ExamBro
ExamBro
MHT CET · Physics · Wave 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

  1. A \(\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right)\)
  2. B \(\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
  3. C \(\sin ^{-1}\left(\sqrt{\frac{3}{2}}\right)\)
  4. D \(\cos ^{-1}\left(\sqrt{\frac{3}{2}}\right)\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

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