Question

The figure shows a ray of light as it travels from medium A to medium B. The refractive index of the medium B relative to medium A will be:

Answer 1

According to snell's law:
$n_{a}sini=n_{b}sinr$
$n_{a}sin{60}^{\circ }=n_{b}sin{45}^{\circ }$
$\frac{n_b}{n_a}=\frac{sin{60}^{\circ }}{sin{45}^{\circ }}$
$\frac{n_b}{n_a}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\frac{\sqrt{3}}{\sqrt{2}}$
$n_{b,a}=\frac{\sqrt{3}}{\sqrt{2}}$