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`

Answer

To solve the problem, we need to calculate the magnifying power of an astronomical telescope for two different cases: when the final image is formed at the least distance of distinct vision (D) and when it is formed at infinity.

### Given Data:
- Focal length of the objective lens (Fo) = 2 m = 200 cm
- Focal length of the eye lens (Fe) = 5 cm
- Least distance of distinct vision (D) = 25 cm

### Step 1: Magnifying Power when the Image is at Least Distance of Distinct Vision

The formula for the magnifying power (M) of an astronomical telescope when the final image is at the least distance of distinct vision is given by:

\[ M = \frac{F_o}{F_e} \left(1 + \frac{F_e}{D}\right) \]

Substituting the values into the formula:

\[ M = \frac{200}{5} \left(1 + \frac{5}{25}\right) \]

### Step 2: Simplifying the Expression

Calculating the magnifying power:

1. Calculate \( \frac{200}{5} = 40 \)
2. Calculate \( \frac{5}{25} = 0.2 \)
3. Therefore, \( 1 + 0.2 = 1.2 \)

So, substituting back:

\[ M = 40 \times 1.2 = 48 \]

Since the image is inverted, we take it as negative:

\[ M = -48 \]

### Step 3: Magnifying Power when the Image is at Infinity

The formula for the magnifying power (M) of an astronomical telescope when the final image is at infinity is given by:

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

Substituting the values:

\[ M = -\frac{200}{5} \]

### Step 4: Calculating the Magnifying Power

Calculating this gives:

\[ M = -40 \]

### Final Answers

1. Magnifying power when the image is at the least distance of distinct vision: \( M = -48 \)
2. Magnifying power when the image is at infinity: \( M = -40 \)

### Summary of Results
- For least distance of distinct vision: \( M = -48 \)
- For image at infinity: \( M = -40 \)

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