Visible light of wavelength `6000 xx 10^(-8) cm` fall normally on a single slit and produces diffraction pattern. It is found that the second diffraction minima is at `60^(@)` from the centre maxima. If the first minima is produced at `theta` is close to

Options

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

Answer

B

Explanation

For second minimum,
`d sin theta= 2lambda`
`:.sin theta=(2lambda)/(d)`
`:. Sin 60^(@)=(2lambda)/(d)`
`:. (sqrt(3))/(2)=(3lambda)/(d)`
`:. (lambda)/(d)=(sqrt(3))/(4) " "...(1)`
For first minimum,
`d sin theta = lambda`
`:. sin theta=(lambda)/(d)`
From equation (1),
`:. sin theta=(sqrt(3))/(4)`
`:. sin theta=0.4330`
From table of sine,
`theta=25.65(@)`
`:. theta~~25^(@)` (Nearer value)