p34_078 - sin N sin f . We see that the L.H.S. of the...

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78. (a) Suppose there are a total of N transparent layers ( N = 5 in our case). We label these layers from left to right with indices 1, 2, . . . , N . Let the index of refraction of the air be n 0 .W ed en o t eth e initial angle of incidence of the light ray upon the air-layer boundary as θ i and the angle of the emerging light ray as θ f . We note that, since all the boundaries are parallel to each other, the angle of incidence θ j at the boundary between the j -th and the ( j + 1)-th layers is the same as the angle between the transmitted light ray and the normal in the j -th layer. Thus, for the Frst boundary (the one between the air and the Frst layer) n 1 n 0 = sin θ i sin θ 1 , for the second boundary n 2 n 1 = sin θ 1 sin θ 2 , and so on. ±inally, for the last boundary n 0 n N = sin θ N sin θ f . Multiplying these equations, we obtain µ n 1 n 0 ¶µ n 2 n 1 ¶µ n 3 n 2 ··· µ n 0 n N = µ sin θ i sin θ 1 ¶µ sin θ 1 sin θ 2 ¶µ sin θ
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Unformatted text preview: sin N sin f . We see that the L.H.S. of the equation above can be reduced to n /n while the R.H.S. is equal to sin i / sin f . Equating these two expressions, we Fnd sin f = n n sin i = sin i , which gives i = f . So for the two light rays in the problem statement, the angle of the emerging light rays are both the same as their respective incident angles. Thus, f = 0 for ray a and f = 20 for ray b . (b) In this case, all we need to do is to change the value of n from 1.0 (for air) to 1.5 (for glass). This does not change the result above. Note that the result of this problem is fairly general. It is independent of the number of layers and the thickness and index of refraction of each layer....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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