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(A) For Young's Double Slit Experiment,the condition for the $n^{th}$ bright fringe is given by the path difference $\Delta x = d \sin 	heta = n \lam

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$\sqrt{2}$

Vedclass · Answer

(A) For Young's Double Slit Experiment,the condition for the $n^{th}$ bright fringe is given by the path difference $\Delta x = d \sin 	heta = n \lambda$.For the first bright fringe $(n = 1)$,the condition becomes $d \sin 	heta = \lambda$.When the apparatus is in air,$\lambda_{air} = \lambda$ and $	heta_1 = 45^{\circ}$,so $d \sin 45^{\circ} = \lambda$.When the apparatus is immersed in a liquid of refractive index $\mu$,the wavelength changes to $\lambda_{liquid} = \frac{\lambda}{\mu}$. The new angle is $	heta_2 = 30^{\circ}$,so $d \sin 30^{\circ} = \frac{\lambda}{\mu}$.Dividing the two equations: $\frac{d \sin 45^{\circ}}{d \sin 30^{\circ}} = \frac{\lambda}{\lambda / \mu} = \mu$.Substituting the values: $\mu = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

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