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Fall 07 Phys 143l HW 2 Solutions

# Fall 07 Phys 143l HW 2 Solutions - Physics 143 Fall Term...

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P a g e | 1 3 Physics 143 Fall Term 2007 HOMEWORK 2 Due: September 10 Problem 1: A very fine wire 7.35 ×10 3 mm in diameter is placed between two flat glass plates, at one edge of the plates parallel to the sides of the plates. The edges of the plates touch on each other on the other side, so that the wire is holding open a thin wedge-shaped film of air between the plates. Light of wavelength 600 nm falls, and is viewed, perpendicular to the plates, and a series of bright and dark bands is seen. How many dark and light bands will there be? Will the area next to the wire be bright or dark? Considering that light falls on the surface of the top plate, one part reflects and the other part passes though it and reflects at the surface of the bottom plate. These two parts of the light interfere at the surface of the top plate. The optical path length difference is, Δ ൌ 2 t െ λ/2 2t is the extra path length that the second light travelling between plates. λ/2 is due to the reflection of the bottom plate(Half Wave Lost). Therefore, the condition for constructive interference is Δ ൌ m λ , ׵ 2t ൌ ቀm ൅ ቁ λ , m=0, 1, 2, 3… And the condition for destructive interference is Δ ൌ ቀm ൅ ቁ λ ׵ 2t ൌ m λ , m=0, 1, 2, 3… At the left end t = 0, which satisfies the condition for destructive interference. Therefore there will be dark at the left end. And at the right end t ൌ d ൌ 7.35 ൈ 10 ିଷ mm n ൌ 2t λ 2 ൈ 7.35 ൈ 10 ିଷ mm 600nm ൌ 24.5 It is a “half integer” which satisfies the condition for constructive interference. Therefore there will be bright in the area next to the wire. And there are twenty-five “half integers” between 0 and 24.5 thus there are 25 bright bands and 25 dark bands from the left end to the right(including the ends) and 24 of each between two ends.

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