ch35-p005 - 5. (a) We take the phases of both waves to be...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

Ask a homework question - tutors are online