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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 (gtﬁbright )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 aircoating 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 antireﬂectance 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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This note was uploaded on 09/13/2009 for the course PHY 102 taught by Professor Luo during the Spring '08 term at SUNY Buffalo.
 Spring '08
 LUO
 Physics

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