Lessons · optics

The diverging lens: one image, always virtual

Regime 3 — an algebra-only domain, and here the honest second register is geometry, not calculus: the ray diagram. The same image distance falls out of similar triangles, and SymPy proves the three principal-ray constructions agree — which IS the thin-lens equation.

Regime 3 · algebra-only Math machine-derived & checked by SymPy 3 modeling assumptions (author-asserted)

A diverging lens (concave — thinner in the middle) with focal length $f = -1$ m. The focal length is negative because parallel rays leave it spreading apart, as if they had come from a focus in front of the lens. An object $h_o = 0.4$ m tall stands a distance $d_o$ in front. Here is the surprise: no matter where you put the object — right up close or far away — a diverging lens makes only one kind of image: virtual, upright, and shrunken, on the same side as the object. Slide the object and watch — the image never flips, never enlarges, never lands on a screen. This is the lens in a door peephole and the front group of a wide-angle view.

Image distance

d_i = d_o f/(d_o-f)

d_o = 1.5

-0.6 m

Magnification

m = -d_i/d_o

d_o = 1.5

m (a pure number)

0.4

Image height

h_i = m\,h_o

d_o = 1.5

h_o = 0.4

0.16 m

Draw three rays from the tip of the object — no equation needed

\text{(1) parallel}\leftrightarrow F \qquad \text{(2) through center}\to\text{straight} \qquad \text{(3) } F' \leftrightarrow \text{parallel}

Geometric optics needs no calculus and, at first, no algebra. From the object tip, the ray parallel to the axis bends outward, as if it had come from the near focus $F$ ; the ray through the lens center goes straight; the ray aimed at the far focus $F'$ leaves parallel. The refracted rays diverge — but traced backward they meet, and that crossing is the (virtual) image.

check $d_i = d_o f/(d_o-f)$ solves $1/d_o+1/d_i=1/f$ ; chief ray $m=-d_i/d_o$ ; parallel & focal rays give the same $m$ ; $m = f/(f-d_o)$

✓ The closed form $d_i = \dfrac{d_o f}{d_o - f}$ satisfies the thin-lens equation $\dfrac{1}{d_o} + \dfrac{1}{d_i} = \dfrac{1}{f}$ — back-substituting leaves zero. [structural]
✓ The ray through the lens center is undeviated, so the object and image triangles about the center are similar: $m = h_i/h_o = -d_i/d_o$ . [structural]
✓ The parallel ray refracts toward (or away from) the focus, giving $m = -(d_i - f)/f$ by similar triangles. It equals the chief-ray value $-d_i/d_o$ — the two independent constructions locate the same image point, and that agreement is exactly the thin-lens equation. [structural]
✓ The focal ray (through a focus, emerging parallel) gives $m = -f/(d_o - f)$ — the same magnification again. All three principal-ray constructions locate one image point. [structural]
✓ The magnification in terms of $d_o, f$ alone: $m = -\dfrac{d_i}{d_o} = \dfrac{f}{f - d_o}$ — its sign sets the orientation (upright when $m>0$ , inverted when $m<0$ ), its size the scale. [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “If you move the object close enough to a diverging lens, you can magnify it or catch a real image on a screen.”

You cannot. A diverging lens has no real focus in front of it — parallel rays spread, as if streaming from a virtual focus on the near side; trace any object's refracted rays backward and they meet at a virtual image that always sits between the lens and the object. So for every object distance the image distance $d_i = d_o f/(d_o - f)$ comes out negative (virtual) and the magnification $|m| < 1$ (reduced and upright). Drag the object from $0.3$ m to $4$ m: the image only ever shrinks toward the focal point, always right-side-up, never on the far side. Only a converging lens ( $f > 0$ ) can magnify or project.

Modeling assumptions — author-asserted, disclosed not discharged

A thin lens: its thickness is negligible next to the focal length, so both refractions happen in one plane at the lens center.
Paraxial rays — all rays stay close to the axis, so a point object maps to a single point image. Real wide lenses blur this.
A single diverging lens, $f < 0$ , with the sign convention that a virtual image (on the object's side, seen through the lens) has $d_i < 0$ and an upright image has $m > 0$ .

The ray diagram, fully annotated

A static rendering (Matplotlib) at the default object distance — the three principal rays converging to the real, inverted image. The interactive version, with object-distance and height sliders, is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):