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Visible light of wavelength 6000 x...

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 :

A

`25^(@)`

B

`30^(@)`

C

`20^(@)`

D

`45^(@)`

Text Solution

Verified by Experts

The correct Answer is:
A
For `2^(nd)` minima
` d sin theta = 2 lambda `
` sin theta = (sqrt(3))/(2)` [`:. theta = 60^(@)` ]
`rArr (lambda)/( d) = (sqrt(3))/( 4)` . . . (i)
So for `1^(st)` minima is
` d sin theta = lambda `
`sin theta = (lambda)/( d) = (sqrt(3))/(4)` [From eq. (i)]
` theta = 25 . 65^(@)` [From sin table]
`theta ~~ 25^(@)`
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