Three lenses of focal length +15cm,+150 cm and +250 cm are available, for making an astronomical telescope. To produced the largest magnification, the focal length of the eye piece should be

Options

A. `+15 cm`
B. `+150cm`
C. `+250cm`
D. None of these

Explanation

To find the focal length of the eyepiece that will produce the largest magnification for an astronomical telescope using the provided lenses, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Magnification Formula**:
The magnification (M) of an astronomical telescope is given by the formula:
\[ M = \frac{f_o}{f_e} \left(1 + \frac{f_e}{d}\right) \]
where:
- \(f_o\) = focal length of the objective lens
- \(f_e\) = focal length of the eyepiece lens
- \(d\) = distance between the two lenses (which we will assume is constant for this problem).

2. **Identify the Focal Lengths**:
The available lenses have the following focal lengths:
- Lens 1: \(f_1 = +15 \, \text{cm}\)
- Lens 2: \(f_2 = +150 \, \text{cm}\)
- Lens 3: \(f_3 = +250 \, \text{cm}\)

3. **Determine the Effect of Eyepiece Focal Length on Magnification**:
From the magnification formula, we can see that the magnification is inversely proportional to the focal length of the eyepiece (\(f_e\)). This means that to maximize magnification, we need to minimize \(f_e\).

4. **Select the Eyepiece with the Smallest Focal Length**:
Among the three lenses, the one with the smallest focal length is:
- \(f_e = 15 \, \text{cm}\)

5. **Conclusion**:
Therefore, to produce the largest magnification for the astronomical telescope, the focal length of the eyepiece should be:
\[ f_e = 15 \, \text{cm} \]

### Final Answer:
The focal length of the eyepiece should be **15 cm**.

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