The focal lengths of the eyepiece and the objective of an astronimical telescope are 2 cm and 100 cm respectively. The magnifying power of the telescope for normal adjustment and the length of the telescope is

Options

A. 50, 102 cm
B. 100, 204 cm
C. 25, 62 cm
D. 75, 125 cm

Explanation

To solve the problem, we will calculate the magnifying power of the astronomical telescope and then determine the length of the telescope.

### Step 1: Identify the given values
- Focal length of the objective lens (F₀) = 100 cm
- Focal length of the eyepiece (Fₑ) = 2 cm

### Step 2: Calculate the magnifying power (M) of the telescope
The formula for the magnifying power of an astronomical telescope in normal adjustment is given by:
\[ M = \frac{F₀}{Fₑ} \]

Substituting the values:
\[ M = \frac{100 \, \text{cm}}{2 \, \text{cm}} \]
\[ M = 50 \]

### Step 3: Calculate the length of the telescope (L)
The length of the telescope in normal adjustment is given by the sum of the focal lengths of the objective and the eyepiece:
\[ L = F₀ + Fₑ \]

Substituting the values:
\[ L = 100 \, \text{cm} + 2 \, \text{cm} \]
\[ L = 102 \, \text{cm} \]

### Final Answers
- Magnifying power (M) = 50
- Length of the telescope (L) = 102 cm

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