Thin Lenses & Image Formation

You’ll be able to

Distinguish converging from diverging lenses and their focal lengths
Apply the thin-lens equation and magnification to locate and describe images
Predict the image formed by a converging lens for different object positions

Converging and diverging lenses

A converging (convex) lens is thicker in the middle; it bends parallel rays inward to a real focal point, and its focal length f is positive. A diverging (concave) lens is thinner in the middle; it spreads parallel rays so they appear to come from a focal point on the incoming side, and its focal length is negative. Lenses obey the same equation as mirrors, but here a real image forms on the opposite side of the lens from the object — the side the light actually reaches.

Thin-lens equation and magnification

1/f = 1/d_o + 1/d_i · m = − d_i / d_o

For a converging lens f > 0; for a diverging lens f < 0. A positive d_i is a real image on the far side of the lens; a negative d_i is a virtual image on the same side as the object.

How the image changes with object distance

For a converging lens, the image depends on where the object sits relative to the focal length. Beyond 2f, the image is real, inverted, and reduced (a camera). At exactly 2f, it is real, inverted, and the same size. Between f and 2f, real, inverted, and enlarged (a projector). Inside the focal length (object closer than f), the image flips to virtual, upright, and enlarged — this is the magnifying-glass mode. A diverging lens, by contrast, always gives a virtual, upright, reduced image.

Worked example

An object is placed 20 cm from a converging lens of focal length 10 cm. Find the image distance and describe the image.

1.Thin-lens equation: 1/d_i = 1/f − 1/d_o = 1/10 − 1/20.
2.Common denominator: 1/d_i = 2/20 − 1/20 = 1/20.
3.Invert: d_i = 20 cm (positive → real image on the far side).
4.Magnification: m = −d_i/d_o = −20/20 = −1 (inverted, same size).

Answer: d_i = 20 cm; the image is real, inverted, and the same size as the object (m = −1) — the object sits at exactly 2f.

Checkpoint

An object is placed 20 cm from a converging lens with a focal length of 10 cm. What is the image distance?

Watch out

Mirrors and lenses use the same equation but opposite geometry for real images: a real mirror image forms in front (same side as the object), while a real lens image forms on the far side (opposite the object). Keep the two straight when interpreting the sign of d_i.

Checkpoint

A diverging lens forms an image of a real object. That image is always:

On the exam

Learn the converging-lens cases by object position (beyond 2f, at 2f, between f and 2f, inside f) — the exam tests all four. And remember the shortcut: diverging lenses always give virtual, upright, diminished images, no computation needed.

Answer the 2 checkpoints as you read.