An astronomical telescope arranged for normal adjustment has a magnification of 6. If the length of the telescope is 35cm, then the focal lengths of objective and eye piece respectively are

Options

A. 30cm,5cm
B. 5cm,30cm
C. 40cm,5cm
D. 30cm,6cm

Explanation

To solve the problem, we need to find the focal lengths of the objective (F₀) and the eyepiece (Fₑ) of an astronomical telescope arranged for normal adjustment. We are given the following information:

- Magnification (M) = 6
- Length of the telescope (L) = 35 cm

### Step-by-Step Solution:

1. **Understand the Magnification Formula:**
The magnification (M) of an astronomical telescope in normal adjustment is given by the formula:
\[ M = \frac{F₀}{Fₑ} \]
Where:
- \( F₀ \) is the focal length of the objective.
- \( Fₑ \) is the focal length of the eyepiece.

2. **Set Up the Equation for Magnification:**
Given that the magnification is 6, we can write:
\[ \frac{F₀}{Fₑ} = 6 \]
This can be rearranged to express \( F₀ \) in terms of \( Fₑ \):
\[ F₀ = 6Fₑ \]

3. **Understand the Length of the Telescope:**
For a telescope in normal adjustment, the length of the telescope (L) is the sum of the focal lengths of the objective and the eyepiece:
\[ L = F₀ + Fₑ \]
Substituting the given length:
\[ 35 = F₀ + Fₑ \]

4. **Substitute for \( F₀ \):**
Now, substitute \( F₀ = 6Fₑ \) into the length equation:
\[ 35 = 6Fₑ + Fₑ \]
This simplifies to:
\[ 35 = 7Fₑ \]

5. **Solve for \( Fₑ \):**
To find \( Fₑ \), divide both sides by 7:
\[ Fₑ = \frac{35}{7} = 5 \text{ cm} \]

6. **Find \( F₀ \):**
Now, substitute \( Fₑ \) back into the equation for \( F₀ \):
\[ F₀ = 6Fₑ = 6 \times 5 = 30 \text{ cm} \]

### Final Answer:
The focal lengths of the objective and eyepiece are:
- \( F₀ = 30 \text{ cm} \)
- \( Fₑ = 5 \text{ cm} \)