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(B) The position of the $n^{th}$ maxima in a Young's double-slit experiment is given by $y_n = \frac{n \lambda D}{d}$.For the first maxima,$n = 1$,so

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(B) The position of the $n^{th}$ maxima in a Young's double-slit experiment is given by $y_n = \frac{n \lambda D}{d}$.For the first maxima,$n = 1$,so $y_1 = \frac{\lambda D}{d}$.To find the speed at which the first maxima moves,we differentiate the position with respect to time $t$:$v = \frac{dy_1}{dt} = \frac{\lambda}{d} \frac{dD}{dt}$.Given values are $\lambda = 500 \, nm = 500 	imes 10^{-9} \, m$,$d = 0.4 \, mm = 0.4 	imes 10^{-3} \, m$,and $\frac{dD}{dt} = 4 \, m/s$.Substituting these values into the equation:$v = \frac{500 	imes 10^{-9} \, m}{0.4 	imes 10^{-3} \, m} 	imes 4 \, m/s$.$v = \frac{500 	imes 10^{-6}}{0.4} 	imes 4 = 500 	imes 10^{-6} 	imes 10 = 5000 	imes 10^{-6} \, m/s = 5 	imes 10^{-3} \, m/s$.Converting to $mm/s$:$v = 5 \, mm/s$.

$A$ light of wavelength $500 \, nm$ is incident on a Young's double-slit. The distance between the slits and the screen is $D = 1.8 \, m$ and the distance between the slits is $d = 0.4 \, mm$. If the screen moves with a speed of $4 \, m/s$,with what speed will the first maxima move? (in $mm/s$)

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