Problem 6.96: A ray is incident on a transparent  slab of refractive index $\sqrt{3}$.   The reflected and the refracted rays are mutually perpendicular.  Find the  angle of incidence.

Solution:

Problem 6.96

With reference to the diagram, $AO$ is the incident ray that makes angle $i$ with the normal. $OB$ is the reflected ray which according to law of reflection, will also make the same angle $i$ with the normal.

$OC$ is the refracted ray that makes angle $r$ with the normal.

From the geometry of the diagram, we get the sum of angles $i$ , $r$ and $90^o$ must be equal to $180^o$.

That is , $i\:+\:r\:+\:90^o\:=\:180^o$.

Or, $i\:+\:r\:=\:90^o$.

Or $r\:=\:(90-i)$.

Using Snell's law for refraction at point of incidence, O, we get

$\frac{sin\:i}{sin\:r}\:=\:\frac{n_1}{n_2}$.

Substituting the values we get,

$\frac{sin\:i}{sin\:(90-i)}\:=\:\frac{n_1}{n_2}$.

Or, $\frac{sin\:i}{cos\:i}\:=\:\frac{n_1}{n_2}$.

Or, $tan\:i\:=\:\frac{n_1}{n_2}$.

Here $n_1$ is the refractive index of air which is equal to 1 and $n_2$ is the refractive of the slab which is $\sqrt{3}$.

Then $tan\:i\:=\:\frac{1}{\sqrt{3}}$.

From this angle of incidence, $i\:=\:30^o$.

The angle of refraction $r\:=\:(90-i)\:=\:(90-30)\:=\:60^o$.