Question

A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is $\sqrt{3}$, then the angle of incidence is

• A. 45°
• B. 90°
• C. 60°
• D. 30°

Answer 1

The correct option is C 60°

Given, refractive index of material of prism $n=\sqrt{3}$, prism angle A = 60°
Method 1
Using prism formula, $n=\frac{sin\frac{A+\delta }{2}}{sin\left(\frac{A}{2}\right)}$
$\Rightarrow \sqrt{3}=\frac{sin\left(\frac{{60}^{\circ }+\delta }{2}\right)}{sin{30}^{\circ }}$
$\Rightarrow sin\left(\frac{{60}^{\circ }+\delta }{2}\right)=sin{60}^{\circ }or\frac{{60}^{\circ }+\delta }{2}={60}^{\circ }$
or angle of minimum deviation $\delta ={60}^{\circ }$
$\Rightarrow i={60}^{\circ }$

Method 2
For minimum deviation, ray should pass symmetrically (i.e. parallel to the base of the equilateral prism)
$\Rightarrow$ From geometry of given figure, we have,

Using Snell's law, $n=\frac{sini}{sinr}$
$sini=nsinr=\sqrt{3}sin{30}^{\circ }$
$\Rightarrow sini=\frac{\sqrt{3}}{2}ori={60}^{\circ }$

Why this question? Concept: The fact that refracted ray is parallel to the base makes the problems related to minimum deviation in a prism short and easy as you can see in method 2.