The minimum magnifying power of a telescope is M. If the focal length of its eyelens is halved, the magnifying power? will become

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding Magnifying Power**:
The minimum magnifying power (M) of a 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 write:
\[ M = \frac{f_o}{f_e} \]

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, denoted as \( f_e' \), is:
\[ f_e' = \frac{f_e}{2} \]

4. **Calculating New Magnifying Power**:
The new magnifying power \( M' \) with the halved focal length of the eyepiece can be calculated using the same formula:
\[ M' = \frac{f_o}{f_e'} \]
Substituting \( f_e' \) into the equation gives:
\[ M' = \frac{f_o}{\frac{f_e}{2}} = \frac{f_o \cdot 2}{f_e} = 2 \cdot \frac{f_o}{f_e} \]

5. **Relating New Magnifying Power to Original**:
Since \( \frac{f_o}{f_e} = M \), we can substitute this into our equation for \( M' \):
\[ M' = 2M \]

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

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

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