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(C) In Fraunhofer diffraction due to a single slit,the angular width of the central maximum is given by $\Delta 	heta = \frac{2\lambda}{a}$.To find t

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$\frac{2f\lambda}{a}$

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(C) In Fraunhofer diffraction due to a single slit,the angular width of the central maximum is given by $\Delta 	heta = \frac{2\lambda}{a}$.To find the linear width $w$ of the central maximum on the screen placed at the focal plane of a lens with focal length $f$,we use the relation $w = f 	imes \Delta 	heta$.Substituting the value of $\Delta 	heta$,we get $w = f 	imes \left( \frac{2\lambda}{a} ight) = \frac{2f\lambda}{a}$.Thus,the linear width of the central maximum is $\frac{2f\lambda}{a}$.

In Fraunhofer diffraction by a single slit,if the slit width is $a$,the wavelength is $\lambda$,and the focal length of the lens is $f$,what is the linear width of the central maximum?

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