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(B) The condition for the $n^{th}$ minima in single-slit diffraction is given by $a \sin 	heta = n \lambda$,where $a$ is the slit width,$\lambda$ is

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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 $a \sin 	heta = n \lambda$,where $a$ is the slit width,$\lambda$ is the wavelength,and $n$ is the order of the minima.For the first minima,$n = 1$,so $a \sin 	heta = \lambda$.Since the angle $	heta$ is very small,$\sin 	heta \approx 	an 	heta = \frac{y}{D}$,where $y$ is the distance from the central maximum and $D$ is the distance to the screen.Thus,$y = \frac{n \lambda D}{a}$.For the first minima $(n = 1)$,$y_1 = \frac{\lambda D}{a}$.The distance between the first minima on either side of the central maximum is $2y_1 = \frac{2 \lambda D}{a}$.Given: $a = 2 \ mm = 2 	imes 10^{-3} \ m$,$\lambda = 500 \ nm = 500 	imes 10^{-9} \ m$,$D = 1 \ m$.Substituting the values: $2y_1 = \frac{2 	imes 500 	imes 10^{-9} 	imes 1}{2 	imes 10^{-3}} = 500 	imes 10^{-6} \ m = 0.5 	imes 10^{-3} \ m = 0.5 \ mm$.

$A$ narrow slit of width $2 \ mm$ is illuminated with a monochromatic light of wavelength $500 \ nm$. If the distance between the slit and the screen is $1 \ m$,then the first minima are separated by a distance of: (in $mm$)

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