An astronomical telescope has an objective of focal length 100 cm and an eye piece of focal length 5 cm. The final image of a star is seen 25 cm from the eyepiece. The magnifying power of the telescope is

Options

A. 20
B. 22
C. 24
D. 26

Answer

To find the magnifying power of the astronomical telescope, we can use the formula for the magnifying power (M) given by:

\[ M = \frac{f_0}{f_e} \left(1 + \frac{f_e}{D}\right) \]

where:
- \(f_0\) is the focal length of the objective lens,
- \(f_e\) is the focal length of the eyepiece,
- \(D\) is the least distance of distinct vision (usually taken as 25 cm).

### Step 1: Identify the given values
- Focal length of the objective lens, \(f_0 = 100 \, \text{cm}\)
- Focal length of the eyepiece, \(f_e = 5 \, \text{cm}\)
- Least distance of distinct vision, \(D = 25 \, \text{cm}\)

### Step 2: Substitute the values into the formula
Substituting the values into the magnifying power formula:

\[ M = \frac{100}{5} \left(1 + \frac{5}{25}\right) \]

### Step 3: Simplify the expression
First, calculate \(\frac{100}{5}\):

\[ \frac{100}{5} = 20 \]

Next, calculate \(\frac{5}{25}\):

\[ \frac{5}{25} = \frac{1}{5} \]

Now, substitute these values back into the formula:

\[ M = 20 \left(1 + \frac{1}{5}\right) \]

### Step 4: Calculate \(1 + \frac{1}{5}\)
\[ 1 + \frac{1}{5} = \frac{5}{5} + \frac{1}{5} = \frac{6}{5} \]

### Step 5: Final calculation of magnifying power
Now substitute this back into the magnifying power equation:

\[ M = 20 \times \frac{6}{5} \]

Calculating this gives:

\[ M = 20 \times 1.2 = 24 \]

### Conclusion
The magnifying power of the telescope is \(M = 24\).