Question

If a transparent medium of refractive index μ = 1.5 and thickness t = 2.5 x 10-5 m is insertedin front of one of the slits of Young's Double Slit experiment, how much will be the shift in theinterference pattern ? The distance between the slits is 5.0 x 10-3 cm and that between slits and

screen is 100 cm.

Answer 1

The formula for lateral shift of fringe is given by :
$$\begin{gathered} y_0=\frac{\beta (\mu -1)t}{\lambda } \\ where\beta =\frac{\lambda D}{d} \\ hence,y_0=\frac{D(\mu -1)t}{d}...................\left(1\right) \\ Given,\mu =1.5 \\ t=2.5X{10}^{-5}m \\ d=5X{10}^{-3}m \\ D=1m \\ Putabovevaluesintheeq.1togetyouranswer. \end{gathered}$$