In Fraunhofer diffraction pattern, slit width is $0.2 \mathrm{~mm}$ and screen is at $2 \mathrm{~m}$ away…
In Fraunhofer diffraction pattern, slit width is $0.2 \mathrm{~mm}$ and screen is at $2 \mathrm{~m}$ away from the lens. If wavelength of light used is $5000 Å$ then the distance between the first minimum on either side of the central maximum is $(\theta$ is small and measured in radian)
$2 \times 10^{-2} \mathrm{~m}$
$10^{-1} \mathrm{~m}$
$10^{-2} \mathrm{~m}$
$10^{-3} \mathrm{~m}$
Solution
Distance between the first minima on either side of the central maximum $=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}=\frac{2 \times 5 \times 10^{-7} \times 2}{0.2 \times 10^{-3}}=10^{-2} \mathrm{~m}$