Frist diffraction minima due to of a single slit diffraction is at `theta = 30^(@)` for a light of wavelength `6000 Å`. The width of slits is

Options

A. `1xx10^(-6)cm`
B. `1.2xx10^(-6)m`
C. `2xx10^(-6)cm`
D. `2.4xx10^(-6)m`

Answer

B

Explanation

`sin theta = (x)/(d) = (lambda)/(b) , a sin theta = n lambda, a = (n lambda)/(sin theta)`