Light of wavelength `6328 Å` is incident normally on a slit having a width of `0.2 mm`. The distance of the screen from the slit is `0.9 m`. The angular width of the central maximum is

Options

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

Answer

D

Explanation

Angular width of central maximum
`theta_(0)=(2lambda)/a=(2xx6328xx10^(-10))/(0.2xx10^(-3))=6.328xx10^(-3) rad`
`=[(6.328xx10^(-3))xx180/pi]^(@)=0.36^(@)`