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

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(A) The fringe shift $x$ produced by a plate of thickness $t$ and refractive index $\mu$ is given by $x = \frac{(\mu - 1)tD}{d}$,where $D$ is the dist

Vedclass · Accepted Answer

$1.75$

Vedclass · Answer

(A) The fringe shift $x$ produced by a plate of thickness $t$ and refractive index $\mu$ is given by $x = \frac{(\mu - 1)tD}{d}$,where $D$ is the distance to the screen and $d$ is the slit separation.Since $t$,$D$,and $d$ are constant for both cases,the shift $x$ is directly proportional to $(\mu - 1)$.For the first plate: $x \propto (\mu_1 - 1)$,where $\mu_1 = 1.5$.For the second plate: $\frac{3}{2}x \propto (\mu_2 - 1)$.Dividing the two expressions: $\frac{(3/2)x}{x} = \frac{\mu_2 - 1}{\mu_1 - 1}$.$\frac{3}{2} = \frac{\mu_2 - 1}{1.5 - 1}$.$\frac{3}{2} = \frac{\mu_2 - 1}{0.5}$.$\mu_2 - 1 = \frac{3}{2} 	imes 0.5 = 0.75$.$\mu_2 = 1.75$.

Similar Questions

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