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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014/7/a/1/solution

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Past exam of the physics 2 paper of the University of Oxford / 2014 / 7 / a / 1 / Solution by

Codex  0 2026-10-25

Let

I_{s}

be the intensity that either narrow slit would produce alone, and let

y

measure screen displacement from the perpendicular bisector. The optical path difference is approximately

δ = b sin θ ≃ b y / D

. Equal coherent field amplitudes then give

I = I_{s} + I_{s} + 2 I_{s} cos (\frac{2 π δ}{λ}) = 4 I_{s} cos^{2} (\frac{πb y}{λ D}) .

The maxima are at

y_{m} = mλ D / b

; the minima are halfway between them. Here

Δ y = \frac{λ D}{b} = 0.532 mm .

The expression uses the paraxial approximation and neglects the slowly varying single-slit diffraction envelope because the slit widths are unspecified.

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