A small fish 0.4 m below the surface of a lake is viewed through a simple converging lens of focal length 3 m.the lens is kep at 0.2 m above the water surface such that the fish lies on the optical axis of the lens. Find the image of the fish seen by the observed. `(mu_(water)=(4)/(3))`

Options

A. A distance of 0.2 m from the water surface
B. A distance of 0.6 m from the water surface
C. A distance of 0.3 m from the water surface
D. The same location of fish

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the Apparent Depth of the Fish
The fish is located 0.4 m (or 40 cm) below the surface of the water. The apparent depth (d') can be calculated using the formula:
\[ d' = \frac{d}{\mu} \]
where \( d \) is the actual depth and \( \mu \) is the refractive index of water.

Given:
- \( d = 40 \, \text{cm} \)
- \( \mu = \frac{4}{3} \)

Calculating the apparent depth:
\[ d' = \frac{40 \, \text{cm}}{\frac{4}{3}} = 40 \times \frac{3}{4} = 30 \, \text{cm} \]

### Step 2: Calculate the Object Distance (U) for the Lens
The lens is positioned 0.2 m (or 20 cm) above the water surface. Therefore, the total distance from the lens to the fish (which is now at the apparent depth of 30 cm) is:
\[ U = \text{apparent depth} + \text{height of lens above water} \]
\[ U = 30 \, \text{cm} + 20 \, \text{cm} = 50 \, \text{cm} \]
Since the object is on the same side as the incoming light, we take \( U \) to be negative:
\[ U = -50 \, \text{cm} \]

### Step 3: Apply the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance,
- \( u \) is the object distance.

Given:
- \( f = 3 \, \text{m} = 300 \, \text{cm} \)

Substituting the values into the lens formula:
\[ \frac{1}{300} = \frac{1}{v} - \frac{1}{-50} \]
This simplifies to:
\[ \frac{1}{300} = \frac{1}{v} + \frac{1}{50} \]

### Step 4: Solve for v
Rearranging the equation gives:
\[ \frac{1}{v} = \frac{1}{300} - \frac{1}{50} \]

Finding a common denominator (which is 300):
\[ \frac{1}{v} = \frac{1}{300} - \frac{6}{300} = \frac{1 - 6}{300} = \frac{-5}{300} \]
Thus,
\[ v = -60 \, \text{cm} \]

### Step 5: Interpret the Result
The negative sign indicates that the image is formed on the same side as the object (the fish). This means that the observer sees the fish at a distance of 60 cm from the lens, which is in the direction opposite to the incoming light.

### Final Answer
The image of the fish seen by the observer is at a distance of 60 cm from the lens, on the same side as the fish.

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