Question

A ray of light suffers a minimum deviation when incident on an equilateral prism of refractive index $\sqrt{2}$ . The angle of incidence is

• A. $30^{\circ }$
• B. $45^{\circ }$
• C. $60^{\circ }$
• D. $50^{\circ }$

Answer 1

Answer: (B)

Given : Refractive index of air, $n_1=1$ Refractive index of prism, $n_2=\sqrt{2}$
Angle of prism, $A=r_1+r_2$
As ray of light suffers minimum deviation, thus $r_1=r_2=r$
$⟹A=2r$
$\therefore r=\frac{A}{2}=\frac{60^{\circ }}{2}=30^{\circ }$
where $A=60^{\circ }$ (equilateral prism)
Using Snell's law, $n_1\sin i=n_2\sin r$
$\therefore 1\times \sin i=\sqrt{2}\times 0.5$
We get: $\sin i=\frac{1}{\sqrt{2}}⟹i=45^{\circ }$