Question:
A light ray is incident from a medium of refractive index \(2\) into a medium of refractive index \(\sqrt{3}\). The critical angle is
A light ray is incident from a medium of refractive index \(2\) into a medium of refractive index \(\sqrt{3}\). The critical angle is
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Critical angle exists only when light travels from denser medium to rarer medium, and it is given by \(\sin C=\frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}}\).
Updated On: Jun 26, 2026
- \(30^\circ\)
- \(45^\circ\)
- \(60^\circ\)
- \(90^\circ\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the formula for critical angle.
When light travels from a denser medium to a rarer medium, the critical angle \(C\) is given by \[ \sin C=\frac{\mu_2}{\mu_1} \] Here, \[ \mu_1=2 \] and \[ \mu_2=\sqrt{3} \]
Step 2: Substitute the values.
\[ \sin C=\frac{\sqrt{3}}{2} \]
Step 3: Find the angle.
We know that \[ \sin 60^\circ=\frac{\sqrt{3}}{2} \] Therefore, \[ C=60^\circ \]
Step 4: Final conclusion.
Hence, the critical angle is \[ \boxed{60^\circ} \]
When light travels from a denser medium to a rarer medium, the critical angle \(C\) is given by \[ \sin C=\frac{\mu_2}{\mu_1} \] Here, \[ \mu_1=2 \] and \[ \mu_2=\sqrt{3} \]
Step 2: Substitute the values.
\[ \sin C=\frac{\sqrt{3}}{2} \]
Step 3: Find the angle.
We know that \[ \sin 60^\circ=\frac{\sqrt{3}}{2} \] Therefore, \[ C=60^\circ \]
Step 4: Final conclusion.
Hence, the critical angle is \[ \boxed{60^\circ} \]
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