MHT CET · Physics · Ray Optics
A prism having refractive index \(\sqrt{2}\) and refracting angle \(30^{\circ}\) has one of the
refracting surfaces silvered. The beam of light incident on the other refracting
surface will retrace its path, if angle of incidence is \(\left[\sin \frac{\pi}{6}=0 \cdot 5\right]\)
- A \(\sin ^{-1}\left(\frac{3}{4}\right)\)
- B \(\sin ^{-1}\left(\frac{1}{2}\right)\)
- C \(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
- D \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Answer & Solution
Correct Answer
(D) \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Step-by-step Solution
Detailed explanation
(D)
\(A=r_{1}+r_{2}\) but \(r_{2}=0\) since is incident normally on the second surface.
\(\therefore \mathrm{r}_{1}=\mathrm{A}=30^{\circ}\)
\(n=\frac{\sin i}{\sin r}\)
\(\therefore \sqrt{2}=\frac{\sin \mathrm{i}}{\sin 30^{\circ}}\)
\(\therefore \sin \mathrm{i}=\sqrt{2} \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}\)
\(\therefore \mathrm{i}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
\(A=r_{1}+r_{2}\) but \(r_{2}=0\) since is incident normally on the second surface.
\(\therefore \mathrm{r}_{1}=\mathrm{A}=30^{\circ}\)
\(n=\frac{\sin i}{\sin r}\)
\(\therefore \sqrt{2}=\frac{\sin \mathrm{i}}{\sin 30^{\circ}}\)
\(\therefore \sin \mathrm{i}=\sqrt{2} \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}\)
\(\therefore \mathrm{i}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
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