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, \( F_o = 50 \, \text{cm} \)
- Focal length of the eyepiece, \( F_e = 5 \, \text{cm} \)
- Least distance of distinct vision, \( D = 25 \, \text{cm} \)
- Distance of the scale from the objective, \( U_o = -200 \, \text{cm} \) (since it is a virtual object)

### Step 1: Calculate the image distance for the objective lens
Using the lens formula for the objective:
\[ \frac{1}{V_o} - \frac{1}{U_o} = \frac{1}{F_o} \]
Substituting the values:
\[ \frac{1}{V_o} - \frac{1}{-200} = \frac{1}{50} \]
\[ \frac{1}{V_o} + \frac{1}{200} = \frac{1}{50} \]
Now, let's find a common denominator and solve for \( V_o \):
\[ \frac{1}{V_o} = \frac{1}{50} - \frac{1}{200} \]
\[ \frac{1}{V_o} = \frac{4 - 1}{200} = \frac{3}{200} \]
Thus,
\[ V_o = \frac{200}{3} \, \text{cm} \]

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

### Step 3: Calculate the image distance for the eyepiece
Using the lens formula for the eyepiece:
\[ \frac{1}{V_e} - \frac{1}{U_e} = \frac{1}{F_e} \]
Where \( U_e \) is the object distance for the eyepiece. Since the image from the objective acts as the object for the eyepiece, we have:
\[ U_e = -V_o = -\frac{200}{3} \, \text{cm} \]
Now substituting the values:
\[ \frac{1}{V_e} - \frac{1}{-\frac{200}{3}} = \frac{1}{5} \]
\[ \frac{1}{V_e} + \frac{3}{200} = \frac{1}{5} \]
Finding a common denominator:
\[ \frac{1}{V_e} = \frac{1}{5} - \frac{3}{200} \]
\[ \frac{1}{V_e} = \frac{40 - 3}{200} = \frac{37}{200} \]
Thus,
\[ V_e = \frac{200}{37} \, \text{cm} \]

### Step 4: Calculate the magnification produced by the eyepiece
The magnification \( M_e \) produced by the eyepiece is given by:
\[ M_e = \frac{V_e}{U_e} \]
Substituting the values:
\[ M_e = \frac{\frac{200}{37}}{-\frac{200}{3}} = -\frac{3}{37} \]

### 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:
\[ M = M_o \times M_e \]
Substituting the values:
\[ M = \left(-\frac{1}{3}\right) \times \left(-\frac{3}{37}\right) = \frac{1}{37} \]

### Step 6: Calculate the separation between the objective and eyepiece
The separation \( L \) between the objective and eyepiece is given by:
\[ L = V_o + |U_e| \]
Substituting the values:
\[ L = \frac{200}{3} + \frac{200}{3} = \frac{400}{3} \approx 133.33 \, \text{cm} \]

### Final Answers:
(a) Magnification produced: \( \frac{1}{37} \)
(b) Separation between objective and eyepiece: \( \frac{400}{3} \, \text{cm} \approx 133.33 \, \text{cm} \)