Lens Formula Calculator: Ray Diagrams and Chromatic Aberration

Hold a magnifying glass just above this line and the letters swell — upright, readable, larger than life. Lift it slowly away and at some point the words smear into nothing, then come back upside down, the whole room now shrunk into a coin of glass. Carry the same lens outside, point it at the sun, and it does a third thing entirely: it burns a single scorching dot into the pavement. Nothing about the glass changed across those three demonstrations. Only the distance did.

One equation covers all of it. The lens formula, $1/v - 1/u = 1/f$, takes the object distance and the focal length and returns exactly where the image sits, which way up it is, and how big it is. This page works through the sign convention that makes it come out right, the magnification rule that is not the one you learned for mirrors, and the five object positions every exam question quietly reduces to. Then it goes after the one thing a lens does that a mirror physically cannot: because glass bends blue light harder than red, a single lens has no single focal length at all. Below is a sheet of film you can slide along the axis behind a lens, hunting for the plane where a point of light lands as a point. With one piece of glass there is no such plane, and finding that out is the point of the exercise.

What Is the Lens Formula and What Do u, v and f Mean?

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Three distances, all measured from the optical centre of the lens — the point on the axis a ray can pass through without being deflected. The minus sign is not decoration. It is the visible half of what separates this from the mirror formula, and it is there because light keeps going through a lens instead of turning round — which puts a real image on the opposite side of the glass from where a mirror puts one.

The formula is useless without a sign convention. The one used here is the New Cartesian convention: distances are measured from the optical centre, and the direction the light is travelling counts as positive.

The line worth memorising above all the others is the one about $v$. For a lens, a real image has $v > 0$. Light passes straight through the glass and carries on, so anything real forms on the far side. That is the exact opposite of a spherical mirror, where real light comes back out the way it went in and a real image has $v < 0$. If you have already worked through the mirror formula and its ray diagrams, this is the single place where carrying your answers across will get every real/virtual verdict backwards.

How Do You Calculate the Image Position Step by Step?

Take a convex lens of focal length 20 cm with an object 25 cm in front of it — a setting you can dial up on the bench further down.

1. Attach the signs. Convex, so $f = +20$ cm. Object on the incoming side, so $u = -25$ cm.
2. Rearrange for v. $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} + \dfrac{1}{-25}$
3. Do the arithmetic. $\dfrac{1}{v} = \dfrac{5}{100} - \dfrac{4}{100} = \dfrac{1}{100}$, so $v = +100$ cm.
4. Read the sign. $v$ is positive, so the image is real, one metre beyond the lens.
5. Magnify. $m = \dfrac{v}{u} = \dfrac{+100}{-25} = -4.00$ — negative, so inverted, and $|m| > 1$, so magnified four times.

Real, inverted, four times life size, projected a metre away from a lens the length of your hand. Every lens question is these five steps.

What Does the Magnification of a Lens Tell You?

$m = \frac{v}{u} = \frac{h'}{h}$

Note that there is no minus sign here, and that this is the most common slip in the whole topic. A spherical mirror uses $m = -v/u$; a thin lens uses $m = +v/u$. Both conventions are built so that a real image comes out inverted, and because the sign of $v$ means opposite things in the two cases, the minus sign has to appear in one and not the other. Put the mirror's minus sign into a lens calculation and every image comes out the wrong way up.

Reading a magnification is two independent questions:

• The sign tells you which way up. $m < 0$ means inverted, $m > 0$ means upright. For a lens, every real image is inverted and every virtual image is upright — no exceptions in either direction.
• The size tells you how big. $|m| > 1$ is magnified, $|m| < 1$ is diminished, $|m| = 1$ is life size.

So $m = -4.00$ is inverted and four times as tall; $m = +3$ is upright and three times as tall; $m = +0.4$ is upright and shrunk to two fifths. Because $m = h'/h$ as well, magnification is the quickest route to an image height: multiply the object height by $m$ and keep the sign.

