normal in the j-th layer. Thus, for the first boundary (the one between the air and the first layer) nni101=sinsin,θfor the second boundary nn2112=sinsin,and so on. Finally, for the last boundary nnNNf0=sinsin,Multiplying these equations, we obtain nnnnnnnnNiNf102132011223FHGIKJFHGIKJFHGIKJFHGIKJ=FHGIKJFHGIKJFHGIKJFHGIKJ""sinsinsinsinsinsinsinsin.We see that the L.H.S. of the equation above can be reduced to n0/n0while the R.H.S. is equal to sini/sinf. Equating these two expressions, we find sinsinsin,θθfiinn=FHGIKJ=00which 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 n0from 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.