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The ratio of angle of minimum deviation ...

The ratio of angle of minimum deviation of a prism in air and when dipped in water will be `(mu_(g)=3//2 and mu_(w)=4//3)`

A

`delta_(2) = 2 delta _(1)`

B

`delta_(2) = (delta_(1))/(2)`

C

`delta_(2) = (delta_(1))/(3)`

D

`delta_(2) = (delta_(1))/(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
Deviation produced by a thin prism `delta = A( n-1).`
In air `delta_(1) = A(""_(a) n_(g) - 1) = A (1.5 - 1) = (A)/(2)`
In water , we consider
`""_(w) n_(g) = (""_(a) n_(g))/(""_(a) n_(w)) = (3//2)/(4//3) = (9)/(8)`
`therefore delta_(2) = A [""_(w) n_(g) - 1] = A[ (9)/(8) - 1] = (A)/(8)`
`therefore (delta_(2))/(delta_(1)) = (A)/(8) xx (2)/(A) = (1)/(4) therefore delta_(2) = (delta_(1))/(4)`
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