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Unformatted text preview: = 3 Ī» = 1800 nm . 3. Now, weād like to ļ¬gure out what the pattern looks like when we ļ¬ll 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 hasnāt changed, so we know that mĪ» water = 2 t , where m is the number of times the pattern cycles all the way back to bright, and Ī» water = Ī»/n water . Weāll say n water = 4 / 3 for ease of calculation. Thus we ļ¬nd m = 2 Ć (1800 nm) Ć (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.
- Spring '08
- Quantum Physics