A plane wavefront (lambda = 6 times 10^{-7} , | vedclass

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(A) For secondary maxima in single slit diffraction,the condition is given by $d \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n$ is the order of th

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(A) For secondary maxima in single slit diffraction,the condition is given by $d \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n$ is the order of the maximum.For the second maximum,$n = 2$,so $d \sin 	heta = \frac{5\lambda}{2}$.Since $	heta$ is very small,$\sin 	heta \approx 	heta = \frac{x}{f}$,where $x$ is the distance from the central axis and $f$ is the focal length.Thus,$d \cdot \frac{x}{f} = \frac{5\lambda}{2} \implies x = \frac{5\lambda f}{2d}$.The linear diameter of the second maximum is the distance between the second maxima on either side of the central axis,which is $2x$.$2x = \frac{5\lambda f}{d} = \frac{5 	imes (6 	imes 10^{-7} \, m) 	imes 0.8 \, m}{0.4 	imes 10^{-3} \, m}$.$2x = \frac{24 	imes 10^{-7}}{0.4 	imes 10^{-3}} = 60 	imes 10^{-4} \, m = 6 	imes 10^{-3} \, m = 6 \, mm$.

$A$ plane wavefront $(\lambda = 6 \times 10^{-7} \, m)$ falls on a slit $0.4 \, mm$ wide. $A$ convex lens of focal length $0.8 \, m$ placed behind the slit focuses the light on a screen. What is the linear diameter of the second maximum in $mm$?

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