Question 96 of 295

Light enters from air into a given medium at an angle of $45^{\circ}$ with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of $15^{\circ}$ from its original direction. The refractive index of the medium is :

A$1.732$B$1.333$C$1.414$D$2.732$

Show answer and solution▾

Answer: (C) $1.414$

Given here, $i=45^{\circ}$
Deviation in light ray, $D=i-r$, here, $r$ is refracted ray.
$\Rightarrow 15^{\circ}=45-r\Rightarrow r=30^{\circ}$
Using Snell's law,
${n}_{1}\sin i={n}_{2}\sin r$
$\Rightarrow 1\sin 45^{\circ}=\mu \sin 30^{\circ}$
$\Rightarrow \dfrac{1}{\sqrt{2}}=\mu \dfrac{1}{2}$
Thus, the refractive index of the medium is $\mu =\sqrt{2}=1.414$