 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/section-a/5/solution

The phase step between adjacent slits is

δ = \frac{2 π d s i n θ}{λ} ≃ \frac{π}{3} .

For equal amplitudes

a_{0}

, the six unmodified contributions form a closed regular hexagon in the complex plane:

S_{0} = a_{0} \sum_{j = 0}^{5} e^{ijπ /3} = 0.

j

's contribution by its negative, so

S = S_{0} - 2 a_{0} e^{ijπ /3}

. Consequently

I_{screen} = ∣ S ∣^{2} = 4∣ a_{0} ∣^{2} = 4 I .

This is independent of which slit is reversed. The result uses the far-field, narrow-slit and small-angle approximations; retaining

sin θ

changes the phase step only by about

7 \times 1 0^{- 7}

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