A ray of light travels from medium A into medium B and refracts at 60° to the normal. If the refractive index of medium B with respect to medium A is n_BA = √(2/3), what is the angle of incidence? (A) 75° (B) 60° (C) 30° (D) 45° (E) 20°

What the problem is asking

You are told the refracted angle in medium B and the relative refractive index $n_{BA}$. The goal is to use refraction (Snell’s law) to back out the incident angle in medium A.

Using the relative refractive index form of Snell’s law

For light going from medium A into medium B, $$n_{BA}=\frac{n_B}{n_A}=\frac{\sin i}{\sin r}.$$ So, $$\sin i = n_{BA}\,\sin r.$$

Substitute the given values

The refracted angle is $r=60^\circ$, so $\sin 60^\circ=\frac{\sqrt{3}}{2}$.

Now compute: $$\sin i = \sqrt{\frac{2}{3}}\cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{2}.$$

Convert back to the angle

$$\sin i = \frac{\sqrt{2}}{2} \Rightarrow i = 45^\circ.$$ So the correct option is D) $45^\circ$.