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 46 1 cm 8.33 10 cm 8.33 10 m 1200 −− == × = × 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 is ( ) tan sin m yL L L d θθ =≈= m . Therefore, the spacing between successive maxima is 1 mm yy y L d + ∆= = ( . The longer wavelength in the light is found to be ) ( )( ) 36 8.44 10 m 8.33 10 m nm 0.150 m ×× 469 = long yd L 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 01 long L dd + =    or () 1 469 nm 8.1 nm 6 short 7 24.43 The grating spacing is 1 mm 2.50 10 mm 2.50 10 m 400 × = × d From sin dm = , the angle of the second-order diffracted ray is ( ) 1 sin 2 d θλ = . (a) When the grating is surrounded by air, the wavelength is
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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.

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