p34_099 - L v 1 cos = n 1 L c r 1 1 n 1 sin 2 using results...

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99. (a) In our solution here, we assume the reader has looked at our solution for problem 98. A light ray traveling directly along the central axis reaches the end in time t direct = L v 1 = n 1 L c . For the ray taking the critical zig-zag path, only its velocity component along the core axis direction contributes to reaching the other end of the ±ber. That component is v 1 cos θ 0 , so the time of travel for this ray is t zig zag =
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Unformatted text preview: L v 1 cos = n 1 L c r 1 1 n 1 sin 2 using results from the previous solution. Plugging in sin = p n 2 1 n 2 2 and simplifying, we obtain t zig zag = n 1 L c ( n 2 /n 1 ) = n 2 1 L n 2 c . The dierence t zig zag t direct readily yields the result shown in the problem statement. (b) With n 1 = 1 . 58 , n 2 = 1 . 53 and L = 300 m, we obtain t = 52 ns....
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