HW_sol_chapter24

HW_sol_chapter24 - 24.3 (a) The distance between the...

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Unformatted text preview: 24.3 (a) The distance between the central maximum and the first order bright fringe is M _(546.1><10—g m)(1.20 m) A¢:=__—=2.62x10—3m= d 0.250 ><10’3 m (b) The distance between the first and second dark bands is l L “H, = 7 = as in (a) above. 24.6 (a) The location of the bright fringe of order m (measured from the position of the central maximum) is (y = (AL/mm, m = 0, i1, i 2,. . .. A}: : V J dark \y _ a m:1 ., dark bright )m If (gt-fibright )m:3 — (ybrigm )m, = 5.30 cm, then 3(lL/d) — O = 5.30 cm, and (5,30 cm)d (5.30 ><10’2 m)(0.050 0><10’3 m) A=—=—=5.89><10’7 m: 3L 3(1.50 m) (b) The separation between adjacent bright fringes is Aymgm = AL/d, or (589 x109 m)(1.50 m) 0.050 0><10‘3 m =1.77><10—2 m: Aybrighl : 24.7 Note that, with the conditions given, the small angle approximation does not work well. That is, sine, tanB, and 9 are significantly different. The approach to be used is outlined below. I (a) At the m = 2 maximum, 5 =dsin6= 22., T@ 400 mi , 300m :- ---------------------- --'r——- or i=£s1n9=£ 1000m : 2 2 L2+3r I 12(3031’11) 400m : (IOOOID)2+(4OOID)2 (b) The next minimum encountered is the m = 2 minimum; and at that point, 5=dsin9=[m+ijfl=§l 2 2 0r 9 : 811171: Sinil M : 27_7O 2d 2(3001n) Then, y=(1000 m)tan 27.70 = 524 1n so the car must travel an additional . 24.19 With rimming > nail. and n > rim, light reflecting at the air-coating boundary experiences a phase reversal, but light reflecting from the coating—lens boundary does not. Therefore, the condition for destructive interference in the two reflected waves is coaflng 2nmfingt 2 all where m = 0, 1, 2,... For finite wavelengths, the lowest allowed value of m is m = 1 . Then, if t: 177.4 nm and n = 1.55, the wavelength associated with this lowest order destructive interference is coaflng 211 t ,1, = 2(1.55)(177.4 nm): 24.21 There will be a phase reversal of the radar waves reflecting from both surfaces of the polymer, giving zero net phase change due to reflections. The requirement for destructive interference in the reflected waves is then A 4n 2t=[m+ijln,or t:(2m+1) where 141:0, 1, 2,... 2 fi lm If the film is as thin as possible, then m = 0 and the needed thickness is 2. 3.00 cm I: =—= 0.500 4” 4050) film This anti-reflectance coating could be easily countered by changing the wavelength of the radar — to 1.50 cm — now creating maximum reflection! 24.28 With a phase reversal due to reflection at each surface of the magnesium fluoride layer, there is zero net phase difference caused by reflections. The condition for destructive interference is the] l l 2t=[m+—]/l” =[m+—]i,where m 20,1, 2,... 2, 2 n film For minimum thickness, m = O, and the thickness is I=(2m+1)4: :(1) (550x109 m) —=9.96><10‘8 : 99.6 4038) m film 24.36 At the positions of the minima, sin 9”, : mM/a) and v : Ltan 9m : Lsinem = m[L(l/a)] a I.” Thus _ 2L1 _ 2(0.500 m)(680><10’9 m) — : X m : mm )"3 _}"1 3.00 X 10—3 m a 24.40 (a) The longest wavelength in the Visible spectrum is 700 nm, and the grating spacing is d =1 mm/oOO 21.67 X 10’3 mm = 1.67 X 10’6 m. . 0 1.67 X1041 m sin 90.00 Thus, m : = % , = 2.38 "m 2. 700 X 10") m red so 2 complete orders will be observed. (b) From 2. = (1 sin 6, the angular separation of the red and Violet edges in the first order will be _. [PM] . [AW] . r1 700x104) m , ,1 400x10” m A9=s1n — —sm =S1n —$ —sm —g5 d d l.67><10 m 1.67><lO m or A9 = 24.52 (a) From Malus’s law, the fraction of the incident intensity of the unpolarized light that is trans- mitted by the polarizer is 1' = 10 (cosl 0) = 10 (0.500) The fraction of this intensity incident on the analyzer that will be transmitted is 1 = I’cosz (35.00) = I’(O.67l) = 10 (O.500)(O.671) = 0.33610 Thus, the fraction of the incident unpolan’zed light transmitted is 1/10 : . (b) The fraction of the original incident light absorbed by the analyzer is 1'— 1 _ 0.50010 —0.33610 _ If.) [0 0.164 24.59 From Malus’s law, the intensity of the light transmitted by the first polarizer is I] = 1,. cos2 6]. The plane of polarization of this light is parallel to the axis of the first plate and is incident on the second plate. Malus’s law gives the intensity transmitted by the second plate as 12 = I 1 cos2 (6: — 61) = I 1. cos2 91 cos2 (92 — 191) . This light is polarized parallel to the axis of the second plate and is incident upon the third plate. A final application of Malus’s law gives the transmitted intensity as If 2 I3 cos2 (6’3 —63) = If cos3 6] cos2 (t92 — 6] )cos2 (63 —62) With 61 = 20.00, 62 = 40.00, and 63 = 60.00, this result yields If. 2 (10.0 units)cos2 (20.00)cos2 (20.00)cos2 (20.00) = ...
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HW_sol_chapter24 - 24.3 (a) The distance between the...

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