A convex lens A of focal length 20cm and a concave lens G of focal length 5cm are kept along the same axis with the distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance d in cm will be

Options

A. 25
B. 15
C. 30
D. 50

Explanation

To solve the problem, we need to determine the distance \( d \) between the convex lens \( A \) and the concave lens \( G \) such that a parallel beam of light entering lens \( A \) exits lens \( G \) as a parallel beam.

### Step-by-Step Solution:

1. **Understanding the Lenses**:
- The focal length of the convex lens \( A \) is \( F_1 = +20 \) cm.
- The focal length of the concave lens \( G \) is \( F_2 = -5 \) cm.

2. **Using the Lens Formula**:
The lens formula for two lenses in combination is given by:
\[ \frac{1}{F} = \frac{1}{F_1} + \frac{1}{F_2} - \frac{d}{F_1 F_2} \]
where \( F \) is the effective focal length of the combination, and \( d \) is the distance between the two lenses.

3. **Setting the Condition for Parallel Beams**:
Since the beam of light leaves lens \( G \) as a parallel beam, the effective focal length \( F \) is \( \infty \) (or \( P = 0 \)). Therefore, we can set:
\[ \frac{1}{F} = 0 \]

4. **Substituting into the Lens Formula**:
Plugging \( F = \infty \) into the lens formula gives:
\[ 0 = \frac{1}{20} + \frac{1}{-5} - \frac{d}{20 \times -5} \]

5. **Calculating the Terms**:
- Calculate \( \frac{1}{20} \) and \( \frac{1}{-5} \):
\[ \frac{1}{20} - \frac{1}{5} = \frac{1}{20} - \frac{4}{20} = -\frac{3}{20} \]
- Substitute this back into the equation:
\[ 0 = -\frac{3}{20} + \frac{d}{100} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{d}{100} = \frac{3}{20} \]

7. **Solving for \( d \)**:
Multiply both sides by 100:
\[ d = 100 \times \frac{3}{20} = 15 \text{ cm} \]

### Final Answer:
Thus, the distance \( d \) between the two lenses is \( 15 \) cm.

---