Let the refractive index of a denser medium with respect to a rarer medium be `n_(12)` and its critical angle be `theta_(c)`. At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is `90^(@)`, Angle A given by -

Options

A. `(1)/(cos^(-1)(sintheta_(C)))`
B. `(1)/(tan^(-1)(sintheta_(C)))`
C. `cos^(-1)sintheta_(C)`
D. `tan^(-1)sintheta_(C)`

Answer

D

Explanation

Applying Snell.s law in the above figure :
`mu_(1) sin theta_(C) = mu_(2) sin 90^(@)`
or , `sin theta_(C) = (mu_(2))/(mu_(1))= ""_(1)mu_(2)`
Considering the following diagram :

From the figure we can write
`A + 90^(@) +r=180^(@)`
OR, `A+r= 90^(@)`
Applying Snell.s law
`mu_(1) sin A = mu_(2) sin r`
`sin A = (mu_(2))/(mu_(1)) sin r`
`sin A = (mu_(2))/(mu_(1)) sin (90^(@)-A)`
`(sin A)/(cos A) = (mu_(2))/(mu_(1))`
Using equation (i)
`tan A = sin theta_(C)`
`rArr A = tan^(-1) ( sin theta_(C))`