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(B) The width of the central maximum in a single-slit diffraction pattern is given by:$W = \frac{2 D \lambda}{a}$where $\lambda$ is the wavelength of

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

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(B) The width of the central maximum in a single-slit diffraction pattern is given by:$W = \frac{2 D \lambda}{a}$where $\lambda$ is the wavelength of the incident light,$D$ is the distance between the slit and the screen,and $a$ is the slit width.According to the problem,the slit width $a$ is equal to the width of the central maximum $W$:$a = W$Substituting the expression for $W$:$a = \frac{2 D \lambda}{a}$Rearranging the equation to solve for $D$:$a^2 = 2 D \lambda$$D = \frac{a^2}{2 \lambda}$

Light of wavelength $\lambda$ is incident on a single slit of width $a$ and the distance between the slit and the screen is $D$. In the diffraction pattern,if the slit width is equal to the width of the central maximum,then $D$ is equal to:

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