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

Explanation

To solve the problem step by step, we will use the information provided about the astronomical telescope, its angular magnification, and the separation between the objective and eyepiece.

### Step 1: Understand the relationship between magnification and focal lengths
The angular magnification \( m \) of an astronomical telescope is given by the formula:
\[ m = -\frac{f_o}{f_e} \]
where \( f_o \) is the focal length of the objective and \( f_e \) is the focal length of the eyepiece. Given that the angular magnification \( m = -5 \), we can write:
\[ -5 = -\frac{f_o}{f_e} \]
This simplifies to:
\[ f_o = 5 f_e \quad \text{(Equation 1)} \]

### Step 2: Use the separation between the objective and eyepiece
The separation \( L \) between the objective and the eyepiece is given as \( 36 \, \text{cm} \). This can be expressed as:
\[ L = f_o + f_e \]
Substituting the value of \( L \):
\[ 36 = f_o + f_e \]

### Step 3: Substitute Equation 1 into the separation equation
From Equation 1, we know that \( f_o = 5 f_e \). Substituting this into the separation equation:
\[ 36 = 5 f_e + f_e \]
This simplifies to:
\[ 36 = 6 f_e \]

### Step 4: Solve for the focal length of the eyepiece
To find \( f_e \), we divide both sides by 6:
\[ f_e = \frac{36}{6} = 6 \, \text{cm} \]

### Step 5: Find the focal length of the objective
Now that we have \( f_e \), we can find \( f_o \) using Equation 1:
\[ f_o = 5 f_e = 5 \times 6 = 30 \, \text{cm} \]

### Final Results
Thus, the focal lengths are:
- Focal length of the eyepiece \( f_e = 6 \, \text{cm} \)
- Focal length of the objective \( f_o = 30 \, \text{cm} \)

### Summary
- Focal length of objective, \( f_o = 30 \, \text{cm} \)
- Focal length of eyepiece, \( f_e = 6 \, \text{cm} \)

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