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A ray of light incident at angle of 40^(...

A ray of light incident at angle of `40^(@)` , on a glass slab , is deviated through `15^(@)` , while passing through glass . What is the critical angle for the glass air surface ?

A

`32^(@)`

B

`35^(@)`

C

`38^(@)`

D

`41^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`delta i=r therefore r = i- delta = 40^(@) - 15^(@) = 25^(@)`
`therefore n = (sini )/(sin r) = (sin 40^(@))/(sin 25^(@)) = 1.52`

If `i_(c)` is the critical angle then `sin i_(c) = (1)/(1.52)`
`therefore i_(c) = sin ^(-1) ((1)/(1.52)) = 41^(@)`
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