A telescope has an objective of focal length `50cm` and an eyepiece of focal length `5cm.` The least distance of distinct vision is `25cm.` The telescope is focused for distinct vision on a scale `2m` away from the objective. Calculate (a) magnification produced and (b) separation between objective and eyepiece.

Explanation

To solve the problem step by step, we will calculate the magnification produced by the telescope and the separation between the objective and the eyepiece.

### Given Data:
- Focal length of the objective lens, \( f_0 = 50 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 5 \, \text{cm} \)
- Least distance of distinct vision, \( D = 25 \, \text{cm} \)
- Distance of the scale from the objective, \( U_0 = -200 \, \text{cm} \) (negative because the object is on the same side as the incoming light)

### Step 1: Calculate the image distance for the objective lens
Using the lens formula:
\[ \frac{1}{f_0} = \frac{1}{V_0} - \frac{1}{U_0} \]
Substituting the values:
\[ \frac{1}{50} = \frac{1}{V_0} - \frac{1}{-200} \]
This simplifies to:
\[ \frac{1}{50} = \frac{1}{V_0} + \frac{1}{200} \]
Rearranging gives:
\[ \frac{1}{V_0} = \frac{1}{50} - \frac{1}{200} \]
Finding a common denominator (200):
\[ \frac{1}{V_0} = \frac{4}{200} - \frac{1}{200} = \frac{3}{200} \]
Thus,
\[ V_0 = \frac{200}{3} \, \text{cm} \]

### Step 2: Calculate the magnification of the objective lens
The magnification \( M_0 \) produced by the objective lens is given by:
\[ M_0 = \frac{V_0}{U_0} \]
Substituting the values:
\[ M_0 = \frac{\frac{200}{3}}{-200} = -\frac{1}{3} \]

### Step 3: Calculate the image distance for the eyepiece lens
For the eyepiece, we need to find the object distance \( U_e \). The image formed by the objective acts as the object for the eyepiece. The distance from the objective to the eyepiece is \( L \) which we will find later. For now, we can assume that the image formed by the objective is at a distance \( V_0 \) from the objective, thus:
\[ U_e = L - V_0 \]
Using the lens formula for the eyepiece:
\[ \frac{1}{f_e} = \frac{1}{V_e} - \frac{1}{U_e} \]
The image distance \( V_e \) for the eyepiece is the least distance of distinct vision \( D \):
\[ \frac{1}{5} = \frac{1}{D} - \frac{1}{U_e} \]
Substituting \( D = 25 \, \text{cm} \):
\[ \frac{1}{5} = \frac{1}{25} - \frac{1}{U_e} \]
Rearranging gives:
\[ \frac{1}{U_e} = \frac{1}{25} - \frac{1}{5} \]
Finding a common denominator (25):
\[ \frac{1}{U_e} = \frac{1}{25} - \frac{5}{25} = -\frac{4}{25} \]
Thus,
\[ U_e = -\frac{25}{4} \, \text{cm} \]

### Step 4: Calculate the magnification of the eyepiece
The magnification \( M_e \) produced by the eyepiece is given by:
\[ M_e = \frac{V_e}{U_e} \]
Substituting the values:
\[ M_e = \frac{25}{-\frac{25}{4}} = 4 \]

### Step 5: Calculate the total magnification
The total magnification \( M \) of the telescope is the product of the magnifications of the objective and the eyepiece:
\[ M = M_0 \times M_e = -\frac{1}{3} \times 4 = -\frac{4}{3} \]

### Step 6: Calculate the separation between the objective and the eyepiece
The separation \( L \) between the objective and the eyepiece is given by:
\[ L = V_0 + |U_e| \]
Substituting the values:
\[ L = \frac{200}{3} + \frac{25}{4} \]
Finding a common denominator (12):
\[ L = \frac{800}{12} + \frac{75}{12} = \frac{875}{12} \approx 72.92 \, \text{cm} \]

### Final Answers:
(a) The magnification produced is \( -\frac{4}{3} \).

(b) The separation between the objective and the eyepiece is approximately \( 72.92 \, \text{cm} \).