A point source of light is placed at the bottom of a water lake. If the area of the illuminated circle on the surface is equal to 3 times the square of the depth of the lake. The refractive index of water is

Options

A. `sqrt(pi+1)`
B. `sqrt((pi)/3+1)`
C. `(pi)/3+1`
D. `(pi)/4+1`

Answer

To solve the problem, we need to find the refractive index of water given that the area of the illuminated circle on the surface is equal to three times the square of the depth of the lake.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a point source of light at the bottom of a lake of depth \( h \).
- The area of the illuminated circle on the surface is given by \( A = 3h^2 \).

2. **Relate Area to Radius**:
- The area \( A \) of a circle is given by the formula:
\[ A = \pi r^2 \]
- Setting the two expressions for area equal gives:
\[ \pi r^2 = 3h^2 \]
- From this, we can solve for \( r^2 \):
\[ r^2 = \frac{3h^2}{\pi} \]

3. **Use the Geometry of the Situation**:
- The light rays from the point source that just graze the surface form a right triangle with the depth \( h \) and the radius \( r \) of the illuminated circle.
- By the Pythagorean theorem:
\[ r^2 + h^2 = d^2 \]
- Here, \( d \) is the distance from the source to the point where the light just grazes the surface.

4. **Express the Critical Angle**:
- The critical angle \( \theta_c \) is defined as the angle of incidence at which light is refracted at 90 degrees. The refractive index \( \mu \) is related to the critical angle by:
\[ \mu = \frac{1}{\sin \theta_c} \]

5. **Determine \( \sin \theta_c \)**:
- From the right triangle, we can express \( \sin \theta_c \) as:
\[ \sin \theta_c = \frac{r}{d} \]
- We need to express \( d \) in terms of \( h \) and \( r \):
\[ d = \sqrt{r^2 + h^2} \]

6. **Substituting Values**:
- Substitute \( r^2 = \frac{3h^2}{\pi} \) into the equation for \( d \):
\[ d = \sqrt{\frac{3h^2}{\pi} + h^2} = \sqrt{h^2 \left(\frac{3}{\pi} + 1\right)} = h \sqrt{\frac{3 + \pi}{\pi}} \]

7. **Final Expression for \( \sin \theta_c \)**:
- Now substituting back into the expression for \( \sin \theta_c \):
\[ \sin \theta_c = \frac{r}{d} = \frac{r}{h \sqrt{\frac{3 + \pi}{\pi}}} \]
- We can find \( r \) from \( r^2 = \frac{3h^2}{\pi} \):
\[ r = \sqrt{\frac{3h^2}{\pi}} = h \sqrt{\frac{3}{\pi}} \]
- Thus,
\[ \sin \theta_c = \frac{h \sqrt{\frac{3}{\pi}}}{h \sqrt{\frac{3 + \pi}{\pi}}} = \frac{\sqrt{3}}{\sqrt{3 + \pi}} \]

8. **Calculate the Refractive Index**:
- Now substituting into the refractive index formula:
\[ \mu = \frac{1}{\sin \theta_c} = \frac{\sqrt{3 + \pi}}{\sqrt{3}} \]

### Final Answer:
The refractive index of water is:
\[ \mu = \sqrt{1 + \frac{3}{\pi}} \]