Light of wavelength 5000 , mathring{A} is inc | vedclass

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(A) The condition for the $n^{th}$ minima in a single slit diffraction pattern is given by $d \sin 	heta = n \lambda$. For small angles,$\sin 	heta

Vedclass · Accepted Answer

$0.2$

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(A) The condition for the $n^{th}$ minima in a single slit diffraction pattern is given by $d \sin 	heta = n \lambda$. For small angles,$\sin 	heta \approx 	an 	heta = \frac{x}{D}$. Thus,the position of the $n^{th}$ minima is $x = \frac{n \lambda D}{d}$. Given: $\lambda = 5000 \, \mathring{A} = 5 	imes 10^{-7} \, m$,$x = 5 \, mm = 5 	imes 10^{-3} \, m$,$D = 2 \, m$,and $n = 1$ for the first minima. Substituting these values into the formula: $d = \frac{n \lambda D}{x} = \frac{1 	imes 5 	imes 10^{-7} 	imes 2}{5 	imes 10^{-3}}$. $d = \frac{10 	imes 10^{-7}}{5 	imes 10^{-3}} = 2 	imes 10^{-4} \, m$. Converting to $mm$: $d = 2 	imes 10^{-4} 	imes 10^3 \, mm = 0.2 \, mm$.

Light of wavelength $5000 \, \mathring{A}$ is incident on a single slit such that the first minima is formed at a distance $5 \, mm$ from the centre. If the screen is placed $2 \, m$ away,then find the width of the slit in $mm$.

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