If we see the outside objects by keeping an eye at a depth h inside water of refractive index u, then the value of RADIUS of circle of view for outside objects will be

Options

A. `h/sqrt(u^(2)-1)`
B. `h/sqrt(u^(2)+1)`
C. `(2h)/sqrt(u^(2)-1)`
D. `h/mu^(2)`

Answer

To find the radius of the circle of view for outside objects when observing from a depth \( h \) inside water with a refractive index \( \mu \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a water surface, and an observer is at a depth \( h \) inside the water.
- The refractive index of water is \( \mu \).
- Light rays from outside the water will enter the water and bend due to refraction.

2. **Identifying the Critical Angle**:
- When light travels from a denser medium (water) to a rarer medium (air), it bends away from the normal.
- The maximum angle of refraction occurs when the angle of refraction \( r \) is \( 90^\circ \). This is the critical angle for total internal reflection.

3. **Applying Snell's Law**:
- According to Snell's Law:
\[ \frac{\sin i}{\sin r} = \frac{\mu_2}{\mu_1} \]
- Here, \( \mu_1 = \mu \) (water) and \( \mu_2 = 1 \) (air). When \( r = 90^\circ \), \( \sin r = 1 \).
- Thus, we have:
\[ \sin i = \mu \cdot \sin r = \mu \cdot 1 = \mu \]
- Therefore, \( \sin i = \frac{1}{\mu} \).

4. **Using Geometry**:
- Consider a right triangle formed by the depth \( h \), the radius \( R \) of the circle of view, and the hypotenuse which connects the point of observation to the point where the light exits the water surface.
- From the geometry of the triangle:
\[ \tan i = \frac{R}{h} \]
- For small angles, \( \tan i \approx \sin i \), so:
\[ \sin i = \frac{R}{\sqrt{R^2 + h^2}} \]

5. **Setting Up the Equation**:
- Equating the two expressions for \( \sin i \):
\[ \frac{R}{\sqrt{R^2 + h^2}} = \frac{1}{\mu} \]
- Cross-multiplying gives:
\[ R = \frac{\sqrt{R^2 + h^2}}{\mu} \]

6. **Squaring Both Sides**:
- Squaring both sides to eliminate the square root:
\[ R^2 = \frac{R^2 + h^2}{\mu^2} \]
- Rearranging gives:
\[ R^2 \mu^2 = R^2 + h^2 \]
- Therefore:
\[ R^2 (\mu^2 - 1) = h^2 \]

7. **Solving for \( R \)**:
- Finally, solving for \( R \):
\[ R^2 = \frac{h^2}{\mu^2 - 1} \]
\[ R = \frac{h}{\sqrt{\mu^2 - 1}} \]

### Final Result:
The radius of the circle of view for outside objects is:
\[ R = \frac{h}{\sqrt{\mu^2 - 1}} \]