An astronomical telescope has an eyepiece of focal-length 5cm. If the angular magnification in normal adjustment is 10, when final image is at least distance of distinct vision (25cm) from eye piece, then angular magnification will be :

Options

A. `A. 10`
B. `B. 12`
C. `C. 50`
D. `D. 60`

Explanation

To solve the problem, we need to find the angular magnification of an astronomical telescope when the final image is at the least distance of distinct vision (25 cm) from the eyepiece. We are given the focal length of the eyepiece and the angular magnification in normal adjustment.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Focal length of the eyepiece, \( f_E = 5 \, \text{cm} \)
- Angular magnification in normal adjustment, \( M = 10 \)
- Least distance of distinct vision, \( D = 25 \, \text{cm} \)

2. **Understand the Formula for Angular Magnification:**
The angular magnification when the final image is at the least distance of distinct vision is given by:
\[ M' = M \left(1 + \frac{f_E}{D}\right) \]
where \( M' \) is the new angular magnification.

3. **Substitute the Values into the Formula:**
Plugging in the known values into the formula:
\[ M' = 10 \left(1 + \frac{5}{25}\right) \]

4. **Calculate the Fraction:**
Calculate \( \frac{5}{25} \):
\[ \frac{5}{25} = 0.2 \]

5. **Add 1 to the Fraction:**
Now, add 1 to the result:
\[ 1 + 0.2 = 1.2 \]

6. **Multiply by the Original Magnification:**
Now, multiply this result by the original magnification:
\[ M' = 10 \times 1.2 = 12 \]

7. **Final Result:**
Therefore, the angular magnification when the final image is at the least distance of distinct vision is:
\[ M' = 12 \]

### Conclusion:
The angular magnification will be 12.