Question 333: Waves Optics

Question

A beam of light is incident on a glass plate at an angle of $$60^{\circ}$$. The reflected ray is polarized. If angle of incidence is $$45^{\circ}$$ then angle of refraction is

Accepted Answer

According to Brewster's law,
$$\begin{aligned}
& 	an 	heta_{\mathrm{B}}=\mathrm{n} \
	herefore \quad & 	an 60^{\circ}=\mathrm{n}
\end{aligned}$$
$$\begin{aligned}
&\begin{array}{ll}
	herefore & \mathrm{n}=\sqrt{3} \
& 	ext { Now, } \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mathrm{n} \
	herefore & \sin \mathrm{r}=\frac{\sin \mathrm{i}}{\mathrm{n}} \
	herefore & \sin \mathrm{r}=\frac{\sin 45^{\circ}}{\sqrt{3}} \
	herefore & \sin \mathrm{r}=\frac{1}{\sqrt{6}} \quad \ldots .\left(\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}ight) \
	herefore & \quad \mathrm{r}=\sin ^{-1}\left(\frac{1}{\sqrt{6}}ight)
\end{array}\
\end{aligned}$$

Answer

$$\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right)$$

Answer

$$\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$$

Answer

$$\sin ^{-1}\left(\sqrt{\frac{3}{2}}\right)$$

Answer

$$\cos ^{-1}\left(\sqrt{\frac{3}{2}}\right)$$

Waves Optics

A beam of light is incident on a glass plate at an angle of $60^{\circ}. The reflected ray is polarized. If angle of incidence is 45^{\circ}$ then angle of refraction is

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