In Young's double-slit experiment,the fringes | vedclass

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Question

(B) The fringe shift $\Delta y$ in a Young's double-slit experiment when a plate of thickness $t$ and refractive index $\mu$ is introduced is given by

Vedclass · Accepted Answer

$1.75$

Vedclass · Answer

(B) The fringe shift $\Delta y$ in a Young's double-slit experiment when a plate of thickness $t$ and refractive index $\mu$ is introduced is given by $\Delta y = \frac{(\mu - 1)tD}{d}$.Since the fringe width $\beta = \frac{\lambda D}{d}$,the shift can be written as $x = \frac{(\mu - 1)t\beta}{\lambda}$.For the first plate with refractive index $\mu_1 = 1.5$,the shift is $x = \frac{(1.5 - 1)t\beta}{\lambda} = \frac{0.5 t\beta}{\lambda}$.For the second plate with refractive index $\mu_2$,the shift is $\frac{3}{2}x = \frac{(\mu_2 - 1)t\beta}{\lambda}$.Dividing the second equation by the first,we get: $\frac{(3/2)x}{x} = \frac{(\mu_2 - 1)}{1.5 - 1}$.$\frac{3}{2} = \frac{\mu_2 - 1}{0.5}$.$0.75 = \mu_2 - 1$.$\mu_2 = 1.75$.

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