A narrow slit of width 2 , mm is illuminated | vedclass

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

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$0.5$

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(B) The condition for the $n^{th}$ minima in single slit diffraction is given by $a \sin 	heta = n \lambda$. For small angles,$\sin 	heta \approx 	heta = \frac{y}{D}$.Thus,the position of the first minima $(n=1)$ is $y = \frac{\lambda D}{a}$.The distance between the first minima on either side of the central maximum is $2y = \frac{2 \lambda D}{a}$.Given: $a = 2 \, mm = 2 	imes 10^{-3} \, m$,$\lambda = 500 \, nm = 5 	imes 10^{-7} \, m$,and $D = 1 \, m$.Substituting the values:$2y = \frac{2 	imes (5 	imes 10^{-7} \, m) 	imes (1 \, m)}{2 	imes 10^{-3} \, m} = 5 	imes 10^{-4} \, m$.Converting to millimeters: $5 	imes 10^{-4} \, m = 0.5 \, mm$.

$A$ narrow slit of width $2 \, mm$ is illuminated by monochromatic light of wavelength $500 \, nm$. The distance between the first minima on either side on a screen at a distance of $1 \, m$ is ........ $mm$.

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