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

Explanation

To solve the problem, we need to find the focal lengths of the objective lens (f₀) and the eyepiece lens (fₑ) of a telescope given the magnifying power and the distance between the two lenses.

### Step-by-step Solution:

1. **Understand the Given Information:**
- Magnifying power (M) = 9
- Distance between the objective and eyepiece (L) = 20 cm

2. **Use the Formula for Magnifying Power of a Telescope:**
The magnifying power (M) of a telescope is given by the formula:
\[ M = \frac{f₀}{fₑ} \]
Where:
- f₀ = focal length of the objective lens
- fₑ = focal length of the eyepiece lens

From the problem, we have:
\[ \frac{f₀}{fₑ} = 9 \quad \text{(Equation 1)} \]

3. **Use the Length of the Telescope:**
The total length of the telescope when adjusted for parallel rays is given by:
\[ L = f₀ + fₑ \]
Substituting the given value:
\[ f₀ + fₑ = 20 \quad \text{(Equation 2)} \]

4. **Substitute Equation 1 into Equation 2:**
From Equation 1, we can express f₀ in terms of fₑ:
\[ f₀ = 9fₑ \]
Substitute this into Equation 2:
\[ 9fₑ + fₑ = 20 \]
Simplifying this gives:
\[ 10fₑ = 20 \]

5. **Solve for fₑ:**
Dividing both sides by 10:
\[ fₑ = \frac{20}{10} = 2 \text{ cm} \]

6. **Find f₀ using fₑ:**
Now substitute fₑ back into the expression for f₀:
\[ f₀ = 9fₑ = 9 \times 2 = 18 \text{ cm} \]

7. **Final Answer:**
The focal lengths of the lenses are:
- Focal length of the objective lens (f₀) = 18 cm
- Focal length of the eyepiece lens (fₑ) = 2 cm

### Summary:
- Focal length of the objective lens (f₀) = 18 cm
- Focal length of the eyepiece lens (fₑ) = 2 cm