22_Ch 24 College Physics ProblemCH24 Wave Optics

# 22_Ch 24 College Physics ProblemCH24 Wave Optics - 324...

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324 CHAPTER 24 24.42 The grating spacing is 4 6 1 cm 8.33 10 cm 8.33 10 m 1 200 = = × = × d Using sin m d λ θ = and the small angle approximation, the distance from the central maximum to the maximum of order m for wavelength λ is ( ) tan sin m y L L L d θ θ λ = = m . Therefore, the spacing between successive maxima is 1 m m y y y L d λ + = = ( . The longer wavelength in the light is found to be ) ( )( ) 3 6 8.44 10 m 8.33 10 m nm 0.150 m × × 469 = long y d L λ = = Since the third order maximum of the shorter wavelength falls halfway between the central maximum and the first order maximum of the longer wavelength, we have 3 2 short L 0 1 long L d d λ λ + = or ( ) 1 469 nm 8.1 nm 6 short λ = = 7 24.43 The grating spacing is 3 6 1 mm 2.50 10 mm 2.50 10 m 400 = = × = × d From sin d m θ λ = , the angle of the second-order diffracted ray is ( ) 1 sin 2 d θ λ = . (a) When the grating is surrounded by air, the wavelength is
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