The focal length of an objective of a telescope is 3 meter and diameter 15 cm . Assuming for a normal eye, the diameter of the pupil is 3 mm for its complete use, the focal length of eye piece must be

Options

A. 6 cm
B. 6.3 cm
C. 20 cm
D. 60 cm

Answer

To find the focal length of the eyepiece (FE) of a telescope given the focal length of the objective (FO) and the diameters of the objective and the pupil, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Focal length of the objective (FO) = 3 m
- Diameter of the objective (D) = 15 cm
- Diameter of the pupil (d) = 3 mm

2. **Convert Units:**
- Convert the focal length of the objective from meters to centimeters:
\[ FO = 3 \text{ m} = 300 \text{ cm} \]
- Convert the diameter of the pupil from millimeters to centimeters:
\[ d = 3 \text{ mm} = 0.3 \text{ cm} \]

3. **Use the Formula for Focal Length of the Eyepiece:**
The relationship between the focal lengths and diameters in a telescope is given by:
\[ \frac{FO}{FE} = \frac{D}{d} \]
Rearranging this formula to find FE gives:
\[ FE = \frac{FO \cdot d}{D} \]

4. **Substitute the Values:**
Now substitute the known values into the equation:
\[ FE = \frac{300 \text{ cm} \cdot 0.3 \text{ cm}}{15 \text{ cm}} \]

5. **Calculate FE:**
- First, calculate the numerator:
\[ 300 \cdot 0.3 = 90 \]
- Now divide by the diameter of the objective:
\[ FE = \frac{90}{15} = 6 \text{ cm} \]

6. **Final Result:**
The focal length of the eyepiece (FE) is:
\[ FE = 6 \text{ cm} \]

### Summary:
The focal length of the eyepiece must be 6 cm.

---