Question:
A beam of light is incident from air on the surface of a liquid. The angle of incidence is \(\theta\) and the angle of refraction is \(\alpha\). If the critical angle for the liquid when surrounded by air is \(\theta_c\), then \(\sin\theta_c\) is
A beam of light is incident from air on the surface of a liquid. The angle of incidence is \(\theta\) and the angle of refraction is \(\alpha\). If the critical angle for the liquid when surrounded by air is \(\theta_c\), then \(\sin\theta_c\) is
Show Hint
Remember the two important relations:
\[
n_1\sin i=n_2\sin r
\]
(Snell's law)
and
\[
\sin C=\frac{1}{n}
\]
(for a denser medium to air). Combining these relations often helps in critical angle problems.
Updated On: Jun 26, 2026
- \(\dfrac{\sin\alpha}{\sin\theta}\)
- \(\sin\alpha \times \sin\theta\)
- \(\dfrac{\sin\theta}{\sin\alpha}\)
- \(\dfrac{\sin\alpha}{\cos\theta}\)
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The Correct Option is A
Solution and Explanation
Step 1: Apply Snell's law for refraction from air to liquid.
Let the refractive index of air be \[ n_a \approx 1 \] and the refractive index of the liquid be \[ n_l. \] According to Snell's law, \[ n_a\sin\theta=n_l\sin\alpha \] Since \(n_a=1\), \[ \sin\theta=n_l\sin\alpha \] Therefore, \[ n_l=\frac{\sin\theta}{\sin\alpha} \]
Step 2: Use the definition of critical angle.
For a liquid-air interface, the critical angle \(\theta_c\) satisfies \[ \sin\theta_c=\frac{n_{\text{air}}}{n_{\text{liquid}}} \] Since \[ n_{\text{air}}=1, \] we get \[ \sin\theta_c=\frac{1}{n_l} \]
Step 3: Substitute the value of \(n_l\).
Using \[ n_l=\frac{\sin\theta}{\sin\alpha}, \] we obtain \[ \sin\theta_c = \frac{1}{\dfrac{\sin\theta}{\sin\alpha}} \] \[ \sin\theta_c = \frac{\sin\alpha}{\sin\theta} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\sin\theta_c=\frac{\sin\alpha}{\sin\theta}} \]
Let the refractive index of air be \[ n_a \approx 1 \] and the refractive index of the liquid be \[ n_l. \] According to Snell's law, \[ n_a\sin\theta=n_l\sin\alpha \] Since \(n_a=1\), \[ \sin\theta=n_l\sin\alpha \] Therefore, \[ n_l=\frac{\sin\theta}{\sin\alpha} \]
Step 2: Use the definition of critical angle.
For a liquid-air interface, the critical angle \(\theta_c\) satisfies \[ \sin\theta_c=\frac{n_{\text{air}}}{n_{\text{liquid}}} \] Since \[ n_{\text{air}}=1, \] we get \[ \sin\theta_c=\frac{1}{n_l} \]
Step 3: Substitute the value of \(n_l\).
Using \[ n_l=\frac{\sin\theta}{\sin\alpha}, \] we obtain \[ \sin\theta_c = \frac{1}{\dfrac{\sin\theta}{\sin\alpha}} \] \[ \sin\theta_c = \frac{\sin\alpha}{\sin\theta} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\sin\theta_c=\frac{\sin\alpha}{\sin\theta}} \]
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