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 find the distance \( d \) between a convex lens A and a concave lens G such that a parallel beam of light entering lens A exits lens B as a parallel beam.

### Step-by-Step Solution:

1. **Identify the Focal Lengths:**
- The focal length of the convex lens A, \( f_1 = +20 \) cm (positive because it is a convex lens).
- The focal length of the concave lens G, \( f_2 = -5 \) cm (negative because it is a concave lens).

2. **Understand the Condition for Parallel Beams:**
- For the light beam to exit as a parallel beam after passing through both lenses, the effective focal length \( F \) of the combination must be infinite. This means that the lenses must be arranged such that their combined effect cancels out any convergence or divergence.

3. **Use the Lens Formula:**
- The lens formula for two lenses in contact is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]
- Since we want \( F = \infty \), we set \( \frac{1}{F} = 0 \):
\[ 0 = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]

4. **Substituting the Values:**
- Substitute \( f_1 = 20 \) cm and \( f_2 = -5 \) cm into the equation:
\[ 0 = \frac{1}{20} + \frac{1}{-5} - \frac{d}{20 \cdot (-5)} \]
- Simplifying \( \frac{1}{-5} \) gives \( -\frac{1}{5} = -\frac{4}{20} \):
\[ 0 = \frac{1}{20} - \frac{4}{20} - \frac{d}{-100} \]
- This simplifies to:
\[ 0 = -\frac{3}{20} + \frac{d}{100} \]

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

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

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