ch33-p076 - 76. (a) Suppose there are a total of N...

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normal in the j -th layer. Thus, for the first boundary (the one between the air and the first layer) n n i 1 01 = sin sin , θ for the second boundary n n 2 1 1 2 = sin sin , and so on. Finally, for the last boundary n n N N f 0 = sin sin , Multiplying these equations, we obtain n n n n n n n n N iN f 1 0 2 1 3 2 0 1 1 2 2 3 F H G I K J F H G I K J F H G I K J F H G I K J = F H G I K J F H G I K J F H G I K J F H G I K J "" sin sin sin sin sin sin sin sin . We see that the L.H.S. of the equation above can be reduced to n 0 / n 0 while the R.H.S. is equal to sin i /sin f . Equating these two expressions, we find sin sin sin , θθ f ii n n = F H G I K J = 0 0 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 , (b) and f = 20° for ray b . (c) In this case, all we need to do is to change the value of n 0 from 1.0 (for air) to 1.5 (for glass). This does not change the result above. That is, we still have f = 0 for ray a , (d) and f = 20° for ray b . Note that the result of this problem is fairly general. It is independent of the number of
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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