MHT CET · Physics · Ray Optics
The angle of incidence is found to be twice the angle of refraction when ray of light passes from vacuum into a medium of refractive index ' \(\mu\) '. The angle of incidence will be
- A \(\cos ^{-1}\left(\frac{\mu}{2}\right)\)
- B \(2 \cos ^{-1}\left(\frac{\mu}{2}\right)\)
- C \(\sin ^{-1}\left(\frac{\mu}{2}\right)\)
- D \(2 \sin ^{-1}\left(\frac{\mu}{2}\right)\)
Answer & Solution
Correct Answer
(B) \(2 \cos ^{-1}\left(\frac{\mu}{2}\right)\)
Step-by-step Solution
Detailed explanation
Snell's law, \(\mu=\frac{\sin i}{\sin r}\)
Given: \(\mathrm{i}=2 \mathrm{r}\)
\(\begin{aligned}
\therefore \quad r & =\frac{i}{2} \\
& \mu=\frac{\sin i}{\sin \left(\frac{i}{2}\right)} \\
& \mu=\frac{2 \sin \left(\frac{i}{2}\right) \cos \left(\frac{i}{2}\right)}{\sin \left(\frac{i}{2}\right)} \\
\therefore \quad & \mu=2 \cos \left(\frac{i}{2}\right)
\end{aligned}\)
\(\Rightarrow i=2 \cos ^{-1}\left(\frac{\mu}{2}\right)\)
Given: \(\mathrm{i}=2 \mathrm{r}\)
\(\begin{aligned}
\therefore \quad r & =\frac{i}{2} \\
& \mu=\frac{\sin i}{\sin \left(\frac{i}{2}\right)} \\
& \mu=\frac{2 \sin \left(\frac{i}{2}\right) \cos \left(\frac{i}{2}\right)}{\sin \left(\frac{i}{2}\right)} \\
\therefore \quad & \mu=2 \cos \left(\frac{i}{2}\right)
\end{aligned}\)
\(\Rightarrow i=2 \cos ^{-1}\left(\frac{\mu}{2}\right)\)
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