Question

The figure below shows a ray of light as it travels from media A to media B.The refractive index of media B to media A is

Answer 1

Given data:

* The angle of incidence in the medium A is,

$i=60^{\circ}$

* The angle of refraction in medium B is,

$r=45^{\circ}$

Solution:

According to the Snell's Law,

$\begin{gathered} n_A\sin (i)=n_B\sin (r) \\ \frac{n_B}{n_A}=\frac{\sin (i)}{\sin (r)} \end{gathered}$

where n_B is the refractive index of medium B and n_A is the refractive index of medium B,

Substituting the known values,

$\begin{gathered} \frac{n_B}{n_A}=\frac{\sin (60^{\circ_{}})}{\sin (45^{\circ})} \\ \frac{n_B}{n_A}=\frac{0.866}{0.707} \\ \frac{n_B}{n_A}=1.225 \end{gathered}$

Thus, the ratio of refractive index of medium B to the refractive index of medium A is 1.225.