 Second-order direction

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2/section-b/b8/second-order-direction

The pitch corresponding to

500 m m^{- 1}

D = 2.00 μ m

D sin θ = mλ, θ_{2} = arcsin (\frac{2 ( 400 nm )}{2.00 μ m}) = 23.5 8^{\circ} .

This locates the array-factor maximum; the single-slit envelope must also be nonzero for an observable peak.

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