"The tangent of a polarizing angle is equal to the (relative) refractive index of the transparent medium on which light is incident." If θp is the polarizing angle and μ is the refractive index of the medium, then \(\mu=\tan\left(\theta_p\right)\)
Proof:
Suppose a beam of unpolarized light AB incident with polarizing angle θp one the surface of a transparent surface XY of a transparent medium of refractive index μ. The beam AB is reflected along with BC (plane-polarized) and refracted along BD (unpolarized or partially plane polarized).
When the reflected beam is completely plane-polarized, the corresponding incident angle θp is called the polarizing angle. Let r be the angle of reflection.
From Snell's law,
\(\mu=\frac{\sin\left(i\right)}{\sin\left(r\right)}\)
But, for polariizing angle, i = θp , the reflected ray are perpendicular to each other.
∴ \(\mu=\frac{\sin\left(\theta_p\right)}{\sin\left(r\right)}\) ------- (i)
Since, rays BC and BD are perpendicular to each other,
∴\(\theta_p+r=90^\circ\)
⇒ \(r=90^\circ-\theta_p\)
With this value, equation (i) becomes
\(\mu=\frac{\sin(\theta_p)}{\sin(90-\theta_p)}=\frac{\sin(\theta_p)}{\cos(\theta_p)}\)
∴ \(\mu=\tan(\theta_p)\)
i.e., refractive indxe of a transparent medium is equal to the tangent of angle of polarization for the medium.