Unpolarized light of intensity $$I_0$$ is incident on surface of a block of glass at Brewster's angle. In that case, which one of the following statements is true?

When unpolarized light is incident on any surface, we always begin by remembering that it can be thought of as an equal mixture of two mutually perpendicular linear polarisations. Hence, for an incident intensity $$I_0$$, we may write

$$I_{\parallel}^{\;(inc)}=\dfrac{I_0}{2},\qquad I_{\perp}^{\;(inc)}=\dfrac{I_0}{2}$$

where the $$\parallel$$ component has its electric field vector in the plane of incidence and the $$\perp$$ component has its electric field vector perpendicular to that plane.

Now, we recall Brewster’s law. It states that for a ray travelling from a medium of refractive index $$n_1$$ to one of refractive index $$n_2$$, there exists an angle of incidence $$\theta_B$$, called Brewster’s angle, for which the reflected component of the $$\parallel$$ polarisation vanishes. In symbols, Fresnel’s reflection coefficient for the parallel component is

$$r_{\parallel}=0\quad\text{when}\quad\theta_i=\theta_B.$$

So, at Brewster’s angle,

$$I_{\parallel}^{\;(refl)} = 0.$$

However, for the perpendicular component, the corresponding reflection coefficient $$r_{\perp}$$ is not zero. Its magnitude is less than one, so a certain fraction of the $$\perp$$ component is reflected while the rest is transmitted. Hence

$$I_{\perp}^{\;(refl)} = R_{\perp}\left(\dfrac{I_0}{2}\right),\qquad 0 < R_{\perp} < 1,$$

where $$R_{\perp}=|r_{\perp}|^{\,2}$$ is the reflectance for the perpendicular polarisation. Combining the two reflected intensities, we have

$$I_{\text{reflected}} = I_{\parallel}^{\;(refl)} + I_{\perp}^{\;(refl)} = 0 + R_{\perp}\left(\dfrac{I_0}{2}\right) = \dfrac{R_{\perp}I_0}{2}.$$

Because $$0<R_{\perp}<1,$$ the numerical value of $$\dfrac{R_{\perp}I_0}{2}$$ is strictly less than $$\dfrac{I_0}{2}.$$ Very importantly, the entire reflected beam now contains only the $$\perp$$ polarisation, i.e. only one linear polarisation. Therefore the reflected light is completely plane-polarised.

Putting these two conclusions together:

• The reflected beam is completely (100%) polarised.
• Its intensity is a fraction $$R_{\perp}/2$$ of $$I_0$$, which is always less than $$I_0/2$$.

So the statement that matches these facts is: “reflected light is completely polarized with intensity less than $$\dfrac{I_0}{2}$$.” This is exactly Option D.

Hence, the correct answer is Option D.