A planet is observed by an astronomical refracting telescope having an objective of ofcal length 16m and an eyepiece of focal length 2 cm. Then,

Options

A. the distance between the objective and the eyepiece is 16.02m
B. the angular mangification of the planet is `-800`
C. the image of the planet is inverted
D. the objective is larger than the eyepiece

Explanation

To solve the problem regarding the astronomical refracting telescope, we need to analyze the given information and compute the required values step by step.

### Given:
- Focal length of the objective lens (\(f_o\)) = 16 m
- Focal length of the eyepiece lens (\(f_e\)) = 2 cm = 0.02 m

### Step 1: Calculate the distance between the objective and the eyepiece
The distance \(L\) between the objective and the eyepiece in a telescope is given by the formula:
\[ L = f_o + f_e \]
Substituting the values:
\[ L = 16 \, \text{m} + 0.02 \, \text{m} = 16.02 \, \text{m} \]

### Step 2: Calculate the angular magnification of the telescope
The angular magnification (\(M\)) of a telescope is given by the formula:
\[ M = -\frac{f_o}{f_e} \]
Substituting the values:
\[ M = -\frac{16 \, \text{m}}{0.02 \, \text{m}} = -800 \]

### Step 3: Determine the nature of the image formed
In a refracting telescope, the image formed is always inverted. Therefore, the image is inverted.

### Step 4: Compare the sizes of the objective and eyepiece
In a telescope, the objective lens is always larger than the eyepiece lens. Hence, this statement is also correct.

### Final Summary of Options:
- **Option A**: The distance between the objective and the eyepiece is 16.02 m. **(Correct)**
- **Option B**: The angular magnification of the planet is -800. **(Correct)**
- **Option C**: The image formed is inverted. **(Correct)**
- **Option D**: The objective is larger than the eyepiece. **(Correct)**

### Conclusion:
All options provided in the question are correct.

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