27_Ch 24 College Physics ProblemCH24 Wave Optics

# 27_Ch 24 College Physics ProblemCH24 Wave Optics - Wave...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Wave Optics 24.58 The wavelength is λ = 329 vsound 340 m s = = 0.170 m 2 000 Hz f Maxima occur where d sin θ = mλ , or θ = sin −1 m ( λ d ) for m = 0 , 1, 2,… Since d = 0.350 m , λ d = 0.486 which gives θ = sin −1 ( 0.486 m ) For m = 0 , 1, and 2 , this yields maxima at 0°, 29.1°, and 76.3° No solutions exist for m ≥ 3 since that would imply sin θ > 1 λ Minima occur where d sin θ = ( m + 1 2 ) λ or θ = sin −1 ( 2m + 1) for m = 0 , 1, 2, … 2d With λ d = 0.486 , this becomes θ = sin −1 ( 2m + 1) ( 0.243 ) For m = 0 and 1 , we find minima at 14.1° and 46.8° No solutions exist for m ≥ 2 since that would imply sin θ > 1 24.59 The source and its image, located 1.00 cm below the mirror, act as a pair of coherent sources. This situation may be treated as double-slit interference, with the slits separated by 2.00 cm, if it is remembered that the light undergoes a phase reversal upon reflection from the mirror. The existence of this phase change causes the conditions for constructive and destructive interference to be reversed. Therefore, dark bands (destructive interference) occur where y = m ( λ L d ) for m = 0 , 1, 2,… The m = 0 dark band occurs at y = 0 (that is, at mirror level). The first dark band above the mirror corresponds to m = 1 and is located at −9 λ L ( 500 × 10 m ) ( 100 m ) y = ( 1) = = 2 .50 × 10 −3 m = 2 .50 mm −2 2 .00 × 10 m d ...
View Full Document

## This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

Ask a homework question - tutors are online