A ray of light of refractive index `1.4`approaches the boundary surface between the liquids and air at an angle of incidence whose sine is `0.8`.Which of the following statements is correct about the behaviour of the light

Options

A. It is impossible to predict the behaviour of the light ray on the basis of the information supplied.
B. The sine of the angle of refraction of the emergent ray will be less than 0.8
C. The ray will be internally reflected
D. The sine of the angle of refraction of the emergent ray will be greater than 0.8

Explanation

To solve the problem, we will analyze the behavior of the light ray as it transitions from a liquid with a refractive index of 1.4 to air, which has a refractive index of 1.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Refractive index of the liquid (n1) = 1.4
- Refractive index of air (n2) = 1.0
- Sine of angle of incidence (sin i) = 0.8

2. **Calculate the Angle of Incidence**:
- We can find the angle of incidence (i) using the sine value:
\[ \sin i = 0.8 \implies i = \arcsin(0.8) \]
- However, we don't need the exact angle for this problem; we will use the sine value directly in our calculations.

3. **Apply Snell's Law**:
- Snell's Law states:
\[ n_1 \sin i = n_2 \sin r \]
- Plugging in the values:
\[ 1.4 \cdot 0.8 = 1.0 \cdot \sin r \]
- This simplifies to:
\[ 1.12 = \sin r \]

4. **Analyze the Result**:
- The value of \(\sin r\) cannot exceed 1, as the sine of an angle must lie within the range of -1 to 1. Since we have \(\sin r = 1.12\), this indicates that the angle of refraction (r) cannot exist.

5. **Conclusion**:
- Since \(\sin r > 1\), total internal reflection occurs. Therefore, the light ray will not pass into the air but will be reflected back into the liquid.

### Final Answer:
The correct statement about the behavior of the light is that the ray will be internally reflected.

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