For an eye kept at a depth h inside water is refractive index `mu`, and viewed outside, the radius of circle through which the outer objects can be seen will be

Options

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

Answer

To solve the problem of finding the radius of the circle through which outer objects can be seen from an eye kept at a depth \( h \) inside water with refractive index \( \mu \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Situation**:
- We have an eye located at a depth \( h \) in water.
- The refractive index of water is given as \( \mu \).
- We need to determine the radius of the circle from which objects can be seen outside the water.

2. **Identifying the Critical Angle**:
- The critical angle \( \theta_c \) is the angle of incidence above which total internal reflection occurs.
- The relationship between the critical angle and the refractive index is given by:
\[ \sin \theta_c = \frac{1}{\mu} \]

3. **Relating the Critical Angle to the Radius**:
- At the critical angle, the refracted ray travels along the surface of the water. The geometry of the situation can be analyzed using a right triangle where:
- The height \( h \) is the opposite side.
- The radius \( r \) is the adjacent side.
- We can use the tangent function:
\[ \tan \theta_c = \frac{r}{h} \]

4. **Expressing \( \tan \theta_c \)**:
- We know that:
\[ \tan \theta_c = \frac{\sin \theta_c}{\cos \theta_c} \]
- From the sine relation, we can find \( \cos \theta_c \) using the Pythagorean identity:
\[ \cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - \left(\frac{1}{\mu}\right)^2} = \sqrt{\frac{\mu^2 - 1}{\mu^2}} \]

5. **Substituting into the Tangent Formula**:
- Now substituting \( \sin \theta_c \) and \( \cos \theta_c \) into the tangent formula:
\[ \tan \theta_c = \frac{\frac{1}{\mu}}{\sqrt{\frac{\mu^2 - 1}{\mu^2}}} = \frac{1}{\mu} \cdot \frac{\mu}{\sqrt{\mu^2 - 1}} = \frac{1}{\sqrt{\mu^2 - 1}} \]

6. **Finding the Radius**:
- From the tangent relation:
\[ \tan \theta_c = \frac{r}{h} \implies r = h \tan \theta_c = h \cdot \frac{1}{\sqrt{\mu^2 - 1}} \]
- Therefore, the radius \( r \) can be expressed as:
\[ r = \frac{h}{\sqrt{\mu^2 - 1}} \]

### Final Answer:
The radius of the circle through which outer objects can be seen is:
\[ r = \frac{h}{\sqrt{\mu^2 - 1}} \]