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Unformatted text preview: 5. (a) We take the phases of both waves to be zero at the front surfaces of the layers. The
phase of the first wave at the back surface of the glass is given by φ1 = k1L – ωt, where k1
(= 2π/λ1) is the angular wave number and λ1 is the wavelength in glass. Similarly, the
phase of the second wave at the back surface of the plastic is given by φ2 = k2L – ωt,
where k2 (= 2π/λ2) is the angular wave number and λ2 is the wavelength in plastic. The
angular frequencies are the same since the waves have the same wavelength in air and the
frequency of a wave does not change when the wave enters another medium. The phase
difference is
⎛1
1⎞
φ1 − φ2 = ( k1 − k2 ) L = 2π ⎜
− ⎟ L.
⎝ λ1 λ 2 ⎠
Now, λ1 = λair/n1, where λair is the wavelength in air and n1 is the index of refraction of
the glass. Similarly, λ2 = λair/n2, where n2 is the index of refraction of the plastic. This
means that the phase difference is
2π
φ1 − φ2 =
( n1 − n2 ) L. λair The value of L that makes this 5.65 rad is bφ − φ gλ = 5.65c400 × 10 mh = 3.60 × 10
L=
2 πb n − n g
2 πb1.60 −1.50g
−9 1 2 1 air −6 m. 2 (b) 5.65 rad is less than 2π rad = 6.28 rad, the phase difference for completely
constructive interference, and greater than π rad (= 3.14 rad), the phase difference for
completely destructive interference. The interference is, therefore, intermediate, neither
completely constructive nor completely destructive. It is, however, closer to completely
constructive than to completely destructive. ...
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 Spring '08
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 Physics

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