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

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