A ray of light when falls on a liquid A-air interface at angle `45^(@)`, it just gets total reflection. When a ray of light falls on liquid B -air interface at angle `30^(@)`, it does not emerge out. We can infer that

Options

A. `mu_(A)gtmu_(B)`
B. `mu_(A)ltmu_(B)`
C. `mu_(A)=mu_(B)`
D. can't be inferred

Answer

To solve the problem, we need to analyze the conditions for total internal reflection and the critical angles for the two liquids involved (liquid A and liquid B).

### Step-by-Step Solution:

1. **Understanding Total Internal Reflection (TIR)**:
- Total internal reflection occurs when light travels from a denser medium to a rarer medium and the angle of incidence exceeds the critical angle (θc).
- The critical angle can be calculated using the formula:
\[ \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) \]
where \( n_1 \) is the refractive index of the denser medium and \( n_2 \) is the refractive index of the rarer medium (in this case, air).

2. **Analyzing Liquid A**:
- Given that light falls on the liquid A-air interface at an angle of \( 45^\circ \) and experiences total internal reflection, we can conclude that:
\[ \theta_c^A = 45^\circ \]
- Using the critical angle formula:
\[ \sin(45^\circ) = \frac{n_{air}}{n_A} \implies n_A = \frac{n_{air}}{\sin(45^\circ)} = \frac{n_{air}}{\frac{1}{\sqrt{2}}} = n_{air} \cdot \sqrt{2} \]

3. **Analyzing Liquid B**:
- For liquid B, light falls on the liquid B-air interface at an angle of \( 30^\circ \) and does not emerge, which means that the critical angle for liquid B must be equal to or less than \( 30^\circ \):
\[ \theta_c^B \leq 30^\circ \]
- Using the critical angle formula again:
\[ \sin(30^\circ) = \frac{n_{air}}{n_B} \implies n_B = \frac{n_{air}}{\sin(30^\circ)} = \frac{n_{air}}{\frac{1}{2}} = 2 \cdot n_{air} \]

4. **Comparing Refractive Indices**:
- From the calculations:
- \( n_A = n_{air} \cdot \sqrt{2} \)
- \( n_B = 2 \cdot n_{air} \)
- To compare \( n_A \) and \( n_B \):
\[ n_A < n_B \implies n_{air} \cdot \sqrt{2} < 2 \cdot n_{air} \]
- Dividing both sides by \( n_{air} \) (assuming \( n_{air} > 0 \)):
\[ \sqrt{2} < 2 \]
- This inequality holds true.

5. **Conclusion**:
- From the above analysis, we conclude that the refractive index of liquid A is less than that of liquid B:
\[ n_A < n_B \]

### Final Answer:
We can infer that the refractive index of liquid A is less than that of liquid B.