The magnifying power of a telescope is `9.` When it is adjusted for parallel rays the distance between the objective and eyepiece is `20cm`. The focal lengths of lenses are

Options

A. `11cm,9cm`
B. `10cm, 10cm`
C. `15cm, 5cm`
D. `18cm,2cm`

Explanation

To solve the problem, we need to find the focal lengths of the objective lens (F) and the eyepiece lens (FE) of a telescope given its magnifying power and the distance between the two lenses.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Magnifying power (m) of the telescope = 9
- Distance between the objective and eyepiece (L) = 20 cm

2. **Using the Formula for Magnifying Power:**
The magnifying power of a telescope is given by the formula:
\[ m = \frac{F}{FE} \]
where F is the focal length of the objective lens and FE is the focal length of the eyepiece lens.

3. **Using the Length of the Telescope:**
The length of the telescope when adjusted for parallel rays is given by:
\[ L = F + FE \]
Given that L = 20 cm, we can write:
\[ F + FE = 20 \quad \text{(1)} \]

4. **Substituting Magnifying Power into the Length Equation:**
From the magnifying power equation, we can express F in terms of FE:
\[ F = 9 \cdot FE \quad \text{(2)} \]
Now, substitute equation (2) into equation (1):
\[ 9FE + FE = 20 \]
This simplifies to:
\[ 10FE = 20 \]

5. **Solving for FE:**
Dividing both sides by 10 gives:
\[ FE = 2 \, \text{cm} \]

6. **Finding Focal Length of the Objective Lens (F):**
Now, substitute the value of FE back into equation (2):
\[ F = 9 \cdot FE = 9 \cdot 2 = 18 \, \text{cm} \]

7. **Final Result:**
- Focal length of the eyepiece (FE) = 2 cm
- Focal length of the objective lens (F) = 18 cm

### Summary:
The focal lengths of the lenses are:
- Focal length of the objective lens (F) = 18 cm
- Focal length of the eyepiece lens (FE) = 2 cm