Question

A beam of unpolarized light is incident on an air-glass interface derive, using a suitable ray diagram that light reflected from the interface is completely polarised, when $\tan i_p=?$

Answer 1

Step 1: Given data

When unpolarized light is incident on a transparent refracting medium, at a particular angle of incidence the reflected light and refracted light are perpendicular to each other that angle of incident is known as a polarizing angle.

The reflected light is completely plane polarised.

Let $i_p$ be the polarising angle.

Let $r$ be the angle of refraction.

Let the refractive index of air be ${\mu }_1=1$.

Let the refractive index of glass be ${\mu }_2=\mu$.

Therefore, the reflected ray and refracted ray are perpendicular to each other.

Step 2: Formula Used

According to Brewster’s law,

$i_p+r=90°$

$r=90°–i_p$

According to Snell’s law

${\mu }_1\sin i={\mu }_2\sin r$

$\therefore \mu =\frac{\sin i}{\sin r}$

Step 3: Brewster's Angle

Upon substituting the values we get,

$\mu =\frac{\sin i_p}{\sin \left(90°-i_p\right)}$

$\Rightarrow \mu =\frac{\sin i_p}{\cos i_p}$

$\Rightarrow \mu =\tan i_p$

This relation is known as the Brewster angle.