The radius of the sun is `0.75xx10^(8)m` and its distance from the earth is `1.5xx10^(11)m`. Find the diameter of the image of the sun formed by a lens of focal length `40 cm`

Explanation

To find the diameter of the image of the sun formed by a lens of focal length 40 cm, we can follow these steps:

### Step 1: Understand the given data
- Radius of the sun, \( R = 0.75 \times 10^8 \, \text{m} \)
- Distance from the earth to the sun, \( U = 1.5 \times 10^{11} \, \text{m} \)
- Focal length of the lens, \( f = 40 \, \text{cm} = 0.4 \, \text{m} \)

### Step 2: Calculate the diameter of the sun
The diameter \( D \) of the sun is given by:
\[ D = 2R = 2 \times 0.75 \times 10^8 \, \text{m} = 1.5 \times 10^8 \, \text{m} \]

### Step 3: Use the lens formula
The lens formula is given by:
\[ \frac{1}{V} - \frac{1}{U} = \frac{1}{f} \]
Where:
- \( V \) is the image distance,
- \( U \) is the object distance (which is very large, so we can approximate).

### Step 4: Approximate the object distance
Since the distance \( U \) is very large compared to the focal length \( f \), we can ignore \( \frac{1}{U} \) in the lens formula:
\[ \frac{1}{V} \approx \frac{1}{f} \]
Thus,
\[ V \approx f = 0.4 \, \text{m} \]

### Step 5: Calculate the magnification
The magnification \( m \) is given by the ratio of the image distance to the object distance:
\[ m = \frac{V}{U} \]
Substituting \( V \) and \( U \):
\[ m = \frac{0.4}{1.5 \times 10^{11}} \approx \frac{0.4}{1.5 \times 10^{11}} \approx 2.67 \times 10^{-12} \]

### Step 6: Calculate the diameter of the image
The diameter of the image \( D' \) can be found using the magnification:
\[ D' = m \times D \]
Substituting the values:
\[ D' = 2.67 \times 10^{-12} \times 1.5 \times 10^8 \]
Calculating:
\[ D' \approx 4.005 \times 10^{-4} \, \text{m} = 0.0004005 \, \text{m} = 0.4 \, \text{mm} \]

### Final Result
The diameter of the image of the sun formed by the lens is approximately **0.4 mm**.

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