In a Young's double slit experiment,each of t | vedclass

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(A) $(1)$ The position of the $n^{	ext{th}}$ bright fringe is given by $y = n \frac{\lambda D}{d}$.For the $8^{	ext{th}}$ fringe,$y = 8 \frac{\lambd

Vedclass · Accepted Answer

$601.50, 24$

Vedclass · Answer

(A) $(1)$ The position of the $n^{	ext{th}}$ bright fringe is given by $y = n \frac{\lambda D}{d}$.For the $8^{	ext{th}}$ fringe,$y = 8 \frac{\lambda D}{d}$.The extreme positions are $y_{\max} = 8 \frac{\lambda D}{d_{\min}}$ and $y_{\min} = 8 \frac{\lambda D}{d_{\max}}$.The separation is $\Delta y = y_{\max} - y_{\min} = 8 \lambda D \left[ \frac{1}{d_{\min}} - \frac{1}{d_{\max}} ight]$.Given $\lambda = 6000 	imes 10^{-10} \ m$,$D = 1 \ m$,$d_{\max} = 0.84 \ mm = 0.84 	imes 10^{-3} \ m$,and $d_{\min} = 0.76 \ mm = 0.76 	imes 10^{-3} \ m$.$\Delta y = 8 	imes 6000 	imes 10^{-10} 	imes 1 	imes \left[ \frac{1}{0.76 	imes 10^{-3}} - \frac{1}{0.84 	imes 10^{-3}} ight] = 48 	imes 10^{-7} 	imes \left[ \frac{0.08}{0.76 	imes 0.84 	imes 10^{-3}} ight] \approx 601.5 \ \mu m$.$(2)$ The speed is $v = \frac{dy}{dt} = \frac{d}{dt} \left( n \frac{\lambda D}{d} ight) = -n \lambda D \frac{1}{d^2} \frac{dd}{dt}$.Given $d = 0.8 + 0.04 \sin \omega t$,so $\frac{dd}{dt} = 0.04 \omega \cos \omega t$.For maximum speed,$\cos \omega t = 1$,so $\frac{dd}{dt} = 0.04 \omega = 0.04 	imes 0.08 = 0.0032 \ mm/s$ and $d = 0.8 \ mm$.$v_{\max} = \left| -8 	imes 6000 	imes 10^{-10} 	imes 1 	imes \frac{0.0032 	imes 10^{-3}}{(0.8 	imes 10^{-3})^2} ight| = 24 \ \mu m/s$.

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To make the central fringe appear at the centre $O$ of the screen,a mica sheet of refractive index $\mu = 1.5$ is introduced in front of one of the slits. Choose the correct statement.

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