A convex lens of focal length 20 cm and a concave lens of focal length 10 cm are placed 10 cm apart with their principal axes coinciding. A beam of light travelling parallel to the principal axis and having a beam diameter 5.0 mm, is incident on the combination. Show that the emergent beam is parallel to the incident one. Find the beam diameter of the emergent beam.

Explanation

To solve the problem step by step, we will analyze the optical system consisting of a convex lens and a concave lens placed 10 cm apart.

### Step 1: Understand the Configuration
We have:
- A convex lens (focal length \( f_1 = +20 \) cm)
- A concave lens (focal length \( f_2 = -10 \) cm)
- The distance between the two lenses is \( d = 10 \) cm.

### Step 2: Draw the Ray Diagram
1. Draw the principal axis.
2. Mark the position of the convex lens and the concave lens.
3. Indicate the focal points of both lenses:
- For the convex lens, the focal point is at \( +20 \) cm from its center.
- For the concave lens, the focal point is at \( -10 \) cm from its center (to the left of the lens).

### Step 3: Determine the Behavior of Light Rays
A beam of light is incident parallel to the principal axis:
- The rays will converge after passing through the convex lens.
- The rays will then diverge after passing through the concave lens.

### Step 4: Calculate the Image Position after the Convex Lens
Since the incoming rays are parallel to the principal axis, they will converge at the focal point of the convex lens:
- The focal point of the convex lens is at \( 20 \) cm from the lens.
- The distance from the convex lens to the concave lens is \( 10 \) cm, so the image formed by the convex lens is at \( 20 - 10 = 10 \) cm to the right of the concave lens.

### Step 5: Calculate the Effective Focal Length of the System
To find the emergent beam direction, we need to find the effective focal length of the combination of the two lenses:
Using the lens formula:
\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{20} + \frac{1}{-10} = \frac{1}{20} - \frac{2}{20} = \frac{-1}{20} \]
Thus, the effective focal length \( f = -20 \) cm.

### Step 6: Show that the Emergent Beam is Parallel
Since the effective focal length is negative, the system behaves like a diverging lens. The emergent rays from the concave lens will be parallel to the incoming rays, confirming that the emergent beam is parallel to the incident one.

### Step 7: Find the Beam Diameter of the Emergent Beam
1. The diameter of the incident beam is \( 5.0 \) mm, which means the radius \( r = \frac{5.0}{2} = 2.5 \) mm.
2. The emergent beam will maintain the same diameter as the incident beam since the system is designed to keep the rays parallel.

Thus, the diameter of the emergent beam is also \( 5.0 \) mm.

### Final Answer:
The emergent beam is parallel to the incident beam, and the beam diameter of the emergent beam is \( 5.0 \) mm.

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