An astronomical telescope has an angular magnification of magnitude 5 for distant object. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length `f_(0)` of the objective and the focal length `f_(0)` of the eyepiece are

Options

A. `f_(o)=45 cm and f_(e)=-9cm`
B. `f_(o)=-7.2 cm and f_(e)=5cm`
C. `f_(o)=50 cm and f_(e)=10cm`
D. f_(o)=30 cm and f_(e)= 6cm

Explanation

To solve the problem, we need to find the focal lengths of the objective lens (\(f_0\)) and the eyepiece (\(f_e\)) of an astronomical telescope given the following information:

1. The angular magnification (\(M\)) is 5.
2. The separation between the objective and the eyepiece is 36 cm.
3. The final image is formed at infinity.

### Step-by-Step Solution:

**Step 1: Understand the relationship between angular magnification and focal lengths.**
The angular magnification \(M\) for a telescope is given by the formula:
\[ M = \frac{f_0}{f_e} \]
where \(f_0\) is the focal length of the objective lens and \(f_e\) is the focal length of the eyepiece.

**Step 2: Substitute the given magnification into the formula.**
From the problem, we know that \(M = 5\). Therefore, we can write:
\[ \frac{f_0}{f_e} = 5 \]
This implies:
\[ f_0 = 5f_e \quad \text{(Equation 1)} \]

**Step 3: Use the separation between the objective and eyepiece.**
The total distance between the objective and the eyepiece is given as 36 cm:
\[ f_0 + f_e = 36 \quad \text{(Equation 2)} \]

**Step 4: Substitute Equation 1 into Equation 2.**
Now, we can substitute \(f_0\) from Equation 1 into Equation 2:
\[ 5f_e + f_e = 36 \]
This simplifies to:
\[ 6f_e = 36 \]

**Step 5: Solve for \(f_e\).**
Dividing both sides by 6 gives:
\[ f_e = 6 \, \text{cm} \]

**Step 6: Calculate \(f_0\) using the value of \(f_e\).**
Now, we can find \(f_0\) using Equation 1:
\[ f_0 = 5f_e = 5 \times 6 = 30 \, \text{cm} \]

### Final Answer:
- The focal length of the eyepiece (\(f_e\)) is **6 cm**.
- The focal length of the objective lens (\(f_0\)) is **30 cm**.