In a single slit diffraction pattern,a light | vedclass

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(C) For the $n^{	ext{th}}$ minima in a single slit diffraction pattern,the condition is given by:$b \sin 	heta = n \lambda$Since $\lambda$ is very s

Vedclass · Accepted Answer

$2$

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(C) For the $n^{	ext{th}}$ minima in a single slit diffraction pattern,the condition is given by:$b \sin 	heta = n \lambda$Since $\lambda$ is very small,$\sin 	heta \approx 	an 	heta = \frac{y_n}{D}$.Thus,the position of the $n^{	ext{th}}$ minima is $y_n = \frac{n \lambda D}{b}$.The position of the first minima $(n=1)$ is $y_1 = \frac{\lambda D}{b}$.The position of the third minima $(n=3)$ is $y_3 = \frac{3 \lambda D}{b}$.The distance between the first and third minima is $\Delta y = y_3 - y_1 = \frac{2 \lambda D}{b}$.Given $\Delta y = 3 	ext{ mm} = 3 	imes 10^{-3} 	ext{ m}$,$\lambda = 6000 \mathring A = 6000 	imes 10^{-10} 	ext{ m}$,and $D = 50 	ext{ cm} = 0.5 	ext{ m}$.Substituting the values:$3 	imes 10^{-3} = \frac{2 	imes 6000 	imes 10^{-10} 	imes 0.5}{b}$$b = \frac{2 	imes 6000 	imes 10^{-10} 	imes 0.5}{3 	imes 10^{-3}}$$b = \frac{6000 	imes 10^{-10}}{3 	imes 10^{-3}} = 2000 	imes 10^{-7} = 2 	imes 10^{-4} 	ext{ m}$.Therefore,the width of the slit is $2 	imes 10^{-4} 	ext{ m}$. The correct option is $C$.

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