A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length `1.0 cm`. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? The diameter of the moon is `3.5 xx 10^(6)m`. and the radius of the lunar orbit round the earth is `3.8 xx 10^(8) m`.

Options

A. `10 cm`
B. `12.5 cm`
C. `15 cm`
D. `17.5 cm`

Explanation

To find the diameter of the image of the moon formed by the objective of the telescope, we can follow these steps:

### Step 1: Understand the relationship between the diameter of the moon, the distance to the moon, and the angle subtended.
The angle subtended by the moon at the Earth can be calculated using the formula:
\[ \alpha = \frac{d}{R} \]
where \(d\) is the diameter of the moon and \(R\) is the distance from the Earth to the moon.

### Step 2: Plug in the values for the diameter of the moon and the distance to the moon.
Given:
- Diameter of the moon, \(d = 3.5 \times 10^6 \, \text{m}\)
- Radius of the lunar orbit (distance to the moon), \(R = 3.8 \times 10^8 \, \text{m}\)

Substituting these values into the formula gives:
\[ \alpha = \frac{3.5 \times 10^6}{3.8 \times 10^8} \]

### Step 3: Calculate the angle \(\alpha\).
Calculating the above expression:
\[ \alpha = \frac{3.5 \times 10^6}{3.8 \times 10^8} \approx 0.00921 \, \text{radians} \]

### Step 4: Relate the angle subtended by the moon to the diameter of the image formed by the objective.
The diameter of the image \(d'\) formed by the objective can be related to the angle \(\alpha\) and the focal length \(f\) of the objective using the formula:
\[ \alpha = \frac{d'}{f} \]
where \(f\) is the focal length of the objective.

### Step 5: Substitute the known values into the equation.
Given:
- Focal length of the objective, \(f = 19 \, \text{m}\)

We can rearrange the formula to find \(d'\):
\[ d' = \alpha \cdot f \]

Substituting the values:
\[ d' = 0.00921 \cdot 19 \]

### Step 6: Calculate the diameter of the image \(d'\).
Calculating the above expression:
\[ d' \approx 0.175 \, \text{m} \]

### Final Answer:
The diameter of the image of the moon formed by the objective is approximately \(0.175 \, \text{m}\).

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