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 Å`, then the distance between the first minimum on either side of the central maximum is (`theta` is small and measured in radian)

Options

A. `10^(-1)m`
B. `10^(-2)m`
C. `2xx10^(-2)m`
D. `2xx10^(-1)m`

Answer

B

Explanation

The distance between the two minima on either side of the central maximum=width of the central maximum
`=(2lamdaD)/(a)=(2xx5xx10^(-7)xx2)/(0.2xx10^(-3))`
`=10^(-2)m`