Light of wavelength ' $\lambda$ ' is incident on a slit of width 'd'. The resulting diffraction pattern is…
Light of wavelength ' $\lambda$ ' is incident on a slit of width 'd'. The resulting diffraction pattern is observed on a screen at a distance 'D'. The linear width of the principal maximum is then equal to the width of the slit if $D$ equals
$\frac{\mathrm{d}}{\lambda}$
$\frac{\mathrm{d}^2}{2 \lambda}$
$\frac{2 \lambda}{\mathrm{d}}$
$\frac{2 \lambda^2}{\mathrm{~d}}$
Solution
In diffraction of light by single slit, the width of central maximum is given as
$\mathrm{W}_{\mathrm{c}}=\frac{2 \lambda \mathrm{D}}{\mathrm{d}}$
Given: $\mathrm{W}_{\mathrm{c}}=\mathrm{d}$
$\begin{aligned}
\therefore \quad d & =\frac{2 \lambda D}{d} \\
& \Rightarrow D=\frac{d^2}{2 \lambda}
\end{aligned}$