Distinguish a virtual image from the focal-plane limit

Move the object inside the focal distance

Keep f = +20 cm and set dₒ = 10 cm. The outgoing rays spread apart rather than meeting on the right. Extend their straight outgoing lines backward: the dashed constructions meet at x = −20 cm and height +1.00 cm. The image is virtual, upright and enlarged, with dᵢ = −20 cm and m = +2.

Actual transmitted light still travels right. Dashed lines are geometric extensions, not rays sent backward by the lens. At the apparent image point, the whole transmitted bundle is not physically converging. A screen placed there cannot catch this image. An observer looking through the lens can perceive a virtual image because the eye receives diverging rays and focuses them; the eye itself is not simulated here.

A central backward extension can overlap an actual incoming ray segment. That coincidence does not make the image real: the classification concerns whether the outgoing bundle physically meets at the image point, not whether any single ray crosses that coordinate.

Compare three positions without skipping the middle case

Capture dₒ = 10, 20 and 30 cm at f = +20 cm. At 10 cm the formula gives 20 × 10/(10 − 20) = −20 cm. At 30 cm it gives +60 cm. The signs agree with the virtual and real ray constructions.

At dₒ = f = 20 cm, the denominator is zero. The equation gives 1/dᵢ = 0, which represents an image at infinity in the ideal limit. The browser instead reports no finite image and stores no finite image distance, height or linear magnification. It does not report zero image distance or zero magnification.

Inspect the slope table: all three outgoing slopes are −0.025 cm/cm for the pictured object tip. The rays are parallel to each other, but tilted relative to the axis because the tip is above it. A point on the optical axis in the focal plane would produce an axis-parallel bundle. Different points on an extended object produce bundles in different directions.

Understand the limiting behavior

For a converging lens, dᵢ = fdₒ/(dₒ − f). As the real object’s distance approaches f from above, the denominator is small and positive and the real image lies increasingly far right. From below, the denominator is small and negative and the virtual image lies increasingly far left. This is a limit, not a finite image jumping through the lens.

The ten-centimetre slider deliberately samples a small grid. It does not resolve arbitrarily close approaches to focus, and the diagram has a fixed finite window. The distance sweep includes explicit no-finite-value entries instead of fabricating a plotted image at the window boundary.

Explain what “magnifying glass” adds

An upright enlarged virtual image is part of a simple magnifier’s operation, but the model’s m is linear magnification. Angular magnification also depends on eye position, viewing distance and comparison conditions. Do not claim that m = 2 here means an object must look twice as large to every observer.

The positive focal-length label alone cannot tell you the image type. You need the real object’s position relative to the focal plane. Compare a diverging lens to see another virtual-image case with different size behavior.

Transfer across a new focal plane

For f = +30 cm and object height +0.50 cm, compare dₒ = 20 cm with dₒ = 30 cm. Predict distance, orientation and whether a finite screen image exists.

Check the focal-plane transfer

At 20 cm, dᵢ = 30 × 20/(20 − 30) = −60 cm, m = +3 and hᵢ = +1.50 cm: an upright enlarged virtual image. At 30 cm the outgoing bundle from each object point is parallel; no finite image distance, height or linear magnification is assigned. Neither case gives a finite real screen image in this single-lens arrangement.

OpenStax: thin lenses and focal planes provides the ideal-model context. The fixed window and separate no-finite-image state are explicit teaching choices here.