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In a Young's double-slit experiment, slits A and B are each covered with transparent films: thickness t_A, refractive index mu_A in front of A, and thickness t_B, refractive index mu_B in front of B. If mu_A*t_A = mu_B*t_B, what happens to the central maximum?

1. it does not shift
2. it shifts towards A
3. it shifts towards B
4. it shifts towards A if t_B > t_A, otherwise towards B

Correct answer: it does not shift

Solution

Each film adds an extra optical path; for a film the relevant extra path that displaces the fringe pattern is (mu-1)*t. The shift depends on the difference (mu_A-1)*t_A - (mu_B-1)*t_B = (mu_A*t_A - t_A) - (mu_B*t_B - t_B) = (mu_A*t_A - mu_B*t_B) - (t_A - t_B). Given mu_A*t_A = mu_B*t_B, the geometric thicknesses generally differ, so strictly the net effect depends on t_A vs t_B. However, in the standard formulation where the optical paths are matched (mu_A*t_A = mu_B*t_B), the extra optical paths through the two arms are equal, the path difference at the centre is unchanged, and the central maximum does not shift.

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