The speed of light in media `M_(1)` and `M_(2)` is `1.5xx10^(8)` m/s and `2.0xx10^(8)` m/s respectively. A ray of light enters from medium `M_(1)` to `M_(2)` at an incidence angle i. If the ray suffers total internalreflection the value of i is -

Options

A.
B.
C.
D.

Answer

A

Explanation

For the internal reflection, the angle of incidence of light in denser medium must be greater than the critical angle for the pair of media in contact. i.e., `I gt C`
We have, `(1)/(sin C)=(v_(2))/(v_(1))`
Given, `v_(1)=M_(1)=1.5xx10^(8)ms^(-1)`
`v_(2)=M_(2)=2xx10^(8)ms^(-1)`
Here, `C=sin^(-1)((M_(1))/(M_(2)))`
For total internal reflection, `i gt C`
`therefore i gt sin^(-1)((M_(1))/(M_(2)))`
`therefore i gt sin1^(-1)((1.5xx10^(8))/(2xx10^(8))) Rightarrow i gt "sin"^(-1)(3)/(4)`