Physics · Mirrors and Lenses

Lens Imaging

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Wrap-up

What you learned

Recommended next: Open concept testCheck whether the core ideas are ready without leaving this concept.
Read next: MirrorsCompare signed image rules

Key takeaway

Positive d_i means the refracted rays really meet on the far side, so the image is real and screen-ready.
Negative d_i means only backward extensions meet on the object side, so the image is virtual.
The sign of m tells upright versus inverted, while |m| tells how large the image is compared with the object.

Common misconception

A diverging lens still forms an image; it is virtual, upright, and reduced because the refracted rays only appear to meet when extended backward.

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

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Compare with mirrorsMirrors and Lenses

Mirrors

Use plane, concave, and convex mirrors to track equal-angle reflection, signed image distance, and magnification on the same live ray diagram.

See diffraction limitMirrors and Lenses

Optical Resolution / Imaging Limits

Image two nearby point sources through one finite aperture and see why diffraction, wavelength, and aperture diameter limit how sharply an optical system can separate them.

Boundary bending lawOptics

Refraction / Snell's Law

Watch one light ray cross a boundary, connect refractive index to speed change, and see Snell's law set the refracted angle, bending direction, and critical-angle limit on the same live diagram.

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Reference and support

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Equation snapshotSigned lens story · MagnificationShow 3 equations

Keep the image-distance sign and the magnification sign visible while you compare the ray diagram.

Signed lens story
$\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}$
Relates signed focal length, object distance, and image distance for one thin lens.
Magnification
$m = \frac{h _{i}}{h _{o}} = - \frac{d _{i}}{d _{o}}$
The sign tells whether the image is upright or inverted, and the magnitude tells the size ratio.
Sign rule
$d_{i} > 0 \Rightarrow real image on the far side$
Positive image distance means the refracted rays really meet on the far side, so the image can be projected onto a screen.

Why it behaves this way

Explanation

Thin-lens imaging starts with one object point. Rays from that point pass through the lens and are redirected. If the refracted rays really meet on the far side, the image point is real. If the rays spread after the lens but their backward extensions meet on the object side, the image point is virtual.

The ray diagram and the thin-lens equation are two views of the same geometry. On this page, changing lens type, focal length, object distance, or object height updates the stage, the signed image distance, and the magnification together. That lets you connect where the image forms, whether it is upright or inverted, and how large it is without treating those as separate facts.

Key ideas

01For a converging lens, an object outside the focal length gives a real inverted image, while an object inside the focal length gives a virtual upright image.

02A diverging lens still forms an image for a real object on the left, but that image is always virtual, upright, and reduced.

03The sign of the image distance tells which side of the lens the image is on: positive for a real image on the far side, negative for a virtual image on the object side.

04The sign of magnification tells orientation, while its magnitude tells how much the image size has been scaled.

Worked examples

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Frozen walkthrough

Step through the frozen example

Frozen walkthrough

These examples use the current lens, object position, and object height from the live bench, so each calculation matches the ray diagram you are actually viewing.

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Example 1 of 2

Frozen valuesUsing frozen parameters

For the current converging lens, what signed image distance does the thin-lens equation predict, and what does the sign mean?

f

Focal length

0.8 m

d_{o}

Object distance

2.4 m

1. Start with the thin-lens equation

Use

\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}

2. Solve for the signed image distance

\frac{1}{d _{i}} = \frac{1}{f} - \frac{1}{d _{o}} = 1.25 - 0.42 = 0.83

3. Interpret the sign of the result

d_{i} = 1.2 m

Signed image-distance result

d_{i} = 1.2 m

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.

Quick test

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Accessibility

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The simulation shows a thin lens at the center of the axis, an object arrow on the left, and an image arrow that moves according to the signed thin-lens equation. Depending on the setup, the image 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 gives orientation and the magnitude gives the image-size scale without leaving the ray diagram.

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