = n = n y = S S mica polysterene (1 0.43) 0.43 S Screen ark fringe 0 Screen
4 . Given that, = 1 m, = 700 nm = m Since, a = mm, = a = mm = 10 3 m (L loy s mirror eperiment) m 1m Fringe with = = 0.35 mm m 3. Given that, the mirror reflects 64% of energy (intensity) of the light. I So, 1 16 r I 5 r 5 So, I ma (r1 r ) I min (r1 r ) = 1 : It can be seen from the figure that, the apparent istance of the screen from the slits is, = 1 + (1 ) So, Fringe with = 5. Given that, = (400 nm to 700 nm), = 0.5 mm = m, = 50 cm = 0.5 m an on the screen y n = 1 mm = m a) We know that for zero intensity (ark fringe) y n = n 1 n where n = 0, 1,,. =0.5mm 50cm y n 1 mm n = 3 3 n m 10 nm (n 1) n (n 1) (n 1) If n = 1, 1 = (/3) 1000 = 667 nm If n = 1, = (/5) 1000 = 400 nm So, the light waves of wavelengths 400 nm an 667 nm will be absent from the out coming light. b) For strong intensity (bright fringes) at the hole nn yn y n = n n When, n = 1, 1 = 3 3 yn = m 1000nm nm is not present in the range 400 nm 700 nm yn Again, where n =, = = 500 nm So, the only wavelength which will have strong intensity is 500 nm. 6. From the iagram, it can be seen that at point O. ath ifference = (AB + BO) (AC + CO) = (AB AC) [Since, AB = BO an AC = CO] = ( ) For ark fringe, path ifference shoul be o multiple of /. So, ( ) = (n + 1)(/) = + (n + 1) /4 + = + (n+1) /16 + (n + 1) / Neglecting, (n+1) /16, as it is very small We get, = (n 1) A B C O For minimum, putting n = 0 min =. 17.4
5 7. For minimum intensity S = = (n +1) / From the figure, we get Z ( ) Z (n 1) Screen Z 4 Z (n 1) Z(n 1) 4 Z = 4 (n 1) ( / 4) 16 (n 1) (n 1) 4(n 1) (1) utting, n = 0 Z = 15/4 n = 1 Z = 15/4 n = 1 Z = 7/1 n = Z = 9/0 Z = 7/1 is the smallest istance for which there will be minimum intensity.. Since, S are in same phase, at O there will be maimum intensity. Given that, there will be a maimum intensity at. path ifference = = n From the figure, ( ) (S ) = ( X ) ( ( ) X ) = 4 4 = 4 ( is so small an can be neglecte) S = 4 = n S S Z O Screen n (X + ) = 4 = X = 4 n n when n = 1, = 3 (1 st orer) n =, = 0 ( n orer) When X = 3, at there will be maimum intensity. 9. As shown in the figure, ( ) = (X) + ( X) (1) (S ) = (X) + (S X) () From (1) an (), ( ) (S ) = ( X) (S X) = (1.5 + R cos ) (R cos 15 ) = 6 R cos ( S ) = 6 Rcos = 3 cos. R For constructive interference, ( S ) = = 3 cos = n cos = n/3 = cos 1 (n/3), where n = 0, 1,,. = 0, 4., 70.5, 90 an similar points in other quarants. 30. a) As shown in the figure, B 0 A 0 = /3 ( ) / 3 + = + ( / 9) + ()/3 = ( ) / 3 (neglecting the term /9 as it is very small) b) To fin the intensity at 0, we have to consier the interference of light waves coming from all the three slits. R 1.5 O C B A S 0 Here, C 0 A 0 =
6 = 1/ = [using binomial epansion] 3 3 So, the corresponing phase ifference between waves from C an A is, c = 4 (1) Again, B = () 3 3 So, it can be sai that light from B an C are in same phase as they have some phase ifference with respect to A. So, R = = (r) r r r cos( / 3) (using vector metho) 4r r r 3r 0 I K( 3r ) 3Kr 3I As, the resulting amplitue is 3 times, the intensity will be three times the intensity ue to iniviual slits. 