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When light of wavelength lambda on an eq...

When light of wavelength `lambda` on an equilateral prism, kept on its minimum deviation position, it is found that the angle of deviation equals the angle the angle of the prism itself. The refractive index of the material of the prism for the wavelength `lambda` is

A

`sqrt(3)`

B

`(sqrt(3))/(2)`

C

2

D

`sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
A

`mu=(sin((A+delta_(m))/(2)))/(sin ((A)/(2)))=(sin((60^(@)+60^(@))/(2)))/sin ((60^(@))/(2))=sqrt(3)`
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