Question

A glass plate $12\times {10}^3$ mm thick is placed in the path of one of the interfering beams in a Young's double slit interference arrangement using monochromatic light of wavelength 6000. If the central band shifts a distance equal to width of 10 bands. What is the thickness (in $\mu$ m) of the plate of diamond of refractive index 2.5 that has to be introduced in the path of second beam to bring the central band to original position?

Answer 1

Since the central bright fringe has shifted to a distance where there was 10th bright fringe, the path difference introduced=$10\lambda$

Now diamond plate needs to be inserted of such a thickness that the introduced path difference is equal to the one introduced by glass plate.

Thus $10\lambda =t(\mu -1)$

$t(\frac{5}{2}-1)=10\times 6000\times {10}^{-10}m$

$⟹t=4\times {10}^{-6}m$