 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-resit/5/solution

s sin θ = λ

. Differentiating at

λ_{0}

gives

s cos θ_{0} δ θ = δ λ, θ_{0} = sin^{- 1} (λ_{0} / s) .

Therefore the positive first-order peaks are separated by

δ θ ≃ \frac{δ λ}{s cos θ _{0}} = \frac{δ λ}{s ^{2} - λ _{0}^{2}} .

The larger wavelength is diffracted farther from the normal. Both first orders must exist,

λ_{0} + δ λ < s

; extremely close to grazing incidence the finite difference of the two inverse sines should replace this local approximation.

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