If the focal length of the ey piece of the telescope is doubled, then its magnifying power (m) will be

Options

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

Explanation

To solve the problem, we need to understand how the magnifying power (M) of a telescope is affected when the focal length of the eyepiece (Fe) is doubled. The magnifying power of a telescope is given by the formula:

\[ M = \frac{F_o}{F_e} \]

Where:
- \( M \) is the magnifying power,
- \( F_o \) is the focal length of the objective lens,
- \( F_e \) is the focal length of the eyepiece.

### Step 1: Understand the effect of doubling the focal length of the eyepiece
According to the problem, the focal length of the eyepiece is doubled. Therefore, if the original focal length of the eyepiece is \( F_e \), the new focal length \( F_e' \) will be:

\[ F_e' = 2F_e \]

### Step 2: Write the new magnifying power
Now, we can write the new magnifying power \( M' \) using the new focal length of the eyepiece:

\[ M' = \frac{F_o}{F_e'} \]

Substituting \( F_e' \) into the equation:

\[ M' = \frac{F_o}{2F_e} \]

### Step 3: Relate the new magnifying power to the original magnifying power
We know that the original magnifying power \( M \) is given by:

\[ M = \frac{F_o}{F_e} \]

Now we can express the new magnifying power \( M' \) in terms of the original magnifying power \( M \):

\[ M' = \frac{F_o}{2F_e} = \frac{1}{2} \cdot \frac{F_o}{F_e} = \frac{M}{2} \]

### Conclusion
Thus, the new magnifying power \( M' \) when the focal length of the eyepiece is doubled is:

\[ M' = \frac{M}{2} \]

This means the magnifying power is halved.

### Final Answer
The correct option is: **half of M**.

---