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(D) The linear width of the central principal maximum in a single-slit diffraction pattern is given by the formula $\beta = \frac{2 \lambda D}{d}$,whe

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$\frac{d^2}{2 \lambda}$

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(D) The linear width of the central principal maximum in a single-slit diffraction pattern is given by the formula $\beta = \frac{2 \lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance of the screen from the slit,and $d$ is the width of the slit.According to the problem,the linear width of the principal maximum is equal to the width of the slit,so we set $\beta = d$.Substituting this into the formula: $d = \frac{2 \lambda D}{d}$.Rearranging the equation to solve for $D$: $d^2 = 2 \lambda D$.Therefore,$D = \frac{d^2}{2 \lambda}$.The correct option is $D$.

$A$ light wave of wavelength $\lambda$ is incident on a slit of width $d$. The resulting diffraction pattern is observed on a screen at a distance $D$. If the linear width of the principal maximum is equal to the width of the slit,then the distance $D$ is:

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