The magnifying power of an astronomical telescope is 8 and the distance between the two lenses is 54 cm. The focal length of eye lens and objective will be respectively.

Options

A. 48 cm, 6 cm
B. 8 cm, 64 cm
C. 64 cm, 8 cm
D. 6 cm, 48 cm

Answer

To find the focal lengths of the eye lens (FE) and the objective lens (F0) of an astronomical telescope, we can follow these steps:

### Step 1: Understand the relationship between the focal lengths and the distance between the lenses
The total distance between the two lenses in an astronomical telescope is given by:
\[ F_0 + F_E = D \]
where:
- \( F_0 \) = focal length of the objective lens
- \( F_E \) = focal length of the eye lens
- \( D \) = distance between the two lenses (54 cm)

### Step 2: Use the formula for magnifying power
The magnifying power (M) of an astronomical telescope is given by:
\[ M = \frac{F_0}{F_E} \]
According to the problem, the magnifying power is 8:
\[ M = 8 \]

### Step 3: Set up the equations
From the magnifying power equation, we can express \( F_0 \) in terms of \( F_E \):
\[ F_0 = 8 F_E \]

### Step 4: Substitute \( F_0 \) in the distance equation
Now substitute \( F_0 \) into the distance equation:
\[ 8 F_E + F_E = 54 \]
This simplifies to:
\[ 9 F_E = 54 \]

### Step 5: Solve for \( F_E \)
Now, solve for \( F_E \):
\[ F_E = \frac{54}{9} = 6 \, \text{cm} \]

### Step 6: Find \( F_0 \)
Now that we have \( F_E \), we can find \( F_0 \):
\[ F_0 = 8 F_E = 8 \times 6 = 48 \, \text{cm} \]

### Final Result
Thus, the focal lengths of the eye lens and the objective lens are:
- \( F_E = 6 \, \text{cm} \)
- \( F_0 = 48 \, \text{cm} \)

### Summary
The focal lengths of the eye lens and the objective lens are respectively 6 cm and 48 cm.

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