Light from a denser medium 1 passes to a rarer medium 2. When the angle of incidence is `theta` the partially reflected and refracted rays are mutually perpendicular. The critical angle will be

Options

A. `sin^(-1)(cottheta)`
B. `sin^(-1)(tantheta)`
C. `sin^(-1)(costheta)`
D. `sin^(-1)(sectheta)`

Answer

b.

Explanation

`mu_(1)sintheta=mu_(2)xxsin(90^@-theta)`
`rArr (mu_(1))/(mu_(2))=tantheta`
For `theta_(c)rArr mu_(1)xx sintheta_(C)=mu_(2)xxsin(90^@)`
`sintheta_(C)=(mu_(1))/(mu_(2))=tantheta`
`rArr theta_(C)=sin^(-1)(tantheta)`