ch33-p076

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

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

## 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.

Ask a homework question - tutors are online