Visible light of wavelength 6000 ` xx 10 ^( - 8 ) ` cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at ` 60 ^(@) ` from the central maximum. If the first minimum is produced at ` theta _ 1`, then ` theta _ 1 ` is close to :

Options

A. `20^(@)`
B. `30^(@)`
C. `25^(@)`
D. `45^(@)`

Answer

C

Explanation

For `2 ^(nd)` minima,
`d sin theta = 2 lamda `
`sin theta = (sqrt3)/(2),` (given)
`implies (lamda )/(d) = (sqrt3)/(4)...(i)`
So, for `1 ^(st)` minima
`d sin theta = lamda `
`implies sin theta = (lamda )/(d) = (sqrt3)/( 4)`
`implies theta = 25.65 ^(@)`
`implies theta ~~ 25 ^(@)`