Question

An equilateral prism of refractive index $\sqrt{2}$ placed in air. A light ray is incident on its surface normally. The net deviation of the ray due to the prism will be
$A)\text{ }{{\sin }^{-1}}\sqrt{\dfrac{3}{2}}-{{60}^{0}}$
$B)\text{ 1}{{\text{2}}^{0}}$
$C)\text{ 3}{{0}^{0}}$
$D)\text{ }{{60}^{0}}$

Answer 1

Hint: This problem can be solved by using the direct formula for the angle of deviation for a light ray through a prism in terms of the angle of incidence, angle of emergence and the refracting angle of the prism. The angle of incidence and refracting angle is given while the angle of emergence has to be found out using the refraction at the second surface of the prism.
Formula used:
$\delta =i+e-A$
${{r}_{1}}+{{r}_{2}}=A$
${{\theta }_{C}}={{\sin }^{-1}}\left( \dfrac{{{n}_{2}}}{{{n}_{1}}} \right)$

Complete answer:
We will use the formula for the angle of deviation of a light ray passing through a prism.
The angle of deviation $\delta $ of a light ray passing through a prism of refracting angle $A$ is given by
$\delta =i+e-A$ --(1)
Where $i,e$ are the angle of incidence and the angle of emergence for the light ray.
Now, let us analyze the question.
Since it is an equilateral prism, the refracting angle of the prism is ${{60}^{0}}$.
Since, the light ray is incident on the surface of the prism normally, the angle of incidence is $i={{0}^{0}}$.
Let the angle of emergence be $e$.
Let the angle of deviation of the light ray be $\delta $.
The refractive index of the material of the prism is $\sqrt{2}$.
Therefore, using (1), we get
$\delta =0+e-{{60}^{0}}=e-{{60}^{0}}$ --(2)
Now, the relation between the angle of refraction ${{r}_{1}}$ at the first surface and the angle of incidence ${{r}_{2}}$ at the second surface and the refracting angle $A$ of the prism is
${{r}_{1}}+{{r}_{2}}=A$ --(3)
Now, since the angle of incidence is zero for the prism, the light is not refracted at the first surface. Therefore, the angle of refraction for the first surface is ${{r}_{1}}={{0}^{0}}$.
Therefore, using (3), we get the angle of incidence at the second surface as
$0+{{r}_{2}}={{60}^{0}}$
$\Rightarrow {{r}_{2}}={{60}^{0}}$ --(4)
Now, the critical angle ${{\theta }_{C}}$ for total internal reflection for a light ray going from a medium of refractive index ${{n}_{1}}$ to a medium of refractive index ${{n}_{2}}$, where ${{n}_{1}}>{{n}_{2}}$ is given by
${{\theta }_{C}}={{\sin }^{-1}}\left( \dfrac{{{n}_{2}}}{{{n}_{1}}} \right)$ --(5)
Using (5), we get for the second surface when the light ray is at the interface of the prism and air, we get the critical angle ${{\theta }_{C}}$ as
${{\theta }_{C}}={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)={{45}^{0}}$ $\left( \because {{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)={{45}^{0}} \right)$ --(6)
Comparing (4) and (6), we get that the angle of incidence at the second surface is greater than the critical angle. Therefore, the light ray will be totally internally reflected. This ray will come out normal to the base of the prism and hence, the angle of emergence will be $e={{0}^{0}}$.
Putting this information in (2), we get
$\delta =0-60=-{{60}^{0}}$
The magnitude of the angle of deviation will be
$\left| \delta \right|=\left| -{{60}^{0}} \right|={{60}^{0}}$

So, the correct answer is “Option D”.

Note:
Students must not confuse this case with the case for the angle of minimum deviation and use the specific formulae for the angle of minimum deviation for a light ray passing through a prism. Minimum deviation is only a special case that happens for a certain angle of incidence and for which the light ray inside the prism is parallel to the base of the prism. In this situation, that is not happening and therefore, if students use those formulae, they will arrive at a completely wrong answer.