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In a Young's double-slit setup, the region between the slits and the screen is filled with water (refractive index > 1). Compared with air, what happens to the interference pattern?

1. the pattern shifts upward while the fringe width stays the same.
2. the fringe width decreases and the central bright fringe shifts upward.
3. the fringe width increases and the central bright fringe does not shift.
4. the fringe width decreases and the central bright fringe does not shift.

Correct answer: the fringe width decreases and the central bright fringe does not shift.

Solution

In water the wavelength becomes lambda/n, so fringe width beta = (lambda/n)*D/d decreases by factor n. Both paths still travel through the same medium, so zero path difference remains on the central axis; the central fringe does not shift.

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