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Unformatted text preview: 57. Let φ1 be the phase diﬀerence of the waves in the two arms when the tube has air in it, and let φ2 be
the phase diﬀerence when the tube is evacuated. These are diﬀerent because the wavelength in air is
diﬀerent from the wavelength in vacuum. If λ is the wavelength in vacuum, then the wavelength in air
is λ/n, where n is the index of refraction of air. This means
φ1 − φ2 = 2L 2πn 2π
4π (n − 1)L
−
=
λ
λ
λ where L is the length of the tube. The factor 2 arises because the light traverses the tube twice, once on
the way to a mirror and once after reﬂection from the mirror. Each shift by one fringe corresponds to a
change in phase of 2π rad, so if the interference pattern shifts by N fringes as the tube is evacuated,
4π (n − 1)L
= 2N π
λ
and
n=1+ Nλ
60(500 × 10−9 m)
=1+
= 1.00030 .
2L
2(5.0 × 10−2 m) ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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