The diffraction pattern of light of wavelengt | vedclass

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(A) The width of the $n^{	ext{th}}$ secondary maxima in a single-slit diffraction pattern is given by $\beta = \frac{\lambda D}{a}$,where $\lambda$ i

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$2 \ mm$

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(A) The width of the $n^{	ext{th}}$ secondary maxima in a single-slit diffraction pattern is given by $\beta = \frac{\lambda D}{a}$,where $\lambda$ is the wavelength,$D$ is the distance of the screen (focal length of the lens),and $a$ is the slit width.Given:$\lambda = 400 \ nm = 400 	imes 10^{-9} \ m$$a = 0.2 \ mm = 0.2 	imes 10^{-3} \ m$$D = 100 \ cm = 1 \ m$Substituting the values:$	ext{Width} = \frac{400 	imes 10^{-9} \ m 	imes 1 \ m}{0.2 	imes 10^{-3} \ m}$$= \frac{400}{0.2} 	imes 10^{-6} \ m$$= 2000 	imes 10^{-6} \ m$$= 2 	imes 10^{-3} \ m = 2 \ mm$.

The diffraction pattern of light of wavelength $400 \ nm$ diffracting from a slit of width $0.2 \ mm$ is focused on the focal plane of a convex lens of focal length $100 \ cm$. The width of the $1^{\text{st}}$ secondary maxima will be:

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