In a double slit experiment,the two slits are | vedclass

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(A) Given:Distance between slits $d = 1\, mm = 1 	imes 10^{-3}\, m$Distance of screen $D = 1\, m$Wavelength $\lambda = 500\, nm = 500 	imes 10^{-9}\

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$0.2$

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(A) Given:Distance between slits $d = 1\, mm = 1 	imes 10^{-3}\, m$Distance of screen $D = 1\, m$Wavelength $\lambda = 500\, nm = 500 	imes 10^{-9}\, m$The width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2\lambda D}{a}$,where $a$ is the slit width.The width of one fringe in a double-slit interference pattern is $\beta = \frac{\lambda D}{d}$.According to the problem,the width of the central maximum of the single-slit pattern contains $10$ maxima of the double-slit pattern.Therefore,$\frac{2\lambda D}{a} = 10 	imes \frac{\lambda D}{d}$.Simplifying the equation: $\frac{2}{a} = \frac{10}{d}$.$a = \frac{2d}{10} = \frac{d}{5}$.Substituting $d = 1\, mm$:$a = \frac{1\, mm}{5} = 0.2\, mm$.

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