A convex lens of focal length 20cm and a concave lens of focal length f are mounted coaxially 5cm apart. Parallel beam of light incident on the convex lens emerges from the concave lens as a parallel beam. Then, f in cm is

Options

A. 35
B. 25
C. 20
D. 15

Explanation

To solve the problem, we need to find the focal length \( f \) of the concave lens such that a parallel beam of light incident on the convex lens emerges as a parallel beam after passing through the concave lens.

### Step-by-Step Solution:

1. **Identify the Focal Lengths:**
- The focal length of the convex lens \( f_1 = +20 \) cm (positive because it is a converging lens).
- The focal length of the concave lens \( f_2 = f \) (we need to find this value).

2. **Understand the Configuration:**
- The convex lens and concave lens are placed 5 cm apart.
- A parallel beam of light entering the convex lens will converge to a point after passing through it.

3. **Condition for Parallel Emergence:**
- For the light to emerge as a parallel beam after passing through the concave lens, the effective focal length of the combined lens system must be infinite. This means that the rays should not converge or diverge after passing through both lenses.

4. **Use the Lens Formula:**
- The formula for two lenses in contact (or coaxially aligned) is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} \]
where \( d \) is the distance between the two lenses (5 cm in this case).

5. **Set the Effective Focal Length to Infinity:**
- Since the final focal length \( F \) must be infinite, we set:
\[ \frac{1}{F} = 0 \]
Therefore, we have:
\[ 0 = \frac{1}{20} + \frac{1}{f} - \frac{5}{20f} \]

6. **Rearranging the Equation:**
- Rearranging gives:
\[ \frac{1}{f} - \frac{5}{20f} = -\frac{1}{20} \]
- Simplifying the left side:
\[ \frac{1}{f} - \frac{1}{4f} = -\frac{1}{20} \]
- Combine the fractions:
\[ \frac{4 - 1}{4f} = -\frac{1}{20} \]
\[ \frac{3}{4f} = -\frac{1}{20} \]

7. **Cross-Multiplying:**
- Cross-multiplying gives:
\[ 3 \cdot 20 = -4f \]
\[ 60 = -4f \]
- Solving for \( f \):
\[ f = -\frac{60}{4} = -15 \text{ cm} \]

8. **Final Answer:**
- The focal length of the concave lens is \( f = -15 \) cm. The magnitude is \( 15 \) cm, and since it is a concave lens, it is negative.

### Summary:
The focal length \( f \) of the concave lens is \( -15 \) cm.