Question

An equilateral prism is made of a transparent material of refractive index 2.A ray of light AB is incident at 45 as shown. The net derivation in the path of the ray when it comes out of prism is?

A. 135°
B. 120°
C. 30°
D. 150°

Answer 1

Hint: In this question we will use the concept of deviation in the path of a ray due to refraction at the surfaces of a prism. Since the speed of the media changes, the difference in their refractive indices causes the phenomenon of refraction to happen. This alteration in speed causes the ray's direction to change.

Formula used:
The expression of refractive index is,
\[\mu =\dfrac{\sin (i)}{\sin ({{r}_{1}})}\]
Here, $i$ is the angle of incidence and $r_1$ is the angle of refraction.

Complete step by step solution:
Let us name the triangle as PQR. The side where AB ray hits is PQ and the silvered side is PR.
Since the triangle is equilateral all the angles of the triangle are equal to 60°.

In the image above, i=incident angle, $r_1$=refractive angle of surface PQ, $r_2$= refractive angle of surface PR, e=angle of emergence and δ=total angle of deviation.
The incident ray AB when entering the prism, it is traveling from a rarer medium to a denser medium. Therefore, the ray will bend towards the normal of the PQ surface.

The relation between i, $r_1$ and μ(refractive index) is:
\[\mu =\dfrac{\sin (i)}{\sin ({{r}_{1}})}\]
$\Rightarrow \sqrt{2}=\dfrac{\sin (45)}{\sin ({{r}_{1}})}$
$\Rightarrow \sin ({{r}_{1}})=\dfrac{1}{2}$
$\Rightarrow {{r}_{1}}={{30}^{\circ }}$
Since $r_1$=30°, the refractive ray became parallel to the base.
By geometry, $r_1+r_2$=angle of the triangle
$\Rightarrow {{30}^{\circ }}+{{r}_{2}}={{60}^{\circ }}$
$\Rightarrow {{r}_{2}}={{30}^{\circ }}$

In the given question, side PR is silvered. Therefore, the refracted ray parallel to the base of
the triangle will reflect. Since angle of incidence for the mirror is 30°, angle of reflection is also equal to 30°.
Total deviation at the reflective surface = 30°+30°=60°

The ray diagram is given below.

Now, the reflective ray will hit the surface QR at 30° to the normal.
The emergence ray is traveling from a denser medium to rarer medium.
According to the formula:
\[\mu =\dfrac{\sin (i)}{\sin ({{r}_{1}})}\]
$\Rightarrow \sqrt{2}=\dfrac{\sin ({{30}^{\circ }})}{\sin (e)}$
$\Rightarrow \sin (e)=\dfrac{\sin ({{30}^{\circ }})}{\sqrt{2}}$
$\therefore e={{45}^{\circ }}$
Total deviation = (45° -30°) + (180° -60°) + (45° -30°) = 150° clockwise.

Hence, the correct answer is D.

Note: By geometry, $r_1+r_2$=angle of the triangle. The angle of emergence is not equal to angle $r_2$ since refraction will happen. When a ray is traveling from a rarer medium to a denser medium, the ray will bend towards the normal. When a ray is traveling from a denser medium to a rarer medium, the ray will bend away from the normal.