A fish in a lake (refractive index `4//3` for water ) is viewed through a convex lens. From water surface, the lens is placed in air at half of the distance of the fish from the water surface, so that the image is formed at the fish itself. The focal length of the lens is how many times the depth of fish in water?

Options

A. 4
B. 6
C. 7.5
D. 9

Explanation

To solve the problem step by step, we need to analyze the situation involving the fish, the lens, and the refractive index of water.

### Step 1: Define the Variables
Let the distance of the fish from the water surface be \( x \). Therefore, the distance of the lens from the water surface is \( \frac{x}{2} \).

### Step 2: Calculate the Apparent Depth
The apparent depth of the fish when viewed from the air (due to the refractive index of water) can be calculated using the formula:
\[ \text{Apparent Depth} = \frac{\text{Real Depth}}{\mu} \]
Where \( \mu \) is the refractive index of water, which is given as \( \frac{4}{3} \). Thus, the apparent depth \( d_a \) is:
\[ d_a = \frac{x}{\frac{4}{3}} = \frac{3x}{4} \]

### Step 3: Calculate the Total Distance from the Lens to the Fish
The total distance from the lens to the fish is the distance from the lens to the water surface plus the apparent depth:
\[ \text{Total Distance} = \frac{x}{2} + \frac{3x}{4} \]
To add these fractions, we need a common denominator:
\[ \text{Total Distance} = \frac{2x}{4} + \frac{3x}{4} = \frac{5x}{4} \]

### Step 4: Define Object and Image Distances
For the lens, we define:
- Object distance \( u = -\frac{5x}{4} \) (negative because it is on the same side as the incoming light)
- Image distance \( v \) is the distance from the lens to the fish, which is:
\[ v = \frac{x}{2} + x = \frac{3x}{2} \]

### Step 5: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \( u \) and \( v \):
\[ \frac{1}{f} = \frac{1}{\frac{3x}{2}} - \frac{1}{-\frac{5x}{4}} \]
This simplifies to:
\[ \frac{1}{f} = \frac{2}{3x} + \frac{4}{5x} \]

### Step 6: Find a Common Denominator
To combine the fractions, we find a common denominator, which is \( 15x \):
\[ \frac{1}{f} = \frac{10}{15x} + \frac{12}{15x} = \frac{22}{15x} \]

### Step 7: Solve for Focal Length
Thus, we have:
\[ f = \frac{15x}{22} \]

### Step 8: Determine the Ratio of Focal Length to Depth
To find how many times the focal length is compared to the depth of the fish:
\[ \text{Depth of fish} = x \]
So, the ratio is:
\[ \frac{f}{x} = \frac{15}{22} \]

### Conclusion
The focal length of the lens is \( \frac{15}{22} \) times the depth of the fish in water.