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. `2 xx 10^(-2) m`
D. `2 xx 10^(-1) m`

Answer

B

Explanation

Given , `a = 0.2xx 10^(-3)m , D = 2`m
`lambda = 5 xx 10^(-7) m`
As, `y = (lambdaD)/(a) = (5 xx 10^(-7) xx 2)/(0.2 xx 10^(-3))`
`:. Y = (5 xx 10^(-7))/(10^(-4))`
`:. Y = 5 xx 10^(-3) m`
Distance between 1st minima on either side.
`= 5 xx 10^(-3) + 5 xx 10^(-3) = 10 xx 10^(-3) = 10^(-2) m`