A diverging lens (f = −4.0 cm) is positioned 2.0 cm to the left of a converging lens (f = +6.0 cm). A 1.0 mm diameter beam of parallel light rays is incident on the diverging lens from the left. After leaving the converging lens, the outgoing rays

Question

BlackTom AI · Accepted Answer

Parallel rays incident on a diverging lens emerge as if they originate from the lens’s focal point on the incident side. Key concepts: thin-lens equation, virtual images, focal point behavior for diverging lenses, and simple ray-triangle geometry for small angles.

1) Relevant knowledge points
- For a lens with focal length f, parallel incoming rays are focused (or appear to come from) the focal point: for a diverging lens (f = −4.0 cm) parallel rays produce a virtual image at 4.0 cm to the left of that lens.
- That virtual image serves as the object for the second (converging) lens. If that object lies at the converging lens’s focal plane (s = f), the converging lens produces a collimated (parallel) output beam.
- Beam diameter changes between lenses according to ray geometry: a ray at height y0 leaving the first lens with an angle θ such that backward extension meets the focal point satisfies θ ≈ y0/|f1| (small-angle approximation). Over a separation d the height grows by θ d.

2) Step-by-step reasoning and calculations
- Place diverging lens (L1) at x = 0, converging lens (L2) at x = +2.0 cm. L1’s virtual image (from parallel input) is at x = −4.0 cm (since s'1 = f1 = −4.0 cm).
- Distance from that virtual image to L2 = 2.0 cm + 4.0 cm = 6.0 cm, so the object distance for L2 is s2 = 6.0 cm, which equals f2 = +6.0 cm.
- Because the object for L2 lies at its focal plane, L2 produces an outgoing parallel beam (collimated).

3) Intermediate geometric calculation for diameter
- Take a marginal ray entering L1 at height y0 = 0.50 mm (half of 1.0 mm diameter). After L1 the ray angle θ ≈ y0/4.0 cm.
- At L2 (2.0 cm downstream) the ray height is y2 = y0 + θ·(2.0 cm) = y0(1 + 2.0/4.0) = 1.5 y0.
- Thus beam diameter at L2 is 1.5 × 1.0 mm = 1.5 mm. The converging lens turns these rays parallel but does not reduce that physical beam diameter, so the outgoing beam is parallel with diameter D = 1.5 mm > 1.0 mm.

4) Why the correct answer is correct
- The converging lens’s object at its focal plane guarantees collimation; the intermediate divergence from L1 increases the beam radius by factor 1.5 before L2, so the final parallel beam has larger diameter.

5) Why other options are wrong (brief)
- “Converge”: would require the object for L2 to be farther than f2 (not the case).
- “Diverge”: would result if the object for L2 were inside its focal length (not the case).
- “Form a parallel beam of diameter D < 1.0 mm” contradicts the geometric expansion computed above.
- “Travel back toward the light” is not produced by two thin lenses in this arrangement.

Conclusion: The outgoing rays form a parallel beam with diameter D > 1.0 mm (specifically 1.5 mm), confirming the provided answer.

A diverging lens (f = −4.0 cm) is positioned 2.0 cm to the left of a converging lens (f = +6.0 cm). A 1.0 mm diameter beam of parallel light rays is incident on the diverging lens from the left. After leaving the converging lens, the outgoing rays单项选择题

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