Light of wavelength 6328 Å is incident normally on a slit of width 0.2 mm. Calculate the angular width of central maximum on a screen distance 9 m?

Options

A. `0.36^(@)`
B. `0.18^(@)`
C. `0.82^(@)`
D. `0.09^(@)`

Answer

A

Explanation

`a sin theta = nlambda`
Since width of central maxima is twice the distane of `1^(st)` dark bands on both sides,
`2 sin theta=(2xx1xxlambda)/(a)`
`=(2xx6.328xx10^(-7))/(2xx10^(-4)) =0.006328 `
`:. 2 theta= sin^(-1) (0.006328)= 0.36^(@)`