ssm_ch35 - Chapter 35 5 (a) Take the phases of both waves...

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Chapter 35 5 (a) Take the phases of both waves to be zero at the front surfaces of the layers. The phase of the first wave at the back surface of the glass is given by Á 1 = k 1 L ¡ !t ,whe re k 1 (= 2 ¼=¸ 1 )is the angular wave number and ¸ 1 is the wavelength in glass. Similarly, the phase of the second wave at the back surface of the plastic is given by Á 2 = k 2 L ¡ !t ,whe re k 2 (= 2 ¼=¸ 2 ) is the angular wave number and ¸ 2 is the wavelength in plastic. The angular frequencies are the same since the waves have the same wavelength in air and the frequency of a wave does not change when the wave enters another medium. The phase difference is Á 1 ¡ Á 2 =( k 1 ¡ k 2 ) L =2 ¼ μ 1 ¸ 1 ¡ 1 ¸ 2 L: Now ¸ 1 = ¸ air =n 1 , where ¸ air is the wavelength in air and n 1 is the index of refraction of the glass. Similarly, ¸ 2 = ¸ air =n 2 ,where n 2 is the index of refraction of the plastic. This means that the phase difference is Á 1 ¡ Á 2 =(2 ¼=¸ air )( n 1 ¡ n 2 ) L .Thev a lueo f L that makes this 5 : 65 rad is L = ( Á 1 ¡ Á 2 ) ¸ air 2 ¼ ( n 1 ¡ n 2 ) = 5 : 65(400 £ 10 ¡ 9 m) 2 ¼ (1 : 60 ¡ 1 : 50) =3 : 60 £ 10 ¡ 6 m : (b) 5 : 65radi sle s sth an2 ¼ rad (= 6 : 28 rad), the phase difference for completely constructive interference, and greater than ¼ rad (= 3 : 14 rad), the phase difference for completely destruc- tive interference. The interference is therefore intermediate, neither completely constructive nor completely destructive. It is, however, closer to completely constructive than to completely de- structive. 15 Interference maxima occur at angles μ such that d sin μ = ,whe re d is the separation of the sources, ¸ is the wavelength, and m is an integer. Since d =2 : 0m and ¸ =0 : 50 m, this means that sin μ =0 : 25 m . You want all values of m (positive and negative) for which j 0 : 25 m j 1. These are ¡ 4, ¡ 3, ¡ 2, ¡ 1, 0, +1, +2, +3, and +4. For each of these except ¡ 4 and +4, there are two different values for μ . A single value of μ ( ¡ 90 ± ) is associated with m = ¡ 4andas ing le value ( ¡ 90 ± ) is associated with m = +4. There are sixteen different angles in all and therefore sixteen maxima. 17 The angular positions of the maxima of a two-slit interference pattern are given by d sin μ = , where d is the slit separation, ¸ is the wavelength, and m is an integer. If μ
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is ¢ μ = ¸=d .Le t ¸ 0 be the wavelength for which the angular separation is 10 : 0% greater. Then 1 : 10 ¸=d = ¸ 0 =d or ¸ 0 =1 : 10 ¸ =1 : 10(589 nm) = 648 nm.
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This note was uploaded on 01/19/2009 for the course PHYS 171.102 taught by Professor Heckman during the Fall '98 term at Johns Hopkins.

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ssm_ch35 - Chapter 35 5 (a) Take the phases of both waves...

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