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. `15^(@)`
B. `20^(@)`
C. `30^(@)`
D. `35^(@)`

Answer

To find the diameter of the image of the moon as seen through the telescope, we can follow these steps:

### Step 1: Calculate the angle subtended by the moon at the objective lens
The angle subtended by the moon (θ₀) can be calculated using the formula:
\[ \theta_0 = \frac{D}{d} \]
where:
- \(D\) is the diameter of the moon = \(3.5 \times 10^3 \text{ km}\)
- \(d\) is the distance from the Earth to the moon = \(3.8 \times 10^5 \text{ km}\)

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

### Step 2: Convert the angle 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 the 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:
\[ M = \frac{400 \text{ cm}}{10 \text{ cm}} = 40 \]

### Step 4: Calculate the angle subtended by the image at the eyepiece
The angle subtended by the image (θ) can be calculated using:
\[ \theta = M \cdot \theta_0 \]
Substituting the values:
\[ \theta = 40 \cdot 0.528 \approx 21.12 \text{ degrees} \]

### Step 5: Calculate the diameter of the image of the moon
The diameter of the image (D') can be calculated using the formula:
\[ D' = d' \cdot \theta \]
where \(d'\) is the distance from the objective to the image. For a telescope, this distance is approximately equal to the focal length of the objective lens (assuming the image is formed at the focal plane).

Using \(d' = f_0 = 400 \text{ cm}\):
\[ D' = 400 \text{ cm} \cdot \tan(\theta) \]
Since \(\theta\) is small, we can use the small angle approximation where \(\tan(\theta) \approx \theta\) in radians. First, convert θ from degrees to radians:
\[ \theta \text{ (in radians)} = 21.12 \times \frac{\pi}{180} \approx 0.368 \text{ radians} \]
Now calculate \(D'\):
\[ D' = 400 \text{ cm} \cdot 0.368 \approx 147.2 \text{ cm} \]

### Final Answer
Thus, the diameter of the image of the moon will be approximately **147.2 cm**.

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