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Concept module

Lens Imaging

Trace principal rays through converging and diverging lenses, connect the signed thin-lens equation to the diagram, and watch image distance and magnification respond to the same object setup.

Interactive lab

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Stable links

Why it behaves this way

Explanation

Thin-lens imaging works when many rays from one object point leave the lens in a pattern that either meets at one image point or appears to meet there when you extend the rays backward. The ray diagram and the thin-lens equation are two views of that same geometry.

This module keeps the setup compact on purpose. You change lens type, focal length, object distance, and object height, then the signed image distance, image orientation, and magnification update together on the stage and in the response graphs.

Key ideas

01A converging lens can make a real inverted image when the object is outside the focal length, but it flips to a virtual upright image when the object moves inside the focal length.
02A diverging lens always makes a virtual upright reduced image for a real object on the left.
03The thin-lens equation gives the signed image distance, and magnification tells both the image size ratio and whether the image is upright or inverted.

Frozen walkthrough

Step through the frozen example

Frozen walkthrough
Solve the current lens setup, not a detached worksheet. The substitutions follow the live controls, and the same signed values appear in the ray diagram and the graphs.

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View plans
Frozen valuesUsing frozen parameters

For the current converging lens, what signed image distance follows from the thin-lens equation?

Signed focal length

0.8 m

Object distance

2.4 m

1. Start from the thin-lens relation

Use .

2. Rearrange for the signed image distance

.

3. Invert the result

So .

Signed image distance

The positive image distance means the refracted rays actually meet on the far side of the lens, so the image can be projected onto a screen.

Common misconception

A diverging lens spreads light out, so it does not make an image at all.

A diverging lens still makes an image. The refracted rays separate, but their backward extensions meet at a virtual image on the object side.

That is why the image is upright, reduced, and cannot be projected onto a screen even though it is still a legitimate image point in the geometry.

Mini challenge

If you move a converging-lens object inward toward the focal point, what does the ray diagram predict before the object crosses inside the focus?

Make a prediction before you reveal the next step.

Decide what happens to image distance and image size.

Check your reasoning against the live bench.

The real image moves farther away and grows in magnitude.
As the object approaches the focal point from outside, the refracted rays need more distance to meet. The signed image distance shoots outward and the magnitude of the magnification grows.

Quick test

Reasoning

Question 1 of 4

Use the sign of d_i, the sign of m, and the ray behavior together. The goal is to reason from the live lens model, not to recite a slogan.

Which setup can place the image on a screen to the right of the lens?

Use the live bench to test the result before moving on.

Accessibility

The simulation shows a thin lens at the center of the principal axis, an object arrow to the left, and an image arrow that moves according to the signed thin-lens equation. Depending on the setup, the image arrow appears on the far side as a real inverted image or on the object side as a virtual upright image.

Optional overlays show the focal markers, the principal rays, and the distance-and-height guide used in magnification.

Graph summary

The object-image graph plots signed image distance against object distance for the current lens family and focal length.

The magnification graph plots m against object distance, so the sign and magnitude of the image scaling are visible without leaving the ray diagram.