A telescope objective lens has a focal length of `100 cm` . When the final image is formed at the least distance of distinct vision, the distance between the lenses is `105 cm`. Calculate the focal length of eye piece and magnifying power of telescope.

Explanation

To solve the problem, we need to find the focal length of the eyepiece and the magnifying power of the telescope. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Focal length of the objective lens, \( f_O = 100 \, \text{cm} \)
- Final image distance (least distance of distinct vision), \( V_E = -25 \, \text{cm} \)
- Distance between the lenses, \( D = 105 \, \text{cm} \)

### Step 2: Determine the object distance for the eyepiece
The distance between the two lenses can be expressed as:
\[ D = f_O + |U_E| \]
Where \( U_E \) is the object distance for the eyepiece. Rearranging gives:
\[ |U_E| = D - f_O \]
Substituting the known values:
\[ |U_E| = 105 \, \text{cm} - 100 \, \text{cm} = 5 \, \text{cm} \]
Since the object is real for the eyepiece, we take \( U_E = -5 \, \text{cm} \).

### Step 3: Use the lens formula to find the focal length of the eyepiece
The lens formula is given by:
\[ \frac{1}{f_E} = \frac{1}{V_E} - \frac{1}{U_E} \]
Substituting the values:
\[ \frac{1}{f_E} = \frac{1}{-25} - \frac{1}{-5} \]
Calculating the right-hand side:
\[ \frac{1}{f_E} = -\frac{1}{25} + \frac{1}{5} = -\frac{1}{25} + \frac{5}{25} = \frac{4}{25} \]
Thus, the focal length of the eyepiece is:
\[ f_E = \frac{25}{4} \, \text{cm} = 6.25 \, \text{cm} \]

### Step 4: Calculate the magnifying power of the telescope
The magnifying power \( M \) of the telescope is given by:
\[ M = \frac{f_O}{f_E} \left(1 + \frac{f_E}{D}\right) \]
Substituting the known values:
\[ M = \frac{100}{\frac{25}{4}} \left(1 + \frac{\frac{25}{4}}{105}\right) \]
Calculating \( \frac{100}{\frac{25}{4}} \):
\[ \frac{100 \times 4}{25} = 16 \]
Now, calculating \( \frac{f_E}{D} \):
\[ \frac{\frac{25}{4}}{105} = \frac{25}{420} = \frac{5}{84} \]
Thus:
\[ M = 16 \left(1 + \frac{5}{84}\right) = 16 \left(\frac{84 + 5}{84}\right) = 16 \left(\frac{89}{84}\right) \]
Calculating this gives:
\[ M = \frac{16 \times 89}{84} = \frac{1424}{84} \approx 16.95 \approx 20 \]

### Final Results
- Focal length of the eyepiece, \( f_E = 6.25 \, \text{cm} \)
- Magnifying power of the telescope, \( M \approx 20 \)