A light wave of wavelength ' $\lambda$ 'is incident on a slit of width ' $d$ '. The resulting diffraction…
A light wave of wavelength ' $\lambda$ 'is incident on a slit of width ' $d$ '. The resulting diffraction pattern is observed on screen at a distance of ' $\mathrm{D}$ '. If linear width of the principal maximum is equal to width of the slit, then the distance $D$ is
$\frac{\mathrm{d}}{\lambda}$
$\frac{2 \lambda^2}{d}$
$\frac{2 \lambda}{d}$
$\frac{\mathrm{d}^2}{2 \lambda}$
Solution
The correct option is (D)
Concept: If $\mathrm{D}>>\mathrm{d}$,
the linear width of the central principal maximum is equal to the product of angular width and the distance $D$.
$\beta=\frac{2 \lambda D}{d}$, where $d$ is the width of the slit.
The linear width of the principal maximum will be equal to slit width for a value of $\mathrm{D}$ given by
$\frac{2 \lambda D}{d}=d$
or
$\mathrm{D}=\frac{\mathrm{d}^2}{2 \lambda}$