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Question

(A) Given: Slit width $a = 2 \, mm = 2 	imes 10^{-3} \, m$.Wavelength $\lambda = 500 \, nm = 500 	imes 10^{-9} \, m$.Distance of screen $D = 1 \, m$

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$0.5$

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(A) Given: Slit width $a = 2 \, mm = 2 	imes 10^{-3} \, m$.Wavelength $\lambda = 500 \, nm = 500 	imes 10^{-9} \, m$.Distance of screen $D = 1 \, m$.For single slit diffraction,the position of the $n^{th}$ minima is given by $y_n = \frac{n \lambda D}{a}$.The distance between the first minima on both sides is the width of the central maximum,which is given by $x = 2y_1 = \frac{2 \lambda D}{a}$.Substituting the values:$x = \frac{2 	imes (500 	imes 10^{-9} \, m) 	imes (1 \, m)}{2 	imes 10^{-3} \, m}$.$x = 500 	imes 10^{-6} \, m = 0.5 	imes 10^{-3} \, m = 0.5 \, mm$.

If a narrow slit of width $2 \, mm$ is illuminated by monochromatic light of wavelength $500 \, nm$,then the distance between the first minima on both sides on a screen at a distance of $1 \, m$ is ....... $mm$.

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