Problem 6.94: A plane separates a transparent media of refractive index of $\sqrt{3}$ from air. A ray of light coming from air is incident at an angle of $60^o$. Determine the angle of deviation of the ray.

Problem 6.94Solution:

The light ray gets refracted as it enters the medium from air. It deviates from the original direction and follows a new path. Angle between the initial and new path is called angle of deviation.


Problem 6.94-1

In the diagram $AO$ represents the incident ray,.

$OC$ represents the refracted ray.

$OB$ is the initial path of the ray.

$NN$ is the normal at the point of incidence.

Angle between $NN$ and $AO$ is the angle of incidence which is $60^o$.

Angle between $NN$ and $OC$ is the angle of refraction which is $r$.

Angle between $OB$ and $OC$ is the angle of deviation which is $\theta$.

From the geometry, Angle between $OB$ and $NN$ equal to angle of incidence which is $60^o$.

Also, angle of deviation $\theta\:=\:60^o\:-\:r$....$(1)$

Angle of deviation can be calculated by applying Snell’s law at the point of incidence $O$.

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

Or, $\frac{sin\:60^o}{sin\:r}\:=\:\frac{\sqrt{3}}{1}$

Or $sin\:i\:=\:\frac{1}{2}$

So angle of refraction $r\:=\:30^o$

Now from equation $(1)$, we get angle of deviation $\theta\:=\:60^o\:-\:30^o\:=\:30^o$.

So the calculation shows that the ray will suffer a deviation of $30^o$ from its initial path.