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Sol6 - UAHuntsville Spring 2010 PH 113 Homework Solutions...

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UAHuntsville Spring, 2010 PH 113: Homework Solutions No. 6 Solution 35: The dispersion of a grating is given by D = d θ / d λ , where θ is the angular position of a line associated with wavelength λ . The angular position and wavelength are related by d sin θ = m λ , where d is the slit separation (which we made boldfaced in order not to confuse it with the d used in the derivative, below) and m is an integer. We differentiate this expression with respect to θ to obtain d d m θ θ λ d cos , = or D d d m = = θ θ λ d cos . Now m = ( d / λ ) sin θ , so D = = d d sin tan . θ θ θ λ λ cos Solution 36: (a) From the expression for the half-width Δ θ hw (given by Δ θ hw = λ N d cos θ ( ) ) and that for the resolving power R (given by R = N m ), we find the product of Δ θ hw and R to be Δ θ hw R = λ N d cos θ × N m = m λ d cos θ = d sin θ d cos θ = tan θ where we used m λ = d sin θ . (b) For first order m = 1, so the corresponding angle θ 1 satisfies d sin θ 1 = m λ = λ . Thus the product in question is given by ( ) ( ) ( ) 1 1 1 2 2 2 1 1 1 2 sin sin 1 1 tan cos 1 sin 1/sin 1 / 1 1 0.89.

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