The focal length of objective and eye-piece of a compound microscope are 1 cm and 5 cm respectively. The microscope magnification is equal to 50. If the distance between two lenses is increased by 2 cm then the magnification is

Options

A. 31
B. 60
C. 16
D. 83

Explanation

To solve the problem, we need to find the new magnification of a compound microscope when the distance between the objective and eyepiece lenses is increased by 2 cm.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Focal length of the objective lens, \( f_o = 1 \, \text{cm} \)
- Focal length of the eyepiece lens, \( f_e = 5 \, \text{cm} \)
- Initial magnification, \( M_1 = 50 \)
- Increase in distance between lenses, \( \Delta L = 2 \, \text{cm} \)

2. **Understand the Magnification Formula:**
The magnification \( M \) of a compound microscope is given by the formula:
\[ M = \frac{L \cdot D}{f_o \cdot f_e} \]
where \( L \) is the distance between the lenses and \( D \) is the near point distance (usually taken as 25 cm for normal vision).

3. **Calculate Initial Distance \( L_1 \):**
Rearranging the magnification formula for the initial condition:
\[ M_1 = \frac{L_1 \cdot D}{f_o \cdot f_e} \]
Plugging in the values:
\[ 50 = \frac{L_1 \cdot 25}{1 \cdot 5} \]
Simplifying gives:
\[ 50 = \frac{L_1 \cdot 25}{5} \]
\[ 50 = 5L_1 \]
\[ L_1 = 10 \, \text{cm} \]

4. **Calculate New Distance \( L_2 \):**
When the distance is increased by 2 cm:
\[ L_2 = L_1 + 2 = 10 + 2 = 12 \, \text{cm} \]

5. **Calculate New Magnification \( M_2 \):**
Now, we can find the new magnification using the new distance \( L_2 \):
\[ M_2 = \frac{L_2 \cdot D}{f_o \cdot f_e} \]
Plugging in the values:
\[ M_2 = \frac{12 \cdot 25}{1 \cdot 5} \]
Simplifying gives:
\[ M_2 = \frac{300}{5} = 60 \]

6. **Conclusion:**
The new magnification \( M_2 \) when the distance between the lenses is increased by 2 cm is:
\[ \boxed{60} \]