In Fraunhofer diffraction pattern, slit width | vedclass

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(C) Given: Slit width $a = 0.2 \, mm = 0.2 	imes 10^{-3} \, m$.Distance of screen $D = 2 \, m$.Wavelength $\lambda = 5000 \, \mathring{A} = 5000 	im

Vedclass · Accepted Answer

$10^{-2} \, m$

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(C) Given: Slit width $a = 0.2 \, mm = 0.2 	imes 10^{-3} \, m$.Distance of screen $D = 2 \, m$.Wavelength $\lambda = 5000 \, \mathring{A} = 5000 	imes 10^{-10} \, m = 5 	imes 10^{-7} \, m$.The position of the $n^{th}$ minimum in a single slit diffraction pattern is given by $a \sin 	heta = n \lambda$.For small $	heta$, $\sin 	heta \approx 	heta = \frac{y}{D}$.Thus, $y_n = \frac{n \lambda D}{a}$.The distance between the first minima on either side of the central maximum is the width of the central maximum, given by $w = y_1 - (-y_1) = 2 y_1 = \frac{2 \lambda D}{a}$.Substituting the values: $w = \frac{2 	imes (5 	imes 10^{-7} \, m) 	imes (2 \, m)}{0.2 	imes 10^{-3} \, m} = \frac{20 	imes 10^{-7}}{0.2 	imes 10^{-3}} = 100 	imes 10^{-4} \, m = 10^{-2} \, m$.

In Fraunhofer diffraction pattern, slit width is $0.2 \, mm$ and screen is at $2 \, m$ away from the lens. If wavelength of light used is $5000 \, \mathring{A}$, then the distance between the first minimum on either side of the central maximum is (assume $\theta$ is small and measured in radians).

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