 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2/section-b/b7/d/solution

A relative phase

π

changes the sign of one slit field. Their sum becomes

ψ (θ) \propto e^{i q D /2} - e^{- i q D /2} = 2 i sin (q D /2), I (θ) \propto sin^{2} (\frac{π D sin θ}{λ}) .

The central interference fringe is now dark. The maxima move to

D sin θ = (m + 1/2) λ

, while the minima occur at

D sin θ = mλ

. Thus the pattern shifts by half a fringe,

2.00 mrad

near the axis, but its angular fringe separation remains

4.00 mrad

. This assumes the glass changes phase uniformly and introduces negligible absorption or beam deflection.

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