 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2019/paper-2/section-b/b8/b/solution

Put the slit centres at

x_{j} = \pm D /2, \pm 3 D /2

. Each slit has the same width

a

and illumination. Its contribution to the Fraunhofer diffraction complex amplitude is

\int_{x_{j} - a /2}^{x_{j} + a /2} e^{- i q x} d x = a sinc (q a /2) e^{- i q x_{j}} .

\sum_{j} e^{- i q x_{j}} = e^{i q D /2} + e^{- i q D /2} + e^{3 i q D /2} + e^{- 3 i q D /2} = 2 cos (q D /2) + 2 cos (3 q D /2) .

ψ (q) \propto 2 a [cos (q D /2) + cos (3 q D /2)] sinc (q a /2), q = \frac{2 π sin θ}{λ} .

The common factor

2 a

can be absorbed in the proportionality constant, giving the requested form. In particular

ψ (0) \propto 4 a

, as expected from four equal in-phase slits. Normalizing by the central intensity gives

\frac{I ( q )}{I ( 0 )} = sinc^{2} (q a /2) cos^{2} (q D /2) cos^{2} (q D) .

The last form follows from

cos x + cos 3 x = 2 cos 2 x cos x

and avoids apparent singularities in the usual sine-ratio array formula.

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