Question 61: Wave Optics

Question

When an unpolarized light falls at a particular angle on a glass plate (placed in air), it is observed that the reflected beam is linearly polarized. The angle of refracted beam with respect to the normal is $\_\_\_\_$ .
$\left(	an ^{-1}(1.52)=57.7^{\circ}ight.$, refractive indices of air and glass are 1.00 and 1.52, respectively.)

Accepted Answer

When unpolarized light strikes an interface at a specific angle called the Brewster angle ( $\mathrm{i}_{\mathrm{B}}$ or $\mathrm{i}_{\mathrm{p}}$ ), the reflected light is completely linearly polarized. At this specific angle, the reflected ray and the refracted ray are perpendicular $\left(90^{\circ}ight)$ to each other.
1. $\mathrm{i}_{\mathrm{p}}=$ Angle of incidence (polarizing angle)
2. $	heta_{\mathrm{r}}=$ Angle of reflection
3. $	heta_{\mathrm{r}^{\prime}}=$ Angle of refraction
Since the angle of reflection equals the angle of incidence :
$$ 	heta_{\mathrm{r}}=\mathrm{i}_{\mathrm{p}} $$
Since the angle between the reflected and refracted ray is $90^{\circ}$ :
$$ \mathrm{i}_{\mathrm{p}}+90^{\circ}+	heta_{\mathrm{r}^{\prime}}=180^{\circ} $$
$$\Rightarrow $$ $	heta_{\mathrm{r}^{\prime}}=90^{\circ}-\mathrm{i}_{\mathrm{p}}$
Applying Snell's Law
$$ \mu_1 \sin \left(\mathrm{i}_{\mathrm{p}}ight)=\mu_2 \sin \left(	heta_{\mathrm{r}^{\prime}}ight) $$
$$\Rightarrow $$ $\mu_1 \sin \left(\mathrm{i}_{\mathrm{p}}ight)=\mu_2 \sin \left(90^{\circ}-\mathrm{i}_{\mathrm{p}}ight)$
$$\Rightarrow $$ $\mu_1 \sin \left(\mathrm{i}_{\mathrm{p}}ight)=\mu_2 \cos \left(\mathrm{i}_{\mathrm{p}}ight)$
$$\Rightarrow $$ $	an \left(\mathrm{i}_{\mathrm{p}}ight)=\frac{\mu_2}{\mu_1}$
If the first medium is air ( $\mu_1 \approx 1$ ), the formula simplifies to :
$$ 	an \left(\mathrm{i}_{\mathrm{p}}ight)=\mu $$
Substituting the values:
$$ 	an \left(\mathrm{i}_{\mathrm{p}}ight)=\frac{1.52}{1.00}=1.52 $$
$$\Rightarrow $$ $\mathrm{i}_{\mathrm{p}}=	an ^{-1}(1.52)=57.7^{\circ}$
As established by the geometry of Brewster's condition :
$$ \mathrm{i}_{\mathrm{p}}+	heta_{\mathrm{r}^{\prime}}=90^{\circ} $$
$$\Rightarrow $$ $	heta_{\mathrm{r}^{\prime}}=90^{\circ}-\mathrm{i}_{\mathrm{p}}$
$$\Rightarrow $$ $	heta_{\mathrm{r}^{\prime}}=90^{\circ}-57.7^{\circ}$
$$ 	heta_{\mathrm{r}^{\prime}}=32.3^{\circ} $$
$$\Rightarrow $$ The angle of the refracted beam with respect to the normal is $32.3^{\circ}$. Hence, the correct option is (D).

Answer

$36.3^{\circ}$

Answer

$39.6^{\circ}$

Answer

$42.6^{\circ}$

Answer

$32.3^{\circ}$

Wave Optics

Choose the correct answer: