`P_(1),P_(2)` are the power of convex and concave lens respectively. The distance between these lenses is d. Effective power of the system of lens is:

Options

A. `P_(1) - P_(2) + dP_(1)P_(2)`
B. `P_(1) + P_(2) - dP_(1)P_(2)`
C. `P_(1) + P_(2)-d`
D. `P_(1) - P_(2) - dP_(1)P_(2)`

Answer

To find the effective power of a system of a convex lens and a concave lens separated by a distance \(d\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Powers of the Lenses:**
- Let \(P_1\) be the power of the convex lens (which is positive).
- Let \(P_2\) be the power of the concave lens (which is negative).

2. **Understand the Focal Lengths:**
- The focal length \(f_1\) of the convex lens is positive, and it is related to its power by the formula:
\[ P_1 = \frac{1}{f_1} \]
- The focal length \(f_2\) of the concave lens is negative, and it is related to its power by the formula:
\[ P_2 = \frac{1}{f_2} \]

3. **Use the Lens Formula for Effective Power:**
- The formula for the effective focal length \(F\) of two lenses separated by a distance \(d\) is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]

4. **Substitute the Values of Powers:**
- Since \(P_1 = \frac{1}{f_1}\) and \(P_2 = \frac{1}{f_2}\), we can rewrite the equation as:
\[ \frac{1}{F} = P_1 + P_2 - \frac{d}{f_1 f_2} \]

5. **Express \(f_1 f_2\) in Terms of Powers:**
- We know that:
\[ f_1 = \frac{1}{P_1} \quad \text{and} \quad f_2 = \frac{1}{P_2} \]
- Therefore:
\[ f_1 f_2 = \frac{1}{P_1} \cdot \frac{1}{P_2} = \frac{1}{P_1 P_2} \]

6. **Substitute \(f_1 f_2\) Back into the Equation:**
- Now, substituting \(f_1 f_2\) into the effective power equation gives:
\[ \frac{1}{F} = P_1 + P_2 - d \cdot P_1 P_2 \]

7. **Find the Effective Power \(P\):**
- The effective power \(P\) is given by:
\[ P = \frac{1}{F} \]
- Thus:
\[ P = P_1 + P_2 - d \cdot P_1 P_2 \]

### Final Result:
The effective power of the system of lenses is:
\[ P = P_1 + P_2 - d \cdot P_1 P_2 \]