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Young's double slit experiment is performed in a medium of refractive index mu1 = 4/3. A thin transparent slab of refractive index mu2 = 3/2 and thickness t = 8 micrometers is placed in front of slit S2. The magnitude of the optical path difference (with respect to the liquid medium) at the central point O is:

1. 1 micrometer
2. 2 micrometers
3. 3 micrometers
4. 0.5 micrometers

Correct answer: 1 micrometer

Solution

When the experiment is conducted in a medium of refractive index mu1, the optical path length through the slab (in terms of the medium) is t*(mu2/mu1). The optical path through the same thickness of liquid would be t. So the extra OPD = t*(mu2/mu1 - 1) = 8*(9/8 - 1) = 8*(1/8) = 1 micrometer.

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