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In a Young's double-slit setup, a transparent slab whose refractive index mu can be varied (mu >= 1) is placed in front of one slit. The slits are of equal width and the slab does not absorb light. How does the resultant intensity at the central point of the screen vary as mu is increased from 1?

1. Intensity falls linearly from I0 at mu=1 down to zero
2. Intensity rises linearly from zero at mu=1
3. Intensity oscillates periodically with mu, starting at I0 when mu=1
4. Intensity rises from zero near mu=1 to a single peak then falls

Correct answer: Intensity oscillates periodically with mu, starting at I0 when mu=1

Solution

Inserting the slab gives an extra optical path difference delta = (mu - 1)*t at the central point, producing phase difference phi = (2*pi/lambda)*(mu - 1)*t. The resultant intensity is I = I_max*cos²(phi/2), which oscillates periodically as mu (and hence phi) increases. At mu = 1 there is no path difference, so phi = 0 and I = I_max = I0. As mu increases, the intensity cycles between maxima and minima.

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