Light of wavelength 600nm is incident normally on a slit of width 0.2 mm. The anuglular width of central maxima in the diffraction pattern is (measured from miminum to minimum )

Options

A. `6xx10^(-3)`rad
B. `4xx10^(-3)`rad
C. `2.4xx10^(-3)`rad
D. `4.5xx10^(-3)`rad

Answer

a

Explanation

Here , `lambda=600nm=6xx10^(-7)m`
`a=0.2mm=2xx10^(-4)m,theta=?`
Angular width of central maxima,
`theta = (2lambda)/(a) = (2xx6xx10^(-7))/(2xx10^(-4))=6xx10^(-3)`rad