MHT CET · Physics · Ray Optics
The Brewster's angle for the glass-air interface is \((54 \cdot 74)^{\circ} .\) If a ray of light passing
from air to glass strikes at an angle of incidence \(45^{\circ}\), then the angle of refraction is
\(\left[\tan (54 \cdot 74)^{\circ}=\sqrt{2}, \quad \sin 45^{\circ}=\frac{1}{\sqrt{2}}\right]\)
- A \(\sin ^{-1}(0 \cdot 5)\)
- B \(\sin ^{-1}(1)\)
- C \(\sin ^{-1}(\sqrt{2})\)
- D \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(A) \(\sin ^{-1}(0 \cdot 5)\)
Step-by-step Solution
Detailed explanation
(D)
\(\mathrm{n}=\tan 54.74=\sqrt{2}\)
\(\sqrt{2}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\sin 45^{\circ}}{\sin \mathrm{r}}\)
\(\therefore \sin \mathrm{i}=\frac{\sin 45^{\circ}}{\sqrt{2}}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2}=0.5\)
\(\therefore \mathrm{r}=\sin ^{-1}(0.5)\)
\(\mathrm{n}=\tan 54.74=\sqrt{2}\)
\(\sqrt{2}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}}=\frac{\sin 45^{\circ}}{\sin \mathrm{r}}\)
\(\therefore \sin \mathrm{i}=\frac{\sin 45^{\circ}}{\sqrt{2}}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2}=0.5\)
\(\therefore \mathrm{r}=\sin ^{-1}(0.5)\)
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