Locate thin-lens images with rays and signed distances

Establish the model and evidence

A converging thin lens has positive focal length under the real-is-positive convention; a diverging lens has negative focal length. In the simple surrounding medium assumed here, parallel incoming rays converge to a focal point for the converging lens or diverge as though from a focal point on the incident side for the diverging lens. A lens has focal behavior on both sides.

For a real object left of the lens, use o>0 and 1/f=1/o+1/i. Unlike the mirror convention’s spatial placement, positive lens image distance is on the transmitted right side. Negative i places a virtual image on the incident left side. Magnification M=−i/o again separates upright positive height from inverted negative height.

The parallel principal ray is redirected through the relevant focal point or appears to originate from it. A ray through the center of the ideal thin lens continues straight. Their actual intersection locates a real image; an intersection only of backward extensions locates a virtual image. Rays are a construction tool, not a claim that only two rays leave an object.

A converging lens with the object inside its focal length gives an upright magnified virtual image, the basic magnifier case. At the focal object location, transmitted rays are parallel and no finite image coordinate exists. A diverging lens with a real object gives an upright reduced virtual image. Thin/paraxial assumptions omit aberration, finite thickness and diffraction, so geometric focus does not certify a perfect real image.

Predict and sketch a third principal ray: a ray through the incident-side focus of a converging lens emerges parallel to the axis. For a diverging lens, an incident ray directed toward the far-side focus emerges parallel. Combine this focal-directed ray with the parallel and central rays already displayed, then check their common image point or backward-extension intersection. With an object inside a converging focal distance, extend the incoming line backward to the incident-side focus to construct the ray; do not mistake that extension for a physical light path.

Worked evidence

For f=+10 cm and o=15 cm, i=+30 cm and M=−2: a real inverted doubled image. At o=5 cm, i=−10 cm and M=+2: a virtual upright doubled image. For f=−10 cm and o=15 cm, i=−6 cm and M=+0.4.

Predict and investigate

Can a converging lens produce a virtual image even though it converges parallel incident rays?

Compare converging cases o=5, 10 and 15 cm with the diverging case at 15 cm. Explain physical rays versus dashed extensions and why the focal-limit state cannot be replaced by a finite screen position.

Transfer to changed conditions

A diverging lens with f=−15 cm has a real object 30 cm away. Find i and M and explain whether a screen on the right can directly capture a sharp image.

Compare the worked transfer

i=−10 cm and M=+1/3. The image is virtual, upright and reduced on the incident side; a right-side screen does not directly capture that virtual image without additional optics.

Explain the boundary

How do lens and mirror positive-image conventions map differently onto the global x coordinate used in the diagrams?

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. Original text, rays and computed/synthetic records, not completed optical measurements. All investigations are virtual: no lasers, intense-light viewing, sunlight focusing, flames or eye tests are requested. Any real practical work needs qualified supervision and independent safety review.