Where Does a Convex Lens Form Its Image at Each Object Distance?

A converging lens gives the light a fixed amount of convergence. Whether the refracted rays actually manage to cross depends on how much divergence they arrived with — which depends entirely on how far away the object is. That single competition produces five distinct outcomes, worked here with $f = +20$ cm, which is what the ray diagram below opens on. Every row in the table is reachable on its $|u|$ slider, so you can walk the object through all five in one sweep.

Read down the $v$ column and it is one continuous motion, not five separate cases. As the object walks in from far away, the image walks out; they meet at 2F, both 40 cm from the lens, at exactly life size. Keep going and the image races off toward infinity, arriving there just as the object reaches F. Push past F and it reappears on the near side, upright and enlarged — that is the magnifying-glass case, and the reason the letters swelled when you held the glass close to the page.

The three rays are three independent constructions that must agree, and any two of them already fix the image point — the third is a check. Two have mirror counterparts; the third does not. A ray aimed at the optical centre carries straight on undeviated, because the two glass faces are parallel there, so the object tip, the centre of the lens and the image tip always lie on one straight line. A mirror has no such ray: its third construction goes through the centre of curvature and retraces itself. That difference, and the full spherical-mirror treatment, is the reflection sibling of this page.

Why does a concave lens always form a virtual, upright, diminished image?

This one is a theorem, not a survey of examples. A concave lens has $f < 0$, and the object is always on the incoming side, so $u < 0$. Then

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

is a sum of two strictly negative terms, and three consequences follow at once and admit no exceptions.

• $1/v < 0$, so $v < 0$ for every object distance. The image is always on the object's side, so always virtual.
• $m = v/u$ is a ratio of two negative numbers, so $m > 0$ always: the image is always upright.
• $|1/v| > |1/u|$, so $|v| < |u|$, and $|m| = |v|/|u| < 1$. It is always diminished.

There is no zone structure to learn, because there are no zones — exactly as for a convex mirror, and for the same algebraic reason.

Why Does a Lens Split White Light Into Colours but a Mirror Never Does?

Everything above quietly assumed that a lens has one focal length. It does not, and the reason is the deepest difference between a lens and a mirror.

A mirror works by reflection: the angle of incidence equals the angle of reflection, at every point, exactly, with no reference to the colour of the light. Feed a mirror red and blue together and they come back along identical paths. A lens works by refraction, which is governed by Snell's law and the refractive index of the glass — and the refractive index is not one number either. It is larger for short wavelengths than for long ones. Blue light meets a stiffer glass than red does, is bent harder at both surfaces, and comes to a focus nearer the lens.

This is exactly the effect that makes a prism throw a rainbow on the wall. A lens is not a prism, but its surfaces are tilted, and every part of it away from the centre does a little prism's worth of splitting. The result has a name: chromatic aberration, a coloured halo around every bright edge, and it is the defect that drove seventeenth-century refracting telescopes to absurd lengths. At a fixed aperture the coloured blur stays roughly the same physical size however long the lens is, so the only way to make it small relative to the picture was to stretch the focal length until the picture was enormous — which is how observatories ended up with objectives strung up on masts tens of metres from the eyepiece.

What is the Abbe number, and why is crown glass better than flint?

Opticians compress a glass's whole dispersion into a single figure, the Abbe number:

$V_d = \frac{n_d - 1}{n_F - n_C}$

The three indices are measured at three standard spectral lines — the blue F line at 486.1 nm, the yellow d line at 587.6 nm, and the red C line at 656.3 nm. The numerator is the glass's refracting power; the denominator is how much that power varies across the visible band. A high Abbe number means low dispersion.

