As shown in the figure,in Young's double slit | vedclass

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(D) The shift in the central maxima in Young's double slit experiment when a thin plate of thickness $t$ and refractive index $\mu$ is introduced in f

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$4$

Vedclass · Answer

(D) The shift in the central maxima in Young's double slit experiment when a thin plate of thickness $t$ and refractive index $\mu$ is introduced in front of one of the slits is given by the formula:$\Delta y = \frac{t(\mu - 1)D}{d}$Since the fringe-width $\beta_0$ is given by $\beta_0 = \frac{\lambda D}{d}$,we can express the shift as:$\Delta y = \frac{t(\mu - 1)}{\lambda} \cdot \frac{\lambda D}{d} = \frac{t(\mu - 1)}{\lambda} \beta_0$Given that the shift is $x\beta_0$,we have $x = \frac{t(\mu - 1)}{\lambda}$.Substituting the given values:$t = 10\,\mu m = 10 	imes 10^{-6}\,m$$\mu = 1.2$$\lambda = 500\,nm = 500 	imes 10^{-9}\,m = 5 	imes 10^{-7}\,m$$x = \frac{10 	imes 10^{-6} 	imes (1.2 - 1)}{5 	imes 10^{-7}}$$x = \frac{10^{-5} 	imes 0.2}{5 	imes 10^{-7}}$$x = \frac{2 	imes 10^{-6}}{5 	imes 10^{-7}} = \frac{20}{5} = 4$Therefore,the value of $x$ is $4$.

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