A convex lens of focal length `25 cm` and a concave lens of focal length `20 cm` are mounted coaxially separated by a distance `d cm` . If the power of the combination is zero, `d` is equal to

Options

A. `45`
B. `30`
C. `15`
D. `5`

Explanation

To solve the problem, we need to find the distance \( d \) between a convex lens and a concave lens such that the power of the combination is zero.

### Step-by-Step Solution:

1. **Identify the Focal Lengths**:
- The focal length of the convex lens (\( f_1 \)) is given as \( +25 \) cm.
- The focal length of the concave lens (\( f_2 \)) is given as \( -20 \) cm (negative because it is a concave lens).

2. **Understand the Power of Lenses**:
- The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{f} \]
- For a combination of lenses, the total power \( P \) is the sum of the powers of individual lenses.

3. **Combined Focal Length Formula**:
- The formula for the combined focal length \( f \) of two lenses separated by a distance \( d \) is:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]

4. **Set the Total Power to Zero**:
- Since the total power is zero, we have:
\[ P = 0 \implies \frac{1}{f} = 0 \]
- This implies that:
\[ \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} = 0 \]

5. **Substituting the Values**:
- Substitute \( f_1 = 25 \) cm and \( f_2 = -20 \) cm into the equation:
\[ \frac{1}{25} + \frac{1}{-20} - \frac{d}{25 \times (-20)} = 0 \]

6. **Finding a Common Denominator**:
- The common denominator for \( \frac{1}{25} \) and \( \frac{1}{-20} \) is \( 100 \):
\[ \frac{4}{100} - \frac{5}{100} - \frac{d}{-500} = 0 \]
- This simplifies to:
\[ -\frac{1}{100} + \frac{d}{500} = 0 \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{d}{500} = \frac{1}{100} \]

8. **Solving for \( d \)**:
- Multiply both sides by \( 500 \):
\[ d = 500 \times \frac{1}{100} = 5 \text{ cm} \]

### Conclusion:
The distance \( d \) between the convex lens and the concave lens such that the power of the combination is zero is \( 5 \) cm.