A small fish `0.4m` below the surface of a lake, is viewed through a simple converging lens of focal length `3m`. The lens is kept at `0.2m` above the water surface such that the fish lies on the optical axis of the lens. Find the distance of the image of the fish from the lens as seen by the observer. The refractive index of water is `4//3`.

Answer

To solve the problem, we need to find the distance of the image of the fish from the lens as seen by the observer. We will follow these steps:

### Step 1: Determine the Real Depth of the Fish
The fish is located 0.4 m (or 40 cm) below the surface of the water.

### Step 2: Calculate the Apparent Depth of the Fish
The apparent depth \( D' \) can be calculated using the formula:
\[ D' = \frac{D}{n} \]
where \( D \) is the real depth and \( n \) is the refractive index of water. Here, \( D = 40 \) cm and \( n = \frac{4}{3} \).

Substituting the values:
\[ D' = \frac{40 \text{ cm}}{\frac{4}{3}} = 40 \times \frac{3}{4} = 30 \text{ cm} \]

### Step 3: Calculate the Distance from the Lens to the Fish
The lens is positioned 0.2 m (or 20 cm) above the water surface. Therefore, the total distance from the lens to the apparent depth of the fish is:
\[ \text{Distance from lens to fish} = D' + \text{height of lens above water} = 30 \text{ cm} + 20 \text{ cm} = 50 \text{ cm} \]

### Step 4: Set Up the Lens Formula
The object distance \( u \) is negative in lens formula convention. Thus:
\[ u = -50 \text{ cm} \]
The focal length \( f \) of the lens is given as 3 m (or 300 cm).

### Step 5: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Substituting the known values:
\[ \frac{1}{v} - \frac{1}{-50} = \frac{1}{300} \]

### Step 6: Solve for \( v \)
Rearranging gives:
\[ \frac{1}{v} + \frac{1}{50} = \frac{1}{300} \]
To combine the fractions, we find a common denominator:
\[ \frac{1}{v} = \frac{1}{300} - \frac{1}{50} \]
Calculating \( \frac{1}{50} \) in terms of 300:
\[ \frac{1}{50} = \frac{6}{300} \]
Thus:
\[ \frac{1}{v} = \frac{1}{300} - \frac{6}{300} = -\frac{5}{300} \]
This simplifies to:
\[ \frac{1}{v} = -\frac{1}{60} \]
Therefore:
\[ v = -60 \text{ cm} \]

### Step 7: Interpret the Result
The negative sign indicates that the image is formed on the same side as the object (the fish), meaning the observer sees the image of the fish at a distance of 60 cm in front of the lens.

### Final Answer
The distance of the image of the fish from the lens as seen by the observer is **60 cm**.

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