The speed of light in media `M_(1) and M_(2)` are `1.5 xx 10^(8) and 2 xx 10^(8) ms^(-1)` respectively. A ray travels from medium `M_(1)` to the medium `M_(2)` with an angle of incidence `theta`. The ray suffers total internal reflection. Then the value of the angle of incidence `theta` is

Options

A. `gt sin^(-1) ((3)/(4))`
B. `lt sin^(-1) ((3)/(4))`
C. `= sin^(-1) ((4)/(3))`
D. `le sin^(-1) ((4)/(3))`

Answer

A

Explanation

Given, `v_(1) = 1.5 xx 10^(8) ms^(-1)`
`v_(2) = 2.0 xx 10^(8) ms^(-1)`
Refractive index for medium `M_(1)` is
`mu_(1) = (c )/(v_(1)) = (3 xx 10^(8) ms^(-1))/(1.5 xx 10^(8) ms^(-1)) =2`
Refractive index for medium `M_(2)` is
`mu_(2) = (c )/(v_(2)) = (3 xx 10^(8) ms^(-1))/(2.0 xx 10^(8) ms^(-1)) = (3)/(2)`
If `theta` is the angle of incidence and C is the critical angle, then for total internal reflection
`sin theta gt sin C "But " sin C = (mu_(2))/(mu_(1))`
`:. sin theta gt (mu_(2))/(mu_(1)) gt (3//2)/(2) or theta gt sin^(-1) ((3)/(4))`