Question
A ray from medium $1$ is refracted below while passing through medium $2.$ Find the refractive index of the second medium with respect to medium $1.$

Answer

Refractive index $\mu$
$=\frac{\sin i}{\sin r} $
$=\frac{\sin 30^{\circ}}{\sin 45^{\circ}}=\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}=\frac{1}{2} \times \frac{\sqrt{2}}{1}$
$=\frac{\sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{1}{1.414}=0.707$
Refractive index $=0.707$

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