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 $r_{1}=r_{2}=r$
$\Longrightarrow A =2 r$
$\therefore r =\frac{ A }{2}=\frac{60^{\circ}}{2}=30^{\circ}$
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: $\sin i =\frac{1}{\sqrt{2}} \Longrightarrow i =45^{\circ}$