1Analysis of Light Paths

The source $S$ emits light in all directions. We must analyze two distinct optical paths:

• Path 1 (Inner Beam): Rays passing through Lens A (aperture 2 cm) and then Lens B.
• Path 2 (Outer Beam): Rays missing Lens A but striking Lens B directly (annular region).

2Path 1: Through Lens A then B

For Lens A: $u_1 = -15 \text{ cm}, f_1 = 7.5 \text{ cm}$.

$$ \frac{1}{v_1} – \frac{1}{-15} = \frac{1}{7.5} \implies \frac{1}{v_1} = \frac{2}{15} – \frac{1}{15} = \frac{1}{15} $$

$$ v_1 = 15 \text{ cm} $$

The image $I_1$ forms 15 cm to the right of A. Since Lens B is at 60 cm from the source (45 cm from A), the object distance for Lens B is $u_2 = -(45 – 15) = -30 \text{ cm}$.

For Lens B: $u_2 = -30 \text{ cm}, f_2 = 20 \text{ cm}$.

$$ \frac{1}{v_2} – \frac{1}{-30} = \frac{1}{20} \implies \frac{1}{v_2} = \frac{3}{60} – \frac{2}{60} = \frac{1}{60} $$

$$ v_2 = 60 \text{ cm} $$

The inner beam converges to a point 60 cm behind Lens B.

Beam Radius at B: By similar triangles (image $I_1$ is at midpoint between A and B), the height at B is double the height at A:

$$ r_{inner} = 2 \text{ cm} \times \frac{30}{15} = 4 \text{ cm} $$

3Path 2: Directly to Lens B

Rays passing outside the 2 cm radius of A hit B directly. Object distance $u = -60 \text{ cm}$, $f_2 = 20 \text{ cm}$.

$$ \frac{1}{v’} – \frac{1}{-60} = \frac{1}{20} \implies \frac{1}{v’} = \frac{3}{60} – \frac{1}{60} = \frac{2}{60} $$

$$ v’ = 30 \text{ cm} $$

The outer beam converges to a point 30 cm behind Lens B.

Inner Boundary at B: The innermost rays of this beam are the ones just grazing Lens A. By similar triangles (Source to B is 4x Source to A), the radius at B is:

$$ r_{grazing} = 2 \text{ cm} \times \frac{60}{15} = 8 \text{ cm} $$

4Minimum Spot Size Calculation

We have an inner cone converging to $z=60$ and an outer cone (expanding after its focus) from $z=30$. The spot size is determined by the intersection of these two envelopes.

• Inner Beam Radius: $r_{in}(z) = 4 (1 – \frac{z}{60})$
• Outer Beam Outer Radius ($R=10$ at lens): $r_{out}(z) = 10 (\frac{z}{30} – 1)$ for $z > 30$

Equating them to find the crossover point:

$$ 4 – \frac{4z}{60} = \frac{10z}{30} – 10 $$

$$ 14 = z \left( \frac{1}{3} + \frac{1}{15} \right) = z \left( \frac{6}{15} \right) $$

$$ z = \frac{14 \times 15}{6} = 35 \text{ cm} $$

The radius at this distance is:

$$ r = 4 \left( 1 – \frac{35}{60} \right) = 4 \left( \frac{25}{60} \right) = \frac{5}{3} \text{ cm} $$