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First diffraction minima due to of a sin...

First diffraction minima due to of a single slit diffraction is at `theta = 30^(@)` for a light of wavelength `6000 Å`. The width of slits is

A

`1xx10^(-6) cm`

B

`1.2xx10^(-6) cm`

C

`2xx10^(-6) cm`

D

`2.4xx10^(-6) cm`

Text Solution

Verified by Experts

The correct Answer is:
B
For `1^(st)` diffraction minima, a `sintheta=nlamda`
`:.a=(1xx6000xx10^(-10))/(sin30^(@))=1.2xx10^(-6)m`
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