Question

The ray of light is incident normally in one surface of a prism of refracting angle ${60^ \circ }$ . The emergent ray grazes the other refracting surface. Find the refractive index of the prism.

Answer 1

To solve this question, one must have a concept of Snell’s law and critical angle and then you can easily solve this question. Here first we have found another angle and then using the concepts of Snell’s law we got the required solution. We know that the critical angle is the angle of incidence with the total internal reflection.

Complete step by step answer:
According to the question we have given that the ray of light is normally incident on the surface of prism. Therefore, we can say that, ${r_1} = 0$. And the angle of refraction by prism is ${60^ \circ }$.So,
${r_1} + {r_2} = {60^ \circ }$
And now substituting the value of ${r_1}$ , we get
${r_2} = {60^ \circ }$
We know from Snell’s law we can say that here,
$\mu = \dfrac{1}{{\sin {\theta _c}}}$

Now, putting the value of $\sin \theta$,
$\mu = \dfrac{1}{{\sin {{60}^ \circ }}} \\\ \Rightarrow \mu = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}} \\\ \Rightarrow \mu = \dfrac{2}{{\sqrt 3 }} \\\$
Now on rationalisation,
$\mu = \dfrac{{2\sqrt 3 }}{3} \\\ \Rightarrow \mu = \dfrac{{3.46}}{3} \\\ \therefore \mu = 1.15 \\\$
Hence the refractive index of the given prism is $1.15$.

Note: The incident ray, the refracted ray, and the normal to the border between the two mediums are all on the same plane. The refractive index of a medium is defined as the sine value of the angle of incidence divided by the sine value of the angle of refraction for a particular pair of media.