 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2022/paper-2/section-a/1/solution

Use Fresnel diffraction, since the detector moves along the axis through successive near-field interference extrema. Write

z

for its distance from the opening and

ρ

for the integration coordinate within the opening. For a uniform incident plane wave, the on-axis complex amplitude, in the paraxial approximation, is proportional to

\frac{e ^{ik z}}{iλ z} \int_{0}^{r} e^{iπ ρ^{2} / (λ z)} 2 π ρ d ρ = e^{ik z} (1 - e^{iπ r^{2} / (λ z)}) .

Thus the light intensity relative to the incident value is

\frac{I ( z )}{I _{inc}} = 4 sin^{2} (\frac{π r ^{2}}{2 λ z}) .

F = r^{2} / (λ z)

is an odd integer at a maximum and an even integer at a minimum. It decreases as the detector moves away. Adjacent extrema therefore differ by one:

\frac{r ^{2}}{λ} (\frac{1}{1.00 m} - \frac{1}{1.25 m}) = 1.

Consequently

r^{2} = 2.25 \times 1 0^{- 6} m^{2}, r = 1.50 \times 1 0^{- 3} m = 1.50 mm .

As a check,

F = 5

z = 1 m

and

F = 4

z = 1.25 m

, so these are indeed a maximum followed by the next minimum. Also

r / z ≃ 1.5 \times 1 0^{- 3}

, consistent with the paraxial approximation.

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