Question 8.19: On the front surface of a dielectric slab as shown in Fig. 8......
On the front surface of a dielectric slab as shown in Fig. 8.19, a uniform plane wave with parallel polarization is incident at Brewster angle \theta _{B}. Show that the wave also experiences no reflection at the rear surface which is parallel to the front one.
Substituting \theta _{B} = 90^{\circ } – \theta _{t}, which is obtained from Eq. (8-154), into Eq. (8-155), we write
\boxed{\Gamma _{\parallel } = \frac{ \tan (\theta _{t} – \theta _{i})}{\tan (\theta _{t} + \theta _{i})}} (8-154)
\frac{n_{2}}{n_{1}} = \frac{ \sin \theta _{B}}{\cos \theta _{B}} = \frac{\cos \theta _{t}}{\sin \theta _{t}} (8-159)
Comparing Eq. (8-159) with Eq. (8-156), we obtain \theta _{t} = \theta^{\prime } _{B}, which implies that the internal wave is incident on the rear surface at Brewster angle. Thus, there is no reflection at the rear surface.
\tan \theta^{\prime }_{B} = \frac{n_{1}}{ n_{2}} (8-156)