The focal length of objective and eye lens of a astronomical telescope are respectively `2m` and `5 cm` . Final image is formed at `(i)` least distance of distinct vision `(ii)` infinity. The magnifying power in both cases will be

Options

A. `-48,-40`
B. `-40, -48`
C. `-40, 48`
D. `-48, 40`

Explanation

To solve the problem of finding the magnifying power of an astronomical telescope for two different cases, we will follow these steps:

### Given Data:
- Focal length of the objective lens, \( F_o = 2 \, \text{m} = 200 \, \text{cm} \)
- Focal length of the eye lens, \( F_e = 5 \, \text{cm} \)

### Case (i): Final Image at Least Distance of Distinct Vision
1. **Identify the formula for magnifying power** when the final image is at the least distance of distinct vision (D = 25 cm):
\[ M = \frac{F_o}{F_e} \left(1 + \frac{F_e}{D}\right) \]

2. **Substitute the values** into the formula:
\[ M = \frac{200}{5} \left(1 + \frac{5}{25}\right) \]

3. **Calculate \( \frac{200}{5} \)**:
\[ \frac{200}{5} = 40 \]

4. **Calculate \( \frac{5}{25} \)**:
\[ \frac{5}{25} = \frac{1}{5} \]

5. **Add 1 to \( \frac{1}{5} \)**:
\[ 1 + \frac{1}{5} = \frac{5}{5} + \frac{1}{5} = \frac{6}{5} \]

6. **Now substitute back into the magnifying power equation**:
\[ M = 40 \times \frac{6}{5} = 48 \]

7. **Since the image is inverted**, the magnifying power will be negative:
\[ M = -48 \]

### Case (ii): Final Image at Infinity
1. **Identify the formula for magnifying power** when the final image is at infinity:
\[ M = -\frac{F_o}{F_e} \]

2. **Substitute the values** into the formula:
\[ M = -\frac{200}{5} \]

3. **Calculate \( -\frac{200}{5} \)**:
\[ M = -40 \]

### Final Answers:
- For the least distance of distinct vision, the magnifying power is \( M = -48 \).
- For the image at infinity, the magnifying power is \( M = -40 \).

### Summary:
- Magnifying power when the final image is at least distance of distinct vision: **-48**
- Magnifying power when the final image is at infinity: **-40**