The rule of thumb that follows is worth carrying: a singlet's blue and red focal lengths differ by roughly $f/V_d$. A 20 cm crown lens should therefore smear its focus over about $20/64.17 = 0.312$ cm, and a 20 cm flint lens over $20/25.76 = 0.776$ cm. The exact figures — computed properly, because $f \propto 1/(n-1)$ is not quite linear in $n$ — are 0.3098 cm and 0.7640 cm. Flint's high index buys you a shorter, more powerful lens out of the same curvature — and if you then match it back to the crown lens's focal length, it charges you two and a half times the colour error for the privilege.

Interactive Lens Dispersion Bench: Watch White Light Split at the Focus

The instrument below is not a ray diagram, and that is deliberate: a ray diagram shows you three neat crossings and invites you to believe one of them is the image. What you get instead is the film. A point source sits in front of the lens, a sheet of film stands behind it, and your job is to slide the film along the axis until the point records as a point. With one piece of glass it never does — the best you can find is a plane that belongs to none of the colours — and hunting for it until you are convinced is the whole of chromatic aberration.

The big black panel is the emulsion, magnified: the coloured discs on it are what actually lands there, composited the way light composites, with the widest colour's diameter printed in the corner. Below it sits the locator strip — the source and the lens sketched in the left-hand gutter, then, past a scale break, an expanded view of the stretch of axis the film is allowed to travel over, with a coloured tick at every wavelength's own focus. Drag that strip sideways to move the film, or step the film slider, which does the same job for keyboard users. The row of seven miniatures underneath is the same exposure taken at seven successive planes across that travel, so you can see the whole hunt at once.

The $|u|$ step is the one worth dwelling on, because it separates two things that are easy to confuse. The focal-length spread, 0.3098 cm, is a property of the glass and the grinding alone — it does not move when you move the object. What moves is how much the imaging geometry magnifies it. At $|u| = 25$ cm the blue image lands at 94.882 cm and the red at 102.456 cm: those same 0.31 cm of focal spread have become 7.575 cm of image separation. Pull $|u|$ back out to 60 cm and the separation falls to 0.695 cm. That is why the film's travel is under three centimetres long at 60 cm and nearly thirty at 25 cm: the instrument is sized to the hunt, and the hunt gets longer as you close in. Chromatic aberration is always there; it is close-up, high-magnification work that makes you look at it.

There is a curiosity hiding at $|u| = 20$ cm, where the object sits exactly at the focal length. Set it there and the hunt disappears: the locator strip reports that no colour converges anywhere on the film's travel, and every plane catches about the same 50 mm flood, near enough the aperture itself. The yellow light is the reason — it leaves the lens perfectly parallel and never converges at all. But blue does not: blue's focal length is 19.787 cm, so the object is outside blue's focus and blue still forms a real image, 18.5 metres away — far beyond anywhere the film can be stood. Red's focal length is 20.096 cm, so the object is inside red's focus and red leaves the lens diverging, as though from a virtual image 41.7 metres back on the incoming side. At that one object distance a single piece of glass is behaving as a converging lens, a collimator and a diverging lens simultaneously, depending only on which colour you ask.

How Does an Achromatic Doublet Cancel Chromatic Aberration?

Chester Moore Hall's fix, from around 1730, is still the standard answer — it is what sits at the front of most binoculars, small telescopes and camera lenses today. Cement two lenses of different glasses together and use one's dispersion to undo the other's.

Two thin lenses in contact simply add their powers, $P = P_1 + P_2$ where $P = 1/f$. Each element's dispersion, meanwhile, is proportional to its own power divided by its own Abbe number. So the pair brings blue and red to the same focus when

$\frac{P_1}{V_1} + \frac{P_2}{V_2} = 0$

Because both Abbe numbers are positive, the two powers must have opposite signs. The only way to cancel the colour error while keeping net converging power is to over-converge with the low-dispersion crown and claw some of it back with a diverging flint element. For a 20 cm doublet from BK7 and SF11 the split comes out at a crown element of $f_1 = +11.9713$ cm and a flint element of $f_2 = -29.8214$ cm — the crown is far stronger than the 20 cm pair it belongs to. That is Example 5 below, worked in full.

Why does an achromat still show a violet fringe?

Because it corrects at exactly two wavelengths, and there are more than two.

The condition above forces the blue F line and the red C line to land together — at $f_d = 20$ cm they both come out at 20.011 cm, and on the bench their two image planes coincide exactly: with the object 60 cm out both converge at 30.025 cm, and standing the film on the d-line plane at 30.000 cm leaves the ⌀ blue, F and ⌀ red, C readouts on the same 41 µm — equally out of focus, which is what being corrected together means. But nothing in the condition says anything about the wavelengths between them. The residual error is a curve in $1/\lambda^2$ whose two roots are the F and C lines, so it must bulge somewhere in the middle, and it does: the yellow d line focuses at 20.000 cm, 0.0110 cm short of the corrected blue-and-red focus. That leftover is called the secondary spectrum, and no cementing of two ordinary glasses removes it.

It is a genuine 28-fold improvement on the crown singlet's 0.3098 cm — but it is not zero, and it is not even across the band. Push past the F line into the violet and the residual grows fast: at 0.40 µm it reaches 0.114 cm, ten times the d-line figure and on the other side. Out at the red end, 0.68 µm, it is only 0.0047 cm. That asymmetry is exactly what you see in a photograph taken through an achromatic lens: the red fringes are gone and a violet one is still there. Killing it needs a third element and an unusual glass — an apochromat, which corrects three wavelengths and costs several times as much.

Keep that 28× away from the other number the bench reports, because they measure different things. Twenty-eight is the correction of the focal lengths. The spot the doublet leaves at its best plane is ⌀ 241 µm against the crown singlet's ⌀ 1.23 mm — a win of about five, not twenty-eight, because the film records every wavelength at once and the uncorrected violet is the one setting the size of the disc. A doublet does far more for where the colours focus than it does for how big the blur is.

What Is the Lens Maker's Equation and How Does It Set the Focal Length?

The lens formula tells you where the image goes once you know $f$. The lens maker's equation tells you what $f$ is in the first place, from the two surface curvatures and the glass:

$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$R_1$ is the radius of curvature of the surface the light meets first and $R_2$ the second; each is positive if its centre of curvature lies on the outgoing side. An equiconvex lens therefore has $R_1 = +R$ and $R_2 = -R$, and its two curvatures add rather than cancel.

Two things follow immediately. First, the geometry and the glass enter as a product, so the same shape ground in flint instead of crown gives a much shorter focal length — 12.74 cm instead of 19.35 cm for the equiconvex lens in Example 4. Second, and this is the whole of the previous two sections in one line: $n$ is the only wavelength-dependent factor, and the bracket is fixed once the glass is ground. So

$f_\lambda = f_d \cdot \frac{n_d - 1}{n_\lambda - 1}$

The shape cancels out. Chromatic aberration is not a manufacturing error you could polish away — it is written into the lens maker's equation itself.

How Is the Lens Formula Different from the Mirror Formula?

The two equations look almost identical, which is exactly the trap.

Two rows do the damage. The sign of $v$ is reversed, and the magnification formula gains or loses its minus sign to compensate. Everything else is bookkeeping. The full spherical-mirror convention, the three construction rays and the five zones of a concave mirror are worked through on the mirror formula and ray diagram page, which is the reflection sibling of this one.

That last row is why no large nighttime research telescope built since Yerkes gathers its light with a lens. Everything on this page about crown, flint and doublets is work a mirror never has to do — and the colour error is only the first of the objections. A metre-wide objective must be cut from a metre-wide flawless blank with two surfaces figured onto it, and it can only be held at its rim, where it sags under its own weight. A mirror needs one good surface, is supported across its whole back, and treats every wavelength identically. The 40-inch refractor at Yerkes, completed in 1897, was the end of that line.

Worked Examples of the Lens Formula for Physics Exams

Lens Formula and Sign Convention Quick Reference