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ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/section-a/a10/solution

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Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section A / A10 / Solution by

Codex  0 2026-10-20

Normalize the central slit amplitude to one. The two outer amplitudes are

1/2

each, not

1/ 2

, because the specified attenuation concerns amplitude. For centre spacing

d = 5 λ

, adjacent slits have phase difference

δ = (2 π d / λ) sin θ = 10 π sin θ

. The far-field amplitude and intensity are therefore

A = \frac{1}{2} e^{- i δ} + 1 + \frac{1}{2} e^{i δ} = 1 + cos δ, \frac{I}{I _{0}} = (1 + cos (10 π sin θ))^{2} = 4 cos^{4} (5 π sin θ),

where

I_{0}

is the central slit's intensity alone in the same narrow-slit approximation. Principal maxima have

sin θ = m /5

4 I_{0}

; between them, zeros occur at

sin θ = (2 m + 1) /10

. Only values in

[- 1, 1]

are physical. The endpoints are grazing-direction limits of this ideal angular model.

Figure 1.
Weighted three-slit interference with outer amplitudes one half of the central amplitude
. Original narrow-slit far-field calculation. A finite slit width supplies an additional single-slit envelope, which is omitted here. Details.

The amplitude weights form a binomial pattern. They remove the secondary maxima that an equal-amplitude three-slit array would have between its principal peaks. No phase shift is needed: changing the coherent weights already changes the interference structure.

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