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

Options

A. (a)`+2D`
B. (b)`-2D`
C. (c)`-3D`
D. (d)`+3D`

Answer

To solve the problem of finding the equivalent power of two convex lenses separated by a distance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Power of the first lens, \( P_1 = 2 \, \text{D} \)
- Power of the second lens, \( P_2 = 3 \, \text{D} \)
- Distance between the lenses, \( d = \frac{1}{3} \, \text{m} \)

2. **Use the Formula for Equivalent Power:**
The formula for the equivalent power \( P_{eq} \) of two lenses separated by a distance \( d \) is given by:
\[ P_{eq} = P_1 + P_2 - \frac{d \cdot P_1 \cdot P_2}{f_1 \cdot f_2} \]
However, since we are using powers directly, we can simplify it to:
\[ P_{eq} = P_1 + P_2 - d \cdot P_1 \cdot P_2 \]

3. **Substitute the Values:**
Substitute the known values into the equation:
\[ P_{eq} = 2 + 3 - \left(\frac{1}{3} \cdot 2 \cdot 3\right) \]

4. **Calculate the Product Term:**
Calculate \( \frac{1}{3} \cdot 2 \cdot 3 \):
\[ \frac{1}{3} \cdot 2 \cdot 3 = \frac{6}{3} = 2 \]

5. **Complete the Calculation:**
Now substitute this back into the equation:
\[ P_{eq} = 2 + 3 - 2 = 3 \]

6. **Final Result:**
Therefore, the equivalent power of the optical system is:
\[ P_{eq} = 3 \, \text{D} \]

### Conclusion:
The power of the optical system formed by the two lenses is \( +3 \, \text{D} \).

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