Question

A ray of light travels from air into a perspex block of refractive index 1.5 at an angle of incidence of ${45}^o$. Calculate the angle of the refraction of the ray at air-to-perspex boundary and at perspex-to-air boundary.

Answer 1

Given:

Refractive index of block ${\mu }_{block}=1.5$

Angle of incidence $i={45}^o$

For refraction from air to block:

We apply Snell's law,

${\mu }_{air}sini={\mu }_{block}sinr$

$1\times sin{45}^o=1.5\times sinr$

$sinr=\frac{sin{45}^o}{1.5}$ . . . (i)

On simplifying,

$sinr=\frac{\left(\frac{1}{\sqrt{2}}\right)}{1.5}$

$sinr=\frac{\sqrt{2}}{3}$

Answer : $r={28.12}^o$ is the angle of the refraction of the ray at air-to-perspex boundary.

Now when the ray will move from block to air, then $r$ is incidence angle.

Applying Snell's law again,

${\mu }_{block}sinr={\mu }_{air}sin{r}_1$

Using equation (i) to substitute the value of $sinr$,

$1\times sin{r}_1={\mu }_{block}\times sinr$

$sin{r}_1=1.5\times \frac{sin{45}^o}{1.5}\Rightarrow sin{r}_1=sin{45}^o$

Answer : $r_1={45}^o$ is the angle of the refraction of the ray at perspex-to-air boundary.