31. Given that, = mm = 10 3 m, = 600 nm = m, I ma = 0.0 W/m, = m For the point, y = 0.5 cm 3 y We know, path ifference = = So, the corresponing phase ifference is, 6 = m 5 10 = = So, the amplitue of the resulting wave at the point y = 0.5 cm is, A = Since, I I r r r cos( / 3) r r r = r ma A [since, maimum amplitue = r] (r) I A r 0. 4r 4r I W/m i) When intensity is half the maimum 4a cos ( / ) 1 4a cos ( / ) 1/ cos( / ) 1/ I 1 I ma / = /4 = / ath ifference, = /4 y = / = /4 ii) When intensity is 1/4 th I 1 of the maimum I 4 4a cos ( / ) 1 4a 4 cos ( / ) 1/ 4 cos( / ) 1/ / = /3 = /3 ath ifference, = /3 y = / = /3 ma
7 33. Given that, = 1 m, = 1 mm = 10 3 m, = 500 nm = m For intensity to be half the maimum intensity. y = (As in problem no. 3) y = y = m The line with of a bright fringe is sometimes efine as the separation between the points on the two sies of the central line where the intensity falls to half the maimum. We know that, for intensity to be half the maimum y = ± 4 Line with = =. 35. i) When, z = /, at S 4, minimum intensity occurs (ark fringe) Amplitue = 0, At S 3, path ifference = 0 Maimum intensity occurs. Amplitue = r. So, on screen, I ma (r 0) I (r 0) min = 1 ii) When, z = /, At S 4, minimum intensity occurs. (ark fringe) Amplitue = 0. At S 3, path ifference = 0 Maimum intensity occurs. Amplitue = r. So, on screen, I ma (r r) I (r 0) min iii) When, z = /4, At S 4, intensity = I ma / Amplitue = r. At S 3, intensity is maimum. Amplitue = r I ma (r r ) I (r r ) min = a) When, z = / So, OS 3 = OS 4 = / ark fringe at S 3 an S 4. At S 3, intensity at S 3 = 0 I 1 = 0 At S 4, intensity at S 4 = 0 I = 0 At, path ifference = 0 hase ifference = 0. I = I 1 + I + I1I cos 0 = = 0 Intensity at = 0. b) Given that, when z = /, intensity at = I Here, OS 3 = OS 4 = y = /4 = y. [Since, = path ifference = y/] 4 Let, intensity at S 3 an S 4 = I At, phase ifference = 0 So, I + I + I cos 0 = I. 4I = I I = 1/4. S S 1 S 4 S 3 S 3 O S 4 z
8 When, z = 3, 3 y = 4 y 3 3 = 4 Let, I be the intensity at S 3 an S 4 when, = 3/ Now comparing, I a a a cos(3 / ) a 1 I = I = I/4. I a a a cos / a Intensity at = I/4 + I/4 + (I/4) cos 0 = I/ + I/ = I. c) When z = / y = OS 3 = OS 4 = / = y. Let, I = intensity at S 3 an S 4 when, =. I a a a cos 4a I a a a cos / a I = I = (I/4) = I/ At, I resultant = I/ + I/ + (I/) cos 0 = I + I = I. So, the resultant intensity at will be I. 37. Given = m For minimum reflection of light, = n n n n 5. = 7 (n) = 0.13 (n) Given that, has a value in between 1. an 1.5. When, n = 5, = = Given that, = m, = For strong reflection, = (n + 1)/ = For minimum thickness, putting n = 0. (n 1) = = = 10 7 m = 100 nm For strong transmission, = n = n Given that, = 1.33, = cm = m. = when, m n n n = 4, 1 = 665 nm n = 5, = 53 nm n = 6, 3 = 443 nm 40. For the thin oil film, = cm = 10 6 m, oil = 1.5 an = 1.50 = m (n 1/ ) n 1 n nm = n 1 For the wavelengths in the region (400 nm 750 nm) When, n = 3, = = nm
9 When, n = 4, = = nm When, n = 5, = = nm For first minimum iffraction, b sin = Here, = 30, b = 5 cm = 5 sin 30 = 5/ =.5 cm. 4. = 560 nm = m, b = 0.0 mm = 10 4 m, = m Since, R = 1. = 1. = M = 0.63 cm. b 4 10 So, iameter = R = 1.37 cm. 43. = 60 nm = m, = 0 cm = 0 10 m, b = cm = 10 m R = So, iameter = R = m 9 = = m 17.9
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