The speed of light in media `M_1` and `M_2` are `1.5 xx 10^8 m//s and 2.0 xx 10^8 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 internal reflection, 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)`