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Question: One of the refractive surfaces of a prism of angle \({{30}^{\circ }}\) is silvered. A ray incident a...

One of the refractive surfaces of a prism of angle 30{{30}^{\circ }} is silvered. A ray incident at angle of 60{{60}^{\circ }} retraces its paths. The refractive index of the material of prism is
A. 2\text{A}\text{. }\sqrt{2}
B. 3\text{B}\text{. }\sqrt{3}
C. 32\text{C}\text{. }\dfrac{3}{2}
A. 2\text{A}\text{. 2}

Explanation

Solution

For the light ray to trace its path, it must hit the reflecting surface perpendicularly. Then with geometry, when the angle of refraction at the refracting surface. Then use Snell’s law and find the refractive index of the prism.

Formula used:
μisini=μrsinr{{\mu }_{i}}\sin i={{\mu }_{r}}\sin r

Complete answer:
It is given that one of the refractive surfaces of the prism is silvered. As a result, this surface for the prism will act as a mirror. It is given that the angle of the prism is 30{{30}^{\circ }}. Let the refractive index of the prism be μ\mu .
Then, it said that a ray of light is incident on the refracting surface (not silvered) at an angle of 60{{60}^{\circ }}. The ray will refract at this surface and fall on the silvered surface. Since, this surface is a mirror. the light will reflect at the surface.

It is said that after reflecting at the second surface, the ray retraces its path. For this, the angle incidence for the reflecting surface must be zero. This means that the rays must hit the reflecting surface perpendicularly (as shown).
From Snell’s law, we get μisini=μrsinr{{\mu }_{i}}\sin i={{\mu }_{r}}\sin r ……. (i),
i and r the angles of incidence and refraction. μi{{\mu }_{i}} and μr{{\mu }_{r}} are the refractive indices of the mediums in which the ray is incident and refracted respectively.
In this case, i = 60{{60}^{\circ }}, μi{{\mu }_{i}}=1 and let μr{{\mu }_{r}}=μ\mu .
Consider the triangle ABC. In this triangle BAC=1803090=60\angle BAC=180-30-90={{60}^{\circ }}.
As you can see in the figure, BAC=90r\angle BAC=90-r.
Therefore, we get that 60=90r60=90-r.
r=30r=30.
Substitute the values in equation (i).
1sin60=μsin30\Rightarrow 1\sin 60=\mu \sin 30
32=μ(12)\Rightarrow \dfrac{\sqrt{3}}{2}=\mu \left( \dfrac{1}{2} \right)
μ=3\Rightarrow \mu =\sqrt{3}.

Hence, the correct option is B.

Note:
Always remember the Snell’s equation properly. Many times students get confused in remembering the correct equation.
Remember that the refractive index of the medium in which the ray is incident and the angle of incidence come on the same side of the equation. Then you will know which quantities will come on the other side of the equation.