Two convex lesses of power 2 D and 5 D are separated by a distance `1/3m` m The power of the optical system formed is

Options

A. `+2D`
B. `-2D`
C. `-3D`
D. `+3D`

Answer

To find the power of the optical system formed by two convex lenses separated by a distance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Powers of the Lenses**:
- Let the power of the first lens \( P_1 = 2D \)
- Let the power of the second lens \( P_2 = 5D \)

2. **Convert Power to Focal Length**:
- The focal length \( f \) of a lens is related to its power \( P \) by the formula:
\[ P = \frac{1}{f} \quad \Rightarrow \quad f = \frac{1}{P} \]
- For the first lens:
\[ f_1 = \frac{1}{P_1} = \frac{1}{2} \text{ m} \]
- For the second lens:
\[ f_2 = \frac{1}{P_2} = \frac{1}{5} \text{ m} \]

3. **Use the Formula for Combined Focal Length**:
- When two lenses are separated by a distance \( d \), the formula for the combined focal length \( f \) is given by:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]
- Here, \( d = \frac{1}{3} \text{ m} \).

4. **Substitute the Values**:
- Substitute \( f_1 \), \( f_2 \), and \( d \) into the formula:
\[ \frac{1}{f} = \frac{1}{\frac{1}{2}} + \frac{1}{\frac{1}{5}} - \frac{\frac{1}{3}}{\left(\frac{1}{2}\right)\left(\frac{1}{5}\right)} \]
- This simplifies to:
\[ \frac{1}{f} = 2 + 5 - \frac{\frac{1}{3}}{\frac{1}{10}} = 2 + 5 - \frac{10}{3} \]

5. **Calculate the Result**:
- Combine the terms:
\[ \frac{1}{f} = 7 - \frac{10}{3} = \frac{21}{3} - \frac{10}{3} = \frac{11}{3} \]
- Therefore, the combined focal length \( f \) is:
\[ f = \frac{3}{11} \text{ m} \]

6. **Calculate the Power of the Optical System**:
- Now, calculate the power of the optical system:
\[ P = \frac{1}{f} = \frac{11}{3} \text{ D} \approx 3.67 \text{ D} \]
- Since we are looking for the nearest whole number, we can round it to \( 4D \).

### Final Answer:
The power of the optical system formed is approximately \( 4D \).