In a Fraunhofer diffraction pattern,the slit | vedclass

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(C) For Fraunhofer diffraction,the position of the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$. For small angles,$\sin 	heta \approx 	h

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$5000$

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(C) For Fraunhofer diffraction,the position of the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$. For small angles,$\sin 	heta \approx 	heta = \frac{y_n}{D}$,where $y_n$ is the distance from the central maximum.Thus,$y_n = \frac{n \lambda D}{a}$.The distance between the first minima on either side of the central maximum is $2y_1 = \frac{2 \lambda D}{a}$.Given: $a = 0.2 \ mm = 2 	imes 10^{-4} \ m$,$D = 2 \ m$,and $2y_1 = 1 \ cm = 10^{-2} \ m$.Substituting the values: $10^{-2} = \frac{2 	imes \lambda 	imes 2}{2 	imes 10^{-4}}$.$10^{-2} = \frac{4 \lambda}{2 	imes 10^{-4}} = 2 	imes 10^4 \lambda$.$\lambda = \frac{10^{-2}}{2 	imes 10^4} = 0.5 	imes 10^{-6} \ m = 5000 	imes 10^{-10} \ m = 5000 \ Å$.

In a Fraunhofer diffraction pattern,the slit width is $0.2 \ mm$ and the screen is at $2 \ m$ away from the lens. If the distance between the first minimum on either side of the central maximum is $1 \ cm$,the wavelength of light used is: (in $Å$)

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