The minimum magnifying power of an astronomical telescope is M. If the focal length of its eye-lens is halved, the minimum magnifying power will become:

Options

A. m/2
B. 2m
C. 3m
D. 4m

Answer

To solve the problem, we need to understand how the magnifying power of an astronomical telescope is affected by changes in the focal length of its eyepiece.

### Step-by-Step Solution:

1. **Understanding the 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. **Given Information**:
We are given that the minimum magnifying power is \( M \). Therefore, we can express this 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. Thus, the new focal length of the eyepiece \( F_e' \) can be expressed as:
\[ F_e' = \frac{F_e}{2} \]

4. **Calculating the New Magnifying Power**:
We need to find the new magnifying power \( M' \) with the new focal length of the eyepiece:
\[ M' = -\frac{F_o}{F_e'} \]
Substituting \( F_e' \) into the equation:
\[ M' = -\frac{F_o}{\frac{F_e}{2}} = -\frac{F_o \cdot 2}{F_e} = 2 \left(-\frac{F_o}{F_e}\right) \]
Since \( -\frac{F_o}{F_e} = M \), we can rewrite the equation as:
\[ M' = 2M \]

5. **Conclusion**:
Therefore, the new minimum magnifying power when the focal length of the eyepiece is halved is:
\[ M' = 2M \]

### Final Answer:
The minimum magnifying power will become \( 2M \).

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