A point source of light is kept at a depth of h in water of refractive index 4/3. The radius of the circle at the surface of water through which light emits is

Options

A. `(3)/(sqrt(7))`h
B. `(sqrt(7))/(3)`h
C. `(sqrt(3))/(7)`h
D. `(7)/(sqrt(3))`h

Explanation

To solve the problem of finding the radius of the circle at the surface of water through which light emits from a point source of light kept at a depth \( h \) in water with a refractive index \( \mu = \frac{4}{3} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- A point source of light is placed at a depth \( h \) in water.
- The light will emit in all directions, but due to refraction, only light that hits the water surface at certain angles will emerge into the air.

2. **Using Snell's Law**:
- According to Snell's Law, when light travels from a medium with a higher refractive index (water) to a lower refractive index (air), the relationship is given by:
\[ \sin \theta_c = \frac{1}{\mu} \]
- Here, \( \theta_c \) is the critical angle, and \( \mu = \frac{4}{3} \) is the refractive index of water.

3. **Calculating the Critical Angle**:
- Substitute the value of \( \mu \):
\[ \sin \theta_c = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]
- Therefore, the critical angle \( \theta_c \) can be calculated using the inverse sine function:
\[ \theta_c = \sin^{-1}\left(\frac{3}{4}\right) \]

4. **Finding the Radius of the Circle**:
- The radius \( r \) of the circle at the water's surface can be found using the geometry of the situation. The radius can be related to the depth \( h \) and the critical angle:
\[ r = h \tan \theta_c \]
- Using the relationship \( \tan \theta_c = \frac{\sin \theta_c}{\cos \theta_c} \):
\[ \tan \theta_c = \frac{\frac{3}{4}}{\sqrt{1 - \left(\frac{3}{4}\right)^2}} = \frac{\frac{3}{4}}{\sqrt{\frac{7}{16}}} = \frac{3}{4} \cdot \frac{4}{\sqrt{7}} = \frac{3}{\sqrt{7}} \]
- Therefore, substituting back into the radius equation:
\[ r = h \cdot \frac{3}{\sqrt{7}} = \frac{3h}{\sqrt{7}} \]

5. **Final Answer**:
- The radius of the circle at the surface of water through which light emits is:
\[ r = \frac{3h}{\sqrt{7}} \]