91Ó°ÊÓ

The refractive index of the material 2 is ${\mathrm{n}}_2=1.5$

The refractive index of the first medium i.e. air is${\mathrm{n}}_1=1$

By using the formula of Brewster’s angle and as the figure say that the layers are parallel so that the refraction regarding the first is the same as the angle of incidence regarding the second surface.

Formula:

Brewster’s angle

${\mathrm{θ}}_{\mathrm{B}}=\tan \left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\right)$

Since the layers are parallel, the angle of refraction of the first surface is the same as the angle of incidence of the second surface. Thus,

${\mathrm{θ}}_2={\left({\mathrm{θ}}_1\right)}_c={90}^{∘}-\mathrm{θ}$

Brewster’s angle is given by

$$\begin{gathered} {\mathrm{θ}}_{\mathrm{B}}={\tan }^{-1}\left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\right) \\ {\mathrm{θ}}_1={\tan }^{-1}\left(\frac{1.5}{1}\right) \\ =56.{31}^{∘} \end{gathered}$$

Therefore,

${\mathrm{θ}}_2={90}^{∘}-56.{31}^o=33.{69}^{∘}$

We can apply Brewster’s law to both refractions. Therefore,

$$\begin{gathered} \left(\frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\right)\left(\frac{{\mathrm{n}}_3}{{\mathrm{n}}_2}\right)=\left({\mathrm{³Ù²¹²Ôθ}}_{1→2}\right)\left({\mathrm{³Ù²¹²Ôθ}}_{2→3}\right) \\ \frac{{\mathrm{n}}_3}{{\mathrm{n}}_1}=\left({\mathrm{³Ù²¹²Ôθ}}_1\right)\left({\mathrm{³Ù²¹²Ôθ}}_2\right) \\ \frac{{\mathrm{n}}_3}{1}=\left(\tan 56.{31}^{∘}\right)\left(\tan 33.{69}^{∘}\right) \\ {\mathrm{n}}_3=1 \end{gathered}$$