The minimum magnifying power of an astronomial telescope is `M`. If the focal length of its eye-lens is halved, how much times the minimum magnifying power will increase.

Options

A. `M/2`
B. 2M
C. 3M
D. 4M

Answer

To solve the problem, we need to understand the relationship between the focal lengths of the lenses in an astronomical telescope and its magnifying power.

### Step-by-Step Solution:

1. **Understanding Magnifying Power**:
The magnifying power (M) of an astronomical telescope is given by the formula:
\[ M = -\frac{F_O}{F_E} \]
where \( F_O \) is the focal length of the objective lens and \( F_E \) is the focal length of the eyepiece lens.

2. **Initial Magnifying Power**:
Let the initial magnifying power be \( M \). Thus, we can express it as:
\[ M = -\frac{F_O}{F_E} \]

3. **Halving the Focal Length of the Eyepiece**:
According to the problem, the focal length of the eyepiece is halved. Therefore, the new focal length of the eyepiece becomes:
\[ F_E' = \frac{F_E}{2} \]

4. **Calculating New Magnifying Power**:
The new magnifying power \( M' \) can be calculated using the new focal length of the eyepiece:
\[ M' = -\frac{F_O}{F_E'} = -\frac{F_O}{\frac{F_E}{2}} = -\frac{2F_O}{F_E} \]

5. **Relating New Magnifying Power to Initial Magnifying Power**:
We can relate the new magnifying power \( M' \) to the initial magnifying power \( M \):
\[ M' = 2 \left(-\frac{F_O}{F_E}\right) = 2M \]

6. **Conclusion**:
Thus, the new magnifying power \( M' \) is twice the initial magnifying power \( M \). Therefore, the minimum magnifying power will increase by a factor of 2.

### Final Answer:
The minimum magnifying power will increase by a factor of **2**.

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