Question

In Fraunhofer diffraction pattern, slit width is $0.2\, mm$ and screen is at $2 \,m$ away from the lens. If wavelength of light used is $5000 \, \mathring A $ then the distance between the first minimum on either side of the central maximum is ($\theta$ is small and measured in radian)

• A. $10^{-1} m $
• B. $10^{-2} m $
• C. $ 2 \times 10^{-2} m $
• D. $ 2 \times 10^{-1} m $

Answer 1

The Correct Option is B

Distance between the first $2$ minimum on other side $=\frac{2 \lambda D }{ d }$
$=\frac{2 \times 5 \times 10^{3} \times 10^{-10} \times 2}{2 \times 10^{-1} \times 10^{-3}}$
$=10 \times \frac{10^{-7}}{10^{-4}}$
$=10^{-2} \,m$