Question

Show that at polarising angle, the reflected and refracted beams of light are at 90° to each other.

Answer 1

Let µ be refractive index of the transparent surface. If i_p is polarising angle and r is the angle of refraction as shown in Fig. then from Snell’ law, we have

µ = \(\frac{sin\ i_p}{\sin r} \) ...........(1)

From Brewster’ law, we have

µ = tan i_p = \(\frac{sin\ i_p}{cos\ i_p} \) .........(2)

From Eqs. (1) & (2), we have    

\(\frac{sin\ i_p}{\sin r} = \frac{sin\ i_p}{cos\ i_p} \)

or sin r = cos i_p    

or sin r = sin (90° - i_p)    

or r = 90° - i_p    

or i_p + r = 90°    

i.e. the reflected and refracted beams of light at polarising angle are perpendicular to each other.