A bird in air looks at a fish directly below it inside in a transparent liquid in a tank. If the distance of the fish as estimated by the bird is `h_1` and that of the bird as estimated by the fish is `h_2,` then the refractive index of the liquid is

Options

A. `h_2/h_1`
B. `h_1/h_2`
C. `(h_1+h_2)/(h_1-h_2)`
D. `(h_1-h_2)/(h_1+h_2)`

Explanation

To find the refractive index of the liquid in which the fish is located, we can analyze the situation using the distances observed by both the bird and the fish. Let's denote the refractive index of the liquid as \( \mu \), the distance of the fish as estimated by the bird as \( h_1 \), and the distance of the bird as estimated by the fish as \( h_2 \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A bird in air looks down at a fish in a tank filled with a transparent liquid. The bird sees the fish at a distance \( h_1 \), while the fish sees the bird at a distance \( h_2 \).
- The light rays from the fish will bend when they pass from the liquid to the air, and the light rays from the bird will bend when they pass from air to the liquid.

2. **Applying Snell's Law**:
- When light travels from one medium to another, Snell's Law states that:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
- Here, \( n_1 \) is the refractive index of air (approximately 1), and \( n_2 \) is the refractive index of the liquid (\( \mu \)).

3. **Distance Observed by the Bird**:
- The distance \( h_1 \) that the bird sees the fish can be expressed as:
\[ h_1 = x + \frac{y}{\mu} \]
- Here, \( x \) is the distance from the surface of the liquid to the fish, and \( y \) is the distance from the bird to the surface of the liquid.

4. **Distance Observed by the Fish**:
- The distance \( h_2 \) that the fish sees the bird can be expressed as:
\[ h_2 = \mu (x + y) \]

5. **Setting Up the Equations**:
- We have two equations:
1. \( h_1 = x + \frac{y}{\mu} \) (Equation 1)
2. \( h_2 = \mu (x + y) \) (Equation 2)

6. **Manipulating the Equations**:
- From Equation 1, we can express \( y \):
\[ y = \mu(h_1 - x) \]
- Substitute \( y \) into Equation 2:
\[ h_2 = \mu \left( x + \mu(h_1 - x) \right) \]
- Simplifying gives:
\[ h_2 = \mu^2 h_1 + \mu x - \mu^2 x \]
- Rearranging yields:
\[ h_2 = \mu^2 h_1 + x(\mu - \mu^2) \]

7. **Finding the Refractive Index**:
- By adding the two equations after multiplying Equation 1 by \( \mu \):
\[ \mu h_1 + h_2 = \mu x + y \]
- Rearranging leads to:
\[ 2\mu x + y = h_2 \]
- From this, we can derive:
\[ \mu h_1 = h_2 \]
- Therefore, we can express the refractive index \( \mu \) as:
\[ \mu = \frac{h_2}{h_1} \]

### Final Answer:
The refractive index of the liquid is given by:
\[ \mu = \frac{h_2}{h_1} \]