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
- A \(\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right)\)
- B \(\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
- C \(\sin ^{-1}\left(\sqrt{\frac{3}{2}}\right)\)
- D \(\cos ^{-1}\left(\sqrt{\frac{3}{2}}\right)\)
Answer & Solution
Correct Answer
(A) \(\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right)\)
Step-by-step Solution
Detailed explanation
According to Brewster's law,
\(\tan \theta_{\mathrm{B}}=\mathrm{n} \)
\( \therefore \tan 60^{\circ}=\mathrm{n}\)
\(\therefore n=\sqrt{3} \)
\( \text { Now, } \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mathrm{n} \)
\( \therefore \sin \mathrm{r}=\frac{\sin \mathrm{i}}{\mathrm{n}} \)
\( \therefore \sin \mathrm{r}=\frac{\sin 45^{\circ}}{\sqrt{3}} \)
\( \therefore \sin \mathrm{r}=\frac{1}{\sqrt{6}} \)
\( \therefore \mathrm{r}=\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right) . \ldots\left(\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right)\)
\(\tan \theta_{\mathrm{B}}=\mathrm{n} \)
\( \therefore \tan 60^{\circ}=\mathrm{n}\)
\(\therefore n=\sqrt{3} \)
\( \text { Now, } \frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\mathrm{n} \)
\( \therefore \sin \mathrm{r}=\frac{\sin \mathrm{i}}{\mathrm{n}} \)
\( \therefore \sin \mathrm{r}=\frac{\sin 45^{\circ}}{\sqrt{3}} \)
\( \therefore \sin \mathrm{r}=\frac{1}{\sqrt{6}} \)
\( \therefore \mathrm{r}=\sin ^{-1}\left(\frac{1}{\sqrt{6}}\right) . \ldots\left(\because \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right)\)
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