Question

The angle of incidence for a ray of light at a refracting surface of a prism is ${45}^{\circ }.$ The angle of prism is ${60}^{\circ }.$ If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism, respectively are

[Assume the surrounding medium to be air]

• A. ${45}^{\circ };\sqrt{2}$
• B. ${30}^{\circ };\frac{1}{\sqrt{2}}$
• C. ${45}^{\circ };\frac{1}{\sqrt{2}}$
• D. ${30}^{\circ };\sqrt{2}$

Answer 1

The correct option is D ${30}^{\circ };\sqrt{2}$

Given,
Angle of incidence, $i={45}^{\circ }$
Angle of prism, $A={60}^{\circ }$

We know that,
Angle of deviation , $\delta =i+e-A$

Since the ray undergoes minimum deviation, angle of emergence $e$ from the second face must be equal to the angle of incidence $i$ on the first face.
So, $e=i={45}^{\circ }$

Substituting the data given in the question, we get minimum deviation

${\delta }_m={45}^{\circ }+{45}^{\circ }-{60}^{\circ }={30}^{\circ }$

At minimum deviation, the refractive index of the material of the prism can be found from the equation,

$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\Rightarrow \mu =\frac{\sin \left(\frac{{60}^{\circ }+{30}^{\circ }}{2}\right)}{\sin \left(\frac{{60}^{\circ }}{2}\right)}$

$\Rightarrow \mu =\frac{\sin {45}^{\circ }}{\sin {30}^{\circ }}$

$\therefore \mu =\sqrt{2}$

Why this Question? This question is based on the concept of application of the relation between refractive index of a prism , minimum angle of deviation and angle of prism.