JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media \(A\) and \(B\) are \(2.4 \times 10^8 m / s\) and \(2.7 \times 10^8 m / s\) respectively, then the value of critical angle is :
- A \( \cot^{-1}(\frac{3}{\sqrt{13}}) \)
- B \( \sin^{-1}(\frac{9}{8}) \)
- C \( \tan^{-1}(\frac{8}{\sqrt{17}}) \)
- D \( \cos^{-1}(\frac{8}{9}) \)
Answer & Solution
Correct Answer
(C) \( \tan^{-1}(\frac{8}{\sqrt{17}}) \)
Step-by-step Solution
Detailed explanation
\(\mu_A \sin c=\mu_B \sin 90\) \(\Rightarrow \sin c =\frac{\mu_{ B }}{\mu_{ A }}=\frac{ v _{ A }}{ v _{ B }}\) \(\therefore \sin c=\frac{2.4 \times 10^8}{2.7 \times 10^8}=\frac{8}{9}\) \(\Rightarrow \tan c=\frac{8}{\sqrt{81-64}}=\frac{8}{\sqrt{17}}\)…
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