The focal lengths of the objective and the eyepiece of an astronomical telescope are 20 cm and 5 cm respectively. If the final image is formed at a distance of 30 cm from the eye piece, find the separation between the lenses for distinct vision:

Options

A. 32.4cm
B. 42.3cm
C. 24.3cm
D. 30.24cm

Answer

To solve the problem, we need to find the separation between the objective and eyepiece lenses of an astronomical telescope for distinct vision. The given data includes:

- Focal length of the objective lens (FO) = 20 cm
- Focal length of the eyepiece lens (FE) = 5 cm
- Distance of the final image from the eyepiece (D) = 30 cm

### Step-by-step Solution:

1. **Understand the arrangement of the telescope**:
The telescope consists of two lenses: the objective lens and the eyepiece lens. The objective lens forms a real image which is then magnified by the eyepiece lens.

2. **Use the formula for separation (L) between the lenses**:
The separation between the lenses for distinct vision can be given by the formula:
\[ L = FO + \left(\frac{D \cdot FE}{D + FE}\right) \]
where:
- \( FO \) = focal length of the objective lens
- \( FE \) = focal length of the eyepiece lens
- \( D \) = distance of the final image from the eyepiece

3. **Substitute the known values into the formula**:
We can substitute the values into the formula:
\[ L = 20 \, \text{cm} + \left(\frac{30 \, \text{cm} \cdot 5 \, \text{cm}}{30 \, \text{cm} + 5 \, \text{cm}}\right) \]

4. **Calculate the denominator**:
First, calculate the denominator:
\[ D + FE = 30 \, \text{cm} + 5 \, \text{cm} = 35 \, \text{cm} \]

5. **Calculate the numerator**:
Now calculate the numerator:
\[ D \cdot FE = 30 \, \text{cm} \cdot 5 \, \text{cm} = 150 \, \text{cm}^2 \]

6. **Calculate the fraction**:
Now calculate the fraction:
\[ \frac{150 \, \text{cm}^2}{35 \, \text{cm}} = \frac{150}{35} \approx 4.2857 \, \text{cm} \]

7. **Add the focal length of the objective lens**:
Now add this result to the focal length of the objective lens:
\[ L = 20 \, \text{cm} + 4.2857 \, \text{cm} \approx 24.2857 \, \text{cm} \]

8. **Round off the answer**:
Rounding off to one decimal place, we get:
\[ L \approx 24.3 \, \text{cm} \]

### Final Answer:
The separation between the lenses for distinct vision is approximately **24.3 cm**.