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(B) Given:Width of aperture $a = 0.02\, cm = 2 	imes 10^{-4}\, m$Wavelength $\lambda = 5 	imes 10^{-5}\, cm = 5 	imes 10^{-7}\, m$Focal length of l

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$0.15$

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(B) Given:Width of aperture $a = 0.02\, cm = 2 	imes 10^{-4}\, m$Wavelength $\lambda = 5 	imes 10^{-5}\, cm = 5 	imes 10^{-7}\, m$Focal length of lens $f = 60\, cm = 0.6\, m$. Since the screen is at the focal plane,the distance $D = f = 0.6\, m$.For Fraunhofer diffraction at a single slit,the condition for the $n^{th}$ dark band (minima) is given by $a \sin 	heta = n\lambda$.For the first dark band,$n = 1$,so $\sin 	heta \approx 	heta = \frac{\lambda}{a}$.The distance from the centre of the screen is $y_1 = D 	heta = \frac{D\lambda}{a}$.Substituting the values:$y_1 = \frac{0.6 	imes 5 	imes 10^{-7}}{2 	imes 10^{-4}} = \frac{3 	imes 10^{-7}}{2 	imes 10^{-4}} = 1.5 	imes 10^{-3}\, m = 0.15\, cm$.

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