In a Young's double-slit experiment, positive y is defined toward the upper slit S₁. A plane wave is incident obliquely so that, before reaching the slits, the optical path to S₁ exceeds that to the lower slit S₂ by d sin i. For small angles, if the normal-incidence fringe width is β, which statement correctly gives the shifted central maximum and the first bright fringe above it?
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Question

In a Young's double-slit experiment, positive y is defined toward the upper slit S₁. A plane wave is incident obliquely so that, before reaching the slits, the optical path to S₁ exceeds that to the lower slit S₂ by d sin i. For small angles, if the normal-incidence fringe width is β, which statement correctly gives the shifted central maximum and the first bright fringe above it?

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G1 Physics · Accepted Answer

Let the extra incident optical path to the upper slit S₁ be d sin i. For a point at height y on the screen, positive toward S₁, the geometrical path difference contributed after the slits is approximately −dy ÷ D. Hence the total path difference is Δ(total) = d sin i−dy ÷ D. For the central maximum, Δ(total) = 0. Therefore, d sin i−dy₀ ÷ D = 0, so y₀ = D sin i. Thus the complete fringe pattern shifts toward the upper slit S₁. Bright fringes remain separated by the same fringe width β. Therefore…