Question

An equilateral prism provides a least deflection angle $60^\circ$ in the air. Find the refractive index of the unknown liquid in which the same prism gives least deflection angle of $30^\circ$.

Answer 1

Determine the angle of refraction at the first surface of the prism. Then you can use Snell’s law to determine the refractive index of the prism. Again use Snell’s law to determine the refractive index of the unknown liquid.

Formula used:
Snell’s law,
${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$
Here, ${n_1}$ is the refractive index of the medium at incident ray and ${n_2}$ is the refractive index of the medium at refracted ray, ${\theta _1}$ is the angle of incident and ${\theta _2}$ is the angle of refraction.

Complete step by step answer:
We have given the least deflection angle of the equilateral prism is $60^\circ$ in air. From this we have to find the angle incident for the prism,
We know the least deflection angle of the prism is given as,
$\delta = 2\alpha - \theta$
Here, $\alpha$ is the angle of incident at the first surface and $\theta$ is the prism angle.
Therefore, we can express the angle incident at the first surface as follows,
$\alpha = \dfrac{{\delta + \theta }}{2}$
We know that the angle of refraction at the first surface of the prism is equal to half the angle of the prism.
We can use Snell’s law to determine the refractive index of the prism as follows,
${n_1}\sin \left( {\dfrac{{\alpha + \theta }}{2}} \right) = {n_2}\sin \left( {\dfrac{\theta }{2}} \right)$
Here, ${n_1}$ is the refractive index of air and ${n_2}$ is the refractive index of the prism.
Substituting $60^\circ$ for $\alpha$, $60^\circ$ for $\theta$ and 1 for ${n_1}$ in the above equation, we get,
$\left( 1 \right)\sin \left( {\dfrac{{60 + 60}}{2}} \right) = {n_2}\sin \left( {\dfrac{{60}}{2}} \right)$
$\Rightarrow {n_2} = \dfrac{{\sin \left( {60} \right)}}{{\sin \left( {30} \right)}}$
$\Rightarrow {n_2} = 1.73$
Now, we have given that the prism is now placed in the liquid whose refractive index is to be determined.
We can use Snell’s law to determine the refractive index of the liquid as follows,
${n_1}\sin \left( {\dfrac{{\alpha + \theta }}{2}} \right) = {n_2}\sin \left( {\dfrac{\theta }{2}} \right)$
Here, ${n_1}$ is the refractive index of liquid and ${n_2}$ is the refractive index of the prism.
Substituting $30^\circ$ for $\alpha$, $60^\circ$ for $\theta$ and 1.73 for ${n_2}$ in the above equation, we get,
$\left( {{n_1}} \right)\sin \left( {\dfrac{{30 + 60}}{2}} \right) = \left( {1.73} \right)\sin \left( {\dfrac{{60}}{2}} \right)$
$\Rightarrow {n_1} = \dfrac{{\left( {1.73} \right)\sin \left( {30} \right)}}{{\sin 45}}$
$\Rightarrow {n_1} = \dfrac{{0.865}}{{0.7071}}$
$\therefore {n_1} = 1.22$

Therefore, the refractive index of the unknown liquid is 1.22.

Note: The angle of the prism is constant and it is $60^\circ$. To answer this type of question, students should remember the refractive index of water and air. The refractive index of pure water is 1.33 and refractive index of air is 1.