The focal length of objective and eyelens of a astronomical telescope are respectively 20 cm and 5 cm. Final image is formed at least distance of distinct vision. The magnifying power will be

Options

A. `-4.8`
B. `-4.0`
C. 4.8
D. `4.0`

Answer

To find the magnifying power of the astronomical telescope, we will follow these steps:

### Step 1: Identify the given values
- Focal length of the objective lens (f_o) = 20 cm
- Focal length of the eyepiece lens (f_e) = 5 cm
- Least distance of distinct vision (D) = 25 cm

### Step 2: Use the formula for magnifying power (M)
The magnifying power of an astronomical telescope when the final image is formed at the least distance of distinct vision is given by the formula:
\[ M = -\frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right) \]

### Step 3: Substitute the values into the formula
Substituting the values we have:
\[ M = -\frac{20 \, \text{cm}}{5 \, \text{cm}} \left(1 + \frac{5 \, \text{cm}}{25 \, \text{cm}}\right) \]

### Step 4: Simplify the expression
First, calculate \(-\frac{20}{5}\):
\[ -\frac{20}{5} = -4 \]

Next, calculate \(\frac{5}{25}\):
\[ \frac{5}{25} = \frac{1}{5} \]

Now, substitute back into the equation:
\[ M = -4 \left(1 + \frac{1}{5}\right) \]

### Step 5: Combine the terms inside the parentheses
\[ 1 + \frac{1}{5} = \frac{5}{5} + \frac{1}{5} = \frac{6}{5} \]

### Step 6: Final calculation of magnifying power
Now substitute this back into the equation:
\[ M = -4 \times \frac{6}{5} = -\frac{24}{5} \]

### Step 7: Convert to decimal
Calculating \(-\frac{24}{5}\):
\[ -\frac{24}{5} = -4.8 \]

### Conclusion
The magnifying power of the astronomical telescope is \(-4.8\).

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