Light of wavelength $5000 Å$ is incident normally on a slit. The first minimum of the diffraction pattern is…
Light of wavelength $5000 Å$ is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at distance of $5 \mathrm{~mm}$ from the central maximum on a screen placed at a distance of $2 \mathrm{~m}$ from the slit. The width of the slit.is
$2 \mathrm{~cm}$
$0.2 \mathrm{~cm}$
$0.02 \mathrm{~cm}$
$0.01 \mathrm{~cm}$
Solution
The slit width is given as:
$\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
$\therefore \quad \mathrm{d}=\frac{2 \times 5000 \times 10^{-10}}{5 \times 10^{-3}}=0.02 \mathrm{~cm}$