Light of wavelength `6328 Å` is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be :

Options

A. `0.36^(@)`
B. `0.18^(@)`
C. `0.72^(@)`
D. `0.09^(@)`

Answer

B

Explanation

The angular half-width of central maxima is given by
`sin theta=(lambda)/(a)`
`rArr theta=(6328xx10^(-10))/(0.2xx10^(-3))` rad
`rArr theta=(6328xx10^(-10)xx180)/(0.2xx10^(-3)xx pi)"degree"=0.180^(@)`