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

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(B) Given: Distance of screen $D = 50 \, cm = 0.5 \, m$,wavelength $\lambda = 6000 \, \mathring{A} = 600 	imes 10^{-9} \, m$,and distance between fir

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$0.2 \, mm$

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(B) Given: Distance of screen $D = 50 \, cm = 0.5 \, m$,wavelength $\lambda = 6000 \, \mathring{A} = 600 	imes 10^{-9} \, m$,and distance between first and third minima $\Delta y = 3 \, mm = 3 	imes 10^{-3} \, m$.The position of the $n^{th}$ minima in a single slit diffraction pattern is given by $y_n = \frac{n D \lambda}{a}$,where $a$ is the slit width.The distance between the first $(n=1)$ and third $(n=3)$ minima is:$\Delta y = y_3 - y_1 = \frac{3 D \lambda}{a} - \frac{1 D \lambda}{a} = \frac{2 D \lambda}{a}$.Substituting the given values:$3 	imes 10^{-3} = \frac{2 	imes 0.5 	imes 600 	imes 10^{-9}}{a}$.Solving for $a$:$a = \frac{2 	imes 0.5 	imes 600 	imes 10^{-9}}{3 	imes 10^{-3}} = \frac{600 	imes 10^{-9}}{3 	imes 10^{-3}} = 200 	imes 10^{-6} \, m = 0.2 	imes 10^{-3} \, m = 0.2 \, mm$.

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

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