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The angle of minimum deviation by prism ...

The angle of minimum deviation by prism is (180^(0)-2A) .Its critival angle will be

A

`sin^(-1)(tan(A/2))`

B

`sin^(-1)(cot(A/2))`

C

`cos^(-1)(cot(A/2))`

D

`cos^(-1)(tan(A/2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`d=180-2A=2i-A`
`mu=(sini)/(sinr)=(sin(90 -A/2))/(sin(A/2)),mu=1/(sinC)`
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