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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 $\lambda$ is the waveleng

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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 $\lambda$ is the wavelength,$D$ is the distance to the screen,and $d$ is the slit width.Given: $D = 50\,cm = 0.5\,m$,$\lambda = 6000 \mathring{A} = 6 	imes 10^{-7}\,m$,and the distance between the first $(n=1)$ and third $(n=3)$ minima is $\Delta y = 3\,mm = 3 	imes 10^{-3}\,m$.The distance between the first and third minima is $\Delta y = y_3 - y_1 = \frac{3 \lambda D}{d} - \frac{1 \lambda D}{d} = \frac{2 \lambda D}{d}$.Substituting the values: $3 	imes 10^{-3} = \frac{2 	imes (6 	imes 10^{-7}) 	imes 0.5}{d}$.$d = \frac{2 	imes 6 	imes 10^{-7} 	imes 0.5}{3 	imes 10^{-3}} = \frac{6 	imes 10^{-7}}{3 	imes 10^{-3}} = 2 	imes 10^{-4}\,m$.Converting to millimeters: $d = 0.2 	imes 10^{-3}\,m = 0.2\,mm$.

$A$ 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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