WeekProb_Feb_16Sol

WeekProb_Feb_16Sol - = 3 = 1800 nm 3 Now we’d like to...

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Solution to Week #4 Exam-type Problem 2 Cindy Keeler October 2, 2004 1. Very near where the slabs touch, the pattern is dark. This is because the ray which bounces off the air/ n = 1 . 2 interface picks up a π -phase shift (it goes from n = 1 to n = 1 . 2). 2. Now, as we increase the path difference (between the ray which bounces off air/( n = 1 . 2) interface and the one which bounces off the ( n = 1 . 5)/air interface), the pattern will get bright, then dark again. It gets dark again exactly when we add Δ x = λ to the path difference of the bottom ray. Looking at the pattern, it starts dark, then comes back to dark 6 times. Thus, the path difference between the two rays at the end of the slabs must be 6 λ . Since Δ x = 2 t , we find t
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Unformatted text preview: = 3 λ = 1800 nm . 3. Now, we’d like to figure out what the pattern looks like when we fill the space between the slabs with water. This time, the interfaces are n = 1 . 5 to 1.33 and n = 1 . 33 to n = 1 . 2 (so neither ray picks up a phase shift). Thus, the far left of the pattern will be bright. Now, the thickness t hasn’t changed, so we know that mλ water = 2 t , where m is the number of times the pattern cycles all the way back to bright, and λ water = λ/n water . We’ll say n water = 4 / 3 for ease of calculation. Thus we find m = 2 × (1800 nm) × (4 / 3) / (600 nm) = 8 . So the pattern in water starts bright, and cycles back to bright 8 times. 1...
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This note was uploaded on 10/05/2009 for the course PHYSICS 7C taught by Professor Lin during the Spring '08 term at Berkeley.

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WeekProb_Feb_16Sol - = 3 = 1800 nm 3 Now we’d like to...

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