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

### Step 1: Calculate the angular size of the moon (θ₀)
The angular size (θ₀) of the moon can be calculated using the formula:
\[ \theta_0 = \tan^{-1}\left(\frac{d}{D}\right) \]
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:
\[ \theta_0 = \tan^{-1}\left(\frac{3.5 \times 10^3}{3.8 \times 10^5}\right) \]

### Step 2: Simplify the argument of the tangent function
Calculating the fraction:
\[ \frac{3.5 \times 10^3}{3.8 \times 10^5} = \frac{3.5}{3.8} \times 10^{-2} \approx 0.921 \times 10^{-2} = 9.21 \times 10^{-3} \]

### Step 3: Calculate θ₀ in radians
Now, we calculate:
\[ \theta_0 \approx \tan^{-1}(9.21 \times 10^{-3}) \approx 9.21 \times 10^{-3} \, \text{radians} \]

### Step 4: Convert θ₀ from radians to degrees
To convert from radians to degrees:
\[ \theta_0 \text{ (in degrees)} = \theta_0 \text{ (in radians)} \times \frac{180}{\pi} \]
Calculating this:
\[ \theta_0 \approx 9.21 \times 10^{-3} \times \frac{180}{\pi} \approx 0.528 \, \text{degrees} \]

### Step 5: Calculate the magnification (M) of the telescope
The magnification of the telescope is given by:
\[ M = \frac{f_o}{f_e} \]
where:
- \(f_o\) is the focal length of the objective = \(4 \, \text{m} = 400 \, \text{cm}\)
- \(f_e\) is the focal length of the eyepiece = \(10 \, \text{cm}\)

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

### Step 6: Calculate the angular size of the image (θ)
The angular size of the image (θ) is given by:
\[ \theta = M \cdot \theta_0 \]
Substituting the values:
\[ \theta = 40 \cdot 0.528 \approx 21.12 \, \text{degrees} \]

### Conclusion
The diameter of the image of the moon as seen through the telescope is approximately \(21.12\) degrees.

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