In a single-slit diffraction experiment,what | vedclass

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(C) The width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2D\lambda}{b}$,where $D$ is the distance to the scree

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$0.2$

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(C) The width of the central maximum in a single-slit diffraction pattern is given by $w = \frac{2D\lambda}{b}$,where $D$ is the distance to the screen,$\lambda$ is the wavelength,and $b$ is the slit width.The fringe width of a double-slit interference pattern is given by $\beta = \frac{D\lambda}{d}$,where $d$ is the slit separation.It is given that $20$ maxima of the double-slit pattern lie within the central maximum of the diffraction pattern. The total width occupied by $20$ interference fringes is $20 	imes \beta = 20 \frac{D\lambda}{d}$.Equating the two widths: $20 \frac{D\lambda}{d} = \frac{2D\lambda}{b}$.Simplifying the equation: $\frac{20}{d} = \frac{2}{b}$.Rearranging for $b$: $b = \frac{2d}{20} = \frac{d}{10}$.Given $d = 2 \, mm$,we get $b = \frac{2 \, mm}{10} = 0.2 \, mm$.

In a single-slit diffraction experiment,what is the width of the slit $(b)$ such that $20$ maxima of a double-slit interference pattern are obtained within the central maximum of the diffraction pattern? (Given: Slit separation for the double-slit arrangement $d = 2 \, mm$)

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