A screen is placed 50,cm from a single slit,w | vedclass

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(B) The position of the $n^{th}$ minima in a single slit diffraction pattern is given by $y_n = \frac{n \lambda D}{d}$,where $d$ is the slit width,$D$

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$0.2$

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(B) The position of the $n^{th}$ minima in a single slit diffraction pattern is given by $y_n = \frac{n \lambda D}{d}$,where $d$ is the slit width,$D$ is the distance to the screen,and $\lambda$ is the wavelength.Given: $D = 50\,cm = 0.5\,m$,$\lambda = 6000\,\mathring{A} = 6 	imes 10^{-7}\,m$,and the distance between the first and third minima $\Delta y = 3\,mm = 3 	imes 10^{-3}\,m$.The distance between the $n_1^{th}$ and $n_2^{th}$ minima is $\Delta y = \frac{(n_2 - n_1) \lambda D}{d}$.Substituting $n_2 = 3$ and $n_1 = 1$,we get $\Delta y = \frac{2 \lambda D}{d}$.Rearranging for $d$: $d = \frac{2 \lambda D}{\Delta y}$.$d = \frac{2 	imes (6 	imes 10^{-7}\,m) 	imes (0.5\,m)}{3 	imes 10^{-3}\,m}$.$d = \frac{6 	imes 10^{-7}}{3 	imes 10^{-3}} = 2 	imes 10^{-4}\,m = 0.2\,mm$.

$A$ screen is placed $50\,cm$ from a single slit,which is illuminated with $6000\,\mathring{A}$ light. If the distance between the first and third minima in the diffraction pattern is $3\,mm,$ the width of the slit is ......$mm$.

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