In short

A thin lens is a piece of transparent glass or plastic with two spherical surfaces, so thin that the distance between the two surfaces is negligible compared to the object and image distances. A convex lens (bulging at the middle) is a converging lens — it bends parallel rays inward to meet at the focus F. A concave lens (thinner at the middle) is a diverging lens — it spreads parallel rays so they appear to come from F on the same side as the object.

For a thin lens of focal length f, an object at signed distance u produces an image at signed distance v satisfying the thin lens formula (Cartesian sign convention, pole at the optical centre):

\boxed{\;\frac{1}{v} - \frac{1}{u} \;=\; \frac{1}{f}\;}

The magnification (ratio of image height to object height) is

\boxed{\;m \;=\; \frac{h'}{h} \;=\; \frac{v}{u}\;}

The power of a lens is defined as

\boxed{\;P \;=\; \frac{1}{f}\;}\qquad\text{(with $f$ in metres; unit: dioptre, D)}

Sign convