From zero to mastery. Interactive simulations, complete derivations, every scenario — for both sign conventions, positive and negative lenses.
A lens is a shaped piece of transparent material that refracts light in a controlled way. The thin lens approximation collapses both refracting surfaces into a single plane — zero thickness, one equation governs everything.
Light travels left to right. The lens sits at the origin.
| Symbol | Quantity | Positive when… | Negative when… |
|---|---|---|---|
| s | Object distance | Real object (left of lens) | Virtual object (right of lens) |
| s′ | Image distance | Real image (right of lens) | Virtual image (left of lens) |
| f | Focal length | Converging / convex | Diverging / concave |
| m | Magnification | Image upright | Image inverted |
Everything follows from similar triangles.
Lens at x=0. Object height h at distance s to the left. Seek image position s′ and image height h′.
Central ray passes undeviated, creating two similar triangles:
Define x = s−f, x′ = s′−f (measured from focal points, not lens):
Object and image focal distances always multiply to f². Symmetric and beautiful.
There are three distinct ways the thin lens equation gets written. Two are algebraically the same rearrangement. The third is genuinely different — and arguably the most physically illuminating.
The symmetric form derived from similar triangles. Both s and s′ appear with positive signs. Most common in physics textbooks. You rearrange to solve for whatever variable you need.
Form A rearranged. Subtract 1/s from both sides. Gives you 1/s′ directly. Algebraically identical — same equation, same answer. Useful when you always compute s′ and want to skip the rearranging step.
Genuinely different convention. Uses signed inputs differently. Seen in some British-curriculum and older texts. This is not Form A rearranged.
In Form C, s is measured as a negative number for a real object. A real object 30 cm to the left of the lens has s = −30 cm in this convention. So when you write:
You compute:
Same numerical answer. But notice what happened: the negative sign on s did the work that the subtraction did in Form B. In Form C, you're not told to subtract — instead, you're forced to think about and supply the correct sign for every quantity.
This convention makes you mentally engage with the direction of each quantity:
The minus sign is baked into the formula. You just plug in |s| = 30. The equation handles the sign for you. Easy, mechanical, but you can get the right answer without thinking about what the sign of s means.
No built-in sign adjustment. You must consciously decide: is this object to the left (negative) or right (positive)? Is this a converging lens (f positive)? You can't sleepwalk through it. Every sign is a deliberate physical statement.
Suppose you have a diverging lens (f = −15 cm) and an object on the left (s = −25 cm):
The negative s′ immediately tells you: virtual image, on the same side as the object. The two negative terms combined and stayed negative — that's physically meaningful. Both f and s were negative here, which makes you ask: why? what does each sign represent? what does it mean for a lens to have negative f? Form C forces those questions.
Compare to Form B with the same situation:
Wait — different answer! That's because in Form B, s is still positive (s = +25), whereas in Form C, s = −25. The forms use different sign conventions for input quantities. Mix them up and you get wrong answers.
Object at 30 cm left of lens, converging lens f = 20 cm.
s = +30 (real object, positive by convention)
s = +30 (same convention as A)
s = −30 (real object is negative in this convention)
Full real-time simulation. Click and drag the object arrow, or click anywhere left of the lens to jump the object there. Drag the focal point markers to change f.
f > 0. Parallel rays converge to a real focal point. Behavior is rich and has a discontinuity at s = f.
f < 0. Always virtual, always upright, always minified. Write f = −F, F > 0.
All key distances, symbolic and numeric (f = 100, h = 40).
| s | s (f=100) | s′ symbolic | s′ numeric | m symbolic | m decimal | h′ | Type | Orient | Size |
|---|
| s | s (F=100) | s′ symbolic | s′ numeric | m symbolic | m decimal | h′ | Type | Orient | Size |
|---|
Positive s′: vertical asymptote at s=f. Negative (virtual) for s<f, positive (real) for s>f. Approaches f from above as s→∞. Crosses s′=2f at s=2f.
Negative s′: smooth monotone curve, always in negative territory, asymptoting to −F.
Positive m: +∞ as s→f⁻, −∞ as s→f⁺, crosses m=−1 at s=2f.
Negative m: 1 at s=0 down to 0 as s→∞. Always positive, always <1.
Real or virtual image depending on s vs f. Magnification spans all of ℝ with discontinuity at s=f. The singularity is the collimated-beam condition.
Key: s=2f → m=−1. f<s<2f → projector. s<f → magnifier.
Always virtual, upright, minified. m = F/(s+F) = 1/(1+s/F). Simple hyperbola. No singularity. No real images.
Key: s=|f| → m=½. s→∞ → m→0.
Swapping s and s′ in 1/s + 1/s′ = 1/f leaves the equation unchanged. An object at s₁ that images to s₁′ means placing an object at s₁′ images back to s₁.
Compound positive system. Real inverted image on retina. Ciliary muscles change f. Myopia: negative corrective lens. Hyperopia: positive.
Object at s≫f. Image at s′≈f, real, inverted. Zoom changes f. Focusing moves lens to adjust s′.
Slide at f<s<2f. Real, inverted, magnified. Slide physically inverted to compensate. Screen at s′.
Myopia: negative lens. Hyperopia: positive. D = 1/f (meters). −2.5 D → f = −40 cm.
Keplerian: pos+pos, M=−f_obj/f_eye. Galilean: pos+neg, erect image.
Object at f+ε. M=−(L/f_obj)×(25/f_eye). Both positive. Up to 1000×.
R positive if center of curvature is to the right. Biconvex: R₁>0, R₂<0 → f>0. Biconcave: f<0.