The optical length of an astronomical telescope with magnifying power of 10, for normal vision is 44cm. What is focal length of the objective?

Options

A. 440 cm
B. 44 cm
C. 40 cm
D. 4 cm

Answer

To solve the problem step by step, we will use the relationships between the optical length of the telescope, the focal lengths of the objective and eyepiece, and the magnifying power.

### Step 1: Understand the given information
We are given:
- Magnifying power (m) = 10
- Optical length (L) = 44 cm

### Step 2: Write the formula for optical length
The optical length of an astronomical telescope is given by the formula:
\[ L = f_o + f_e \]
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eyepiece lens.

### Step 3: Write the formula for magnifying power
The magnifying power of a telescope is given by:
\[ m = \frac{f_o}{f_e} \]
From the given magnifying power, we can express the focal length of the objective in terms of the focal length of the eyepiece:
\[ f_o = m \cdot f_e \]
Substituting the value of \( m \):
\[ f_o = 10 \cdot f_e \]

### Step 4: Substitute \( f_o \) in the optical length equation
Now, we can substitute the expression for \( f_o \) into the optical length equation:
\[ L = f_o + f_e \]
Substituting \( f_o = 10 \cdot f_e \):
\[ 44 = 10f_e + f_e \]
This simplifies to:
\[ 44 = 11f_e \]

### Step 5: Solve for \( f_e \)
Now, we can solve for \( f_e \):
\[ f_e = \frac{44}{11} = 4 \text{ cm} \]

### Step 6: Find \( f_o \)
Now that we have \( f_e \), we can find \( f_o \):
\[ f_o = 10 \cdot f_e = 10 \cdot 4 = 40 \text{ cm} \]

### Final Answer
The focal length of the objective lens is:
\[ f_o = 40 \text{ cm} \]