The diameter of the moon is `3.5xx10^(3)km` and its distance from the earth is `3.8xx10^(5) km`. It is seen by a telescope having the focal length of the objective and the eye-piece as `4m` and `10cm` respectively. The diameter of the image of the moon will be approximately

Options

A. `2^(@)`
B. `20^(@)`
C. `40^(@)`
D. `50^(@)`

Explanation

To solve the problem of finding the diameter of the image of the moon as seen through a telescope, we can follow these steps:

### Step 1: Calculate the angular size of the moon
The angular size (θ₀) of the moon can be calculated using the formula:
\[ \theta_0 = \frac{D}{d} \]
where:
- \(D\) = diameter of the moon = \(3.5 \times 10^3 \, \text{km}\)
- \(d\) = distance from the earth to the moon = \(3.8 \times 10^5 \, \text{km}\)

Substituting the values:
\[ \theta_0 = \frac{3.5 \times 10^3}{3.8 \times 10^5} \approx 9.21 \times 10^{-3} \, \text{radians} \]

### Step 2: Convert angular size from radians to degrees
To convert radians to degrees, we use the conversion factor \( \frac{180}{\pi} \):
\[ \theta_0 \text{ (in degrees)} = 9.21 \times 10^{-3} \times \frac{180}{\pi} \approx 0.528 \, \text{degrees} \]

### Step 3: Calculate the magnification of the telescope
The magnification (M) of a telescope is given by the ratio of the focal lengths of the objective (f₀) and the eyepiece (fₑ):
\[ M = \frac{f_0}{f_e} \]
Given:
- \(f_0 = 4 \, \text{m} = 400 \, \text{cm}\)
- \(f_e = 10 \, \text{cm}\)

Substituting the values:
\[ M = \frac{400}{10} = 40 \]

### Step 4: Calculate the angular size of the image of the moon
The angular size of the image (θ) can be calculated using the magnification:
\[ M = \frac{\theta}{\theta_0} \]
Rearranging gives:
\[ \theta = M \cdot \theta_0 \]
Substituting the values:
\[ \theta = 40 \cdot 9.21 \times 10^{-3} \approx 0.3684 \, \text{radians} \]

### Step 5: Calculate the diameter of the image of the moon
The diameter of the image (Dᵢ) can be calculated using the formula:
\[ D_i = d \cdot \theta \]
where \(d\) is the distance from the objective lens to the image, which can be approximated as the distance to the moon since the image is formed at a distance much larger than the diameter of the moon.

Using \(d = 3.8 \times 10^5 \, \text{km} = 3.8 \times 10^8 \, \text{cm}\):
\[ D_i = 3.8 \times 10^8 \cdot 0.3684 \approx 1.4 \times 10^8 \, \text{cm} = 1.4 \times 10^5 \, \text{km} \]

### Final Answer
The diameter of the image of the moon will be approximately \(1.4 \times 10^5 \, \text{km}\).