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

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
  1. $\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right)$
  2. $\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
  3. $\sin ^{-1}\left(\sqrt{\frac{3}{2}}\right)$
  4. $\cos ^{-1}\left(\sqrt{\frac{3}{2}}\right)$

Solution

According to Brewster's law, $\begin{array}{ll} & \tan \theta_{\mathrm{B}}=\mathrm{n} \\ \therefore \quad & \tan 60^{\circ}=\mathrm{n} \end{array}$ $\begin{array}{ll}\therefore \quad & n=\sqrt{3} \\ & \text { Now, } \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mathrm{n} \\ \therefore \quad & \sin \mathrm{r}=\frac{\sin \mathrm{i}}{\mathrm{n}} \\ \therefore \quad & \sin \mathrm{r}=\frac{\sin 45^{\circ}}{\sqrt{3}} \\ \therefore \quad & \sin \mathrm{r}=\frac{1}{\sqrt{6}} \\ \therefore & \mathrm{r}=\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right) . \ldots\left(\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right)\end{array}$

Asked in: MHT CET 2023 (12 May Shift 1)

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