 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-resit/8/i/solution

ξ

across the slit have relative phase

k ξ sin θ

. With uniform illumination, the field relative to its axial value is

\frac{E ( θ )}{E ( 0 )} = \frac{1}{w} \int_{- w /2}^{w /2} e^{ik ξ s i n θ} d ξ = \frac{s i n β}{β}, β = \frac{π w s i n θ}{λ} .

I (θ) = I (0) (\frac{sin β}{β})^{2} .

On the detector

sin θ = y / d^{2} + y^{2}

; near the axis,

β ≃ π w y / (λ d)

. The central maximum takes the continuous limit

sin β / β \to 1

, and the minima satisfy

w sin θ = mλ

for nonzero integers

m

. The stated far-field condition justifies adding the aperture contributions with a linear phase.

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