Question:
When a ray of light is refracted from one medium to another, then the wavelength changes from \(6000\text{ \AA}\) to \(4000\text{ \AA}\). The critical angle for the interface will be
When a ray of light is refracted from one medium to another, then the wavelength changes from \(6000\text{ \AA}\) to \(4000\text{ \AA}\). The critical angle for the interface will be
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Key:
$\lambda \propto \frac{1}{n}$
Updated On: May 8, 2026
- \(\cos^{-1} \left( \frac{3}{2} \right)\)
- \(\sin^{-1} \left( \frac{2}{\sqrt{3}} \right)\)
- \(\sin^{-1} \left( \frac{2}{3} \right)\)
- \(\cos^{-1} \left( \frac{2}{\sqrt{3}} \right)\)
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The Correct Option is C
Solution and Explanation
Concept: Refractive index relation: \[ \frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2} \]
Step 1: Substitute values. \[ \frac{n_2}{n_1} = \frac{6000}{4000} = \frac{3}{2} \]
Step 2: Critical angle formula. \[ \sin C = \frac{n_1}{n_2} \] \[ \sin C = \frac{2}{3} \]
Step 3: Conclusion.
\[ C = \sin^{-1}\left(\frac{2}{3}\right) \] Final Answer: Option (C)
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