When a telescope is adjusted for parallel light, the distance of the objective from the eye piece is found to be `80 cm`. The magnifying power of the telescope is 19 . The focal length of the lenses are

Options

A. 61 cm, 19 cm
B. 40 cm, 40 cm
C. 76 cm, 4 cm
D. 50 cm, 30 cm

Answer

To solve the problem, we need to find the focal lengths of the objective lens (F₀) and the eyepiece lens (Fₑ) of a telescope when it is adjusted for parallel light. The given data includes:

- The distance between the objective and the eyepiece (length of the telescope) is 80 cm.
- The magnifying power (M) of the telescope is 19.

### Step-by-Step Solution:

1. **Understanding the Magnifying Power Formula**:
The magnifying power (M) of a telescope in normal adjustment is given by the formula:
\[ M = \frac{F₀}{Fₑ} \]
where:
- F₀ = focal length of the objective lens
- Fₑ = focal length of the eyepiece lens

2. **Setting Up the Equation**:
From the magnifying power given:
\[ 19 = \frac{F₀}{Fₑ} \]
This can be rearranged to express F₀ in terms of Fₑ:
\[ F₀ = 19 \times Fₑ \]

3. **Using the Length of the Telescope**:
The total length of the telescope is the sum of the focal lengths of the objective and the eyepiece:
\[ F₀ + Fₑ = 80 \text{ cm} \]

4. **Substituting F₀ in the Length Equation**:
Substitute the expression for F₀ from step 2 into the length equation:
\[ 19Fₑ + Fₑ = 80 \]
This simplifies to:
\[ 20Fₑ = 80 \]

5. **Solving for Fₑ**:
Divide both sides by 20:
\[ Fₑ = \frac{80}{20} = 4 \text{ cm} \]

6. **Finding F₀**:
Now substitute Fₑ back into the equation for F₀:
\[ F₀ = 19 \times Fₑ = 19 \times 4 = 76 \text{ cm} \]

### Final Answer:
The focal lengths of the lenses are:
- Focal length of the eyepiece (Fₑ) = 4 cm
- Focal length of the objective (F₀) = 76 cm