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In a Young's double-slit experiment, the upper slit is covered by a thin sheet of thickness t and refractive index 2*mu, and the lower slit by a sheet of thickness 2t and refractive index mu. Find the position y of the central maximum on the screen (D = slit-to-screen distance, d = slit separation; take downward as negative).

1. Zero
2. tD/d
3. -tD/d
4. None of these

Correct answer: Zero

Solution

Extra optical path from the upper slit's sheet: (2*mu - 1)*t. Extra optical path from the lower slit's sheet: (mu - 1)*(2t) = (2*mu - 2)*t. The net extra path difference introduced between the two slits is (2*mu - 1)t - (2*mu - 2)t = t (not zero), so in general the central maximum would shift. However, comparing the two extra paths: upper = (2*mu - 1)t and lower = (2*mu - 2)t; the upper path is longer by t, meaning the central maximum shifts. Re-evaluating: the path-difference at a point y is delta = (yd/D) + [path added at lower] - [path added at upper]. Setting delta = 0 for the central maximum: yd/D = (2*mu-1)t - (2*mu-2)t = t, giving y = tD/d only if upper has the larger added path. Because (2*mu-1)t > (2*mu-2)t, the central maximum shifts toward the upper slit by y = tD/d in magnitude. The correct nonzero position depends on the sign convention; with the given options the intended balanced design makes the standard answer y = tD/d. Given the ambiguity in sign convention, the magnitude is tD/d.

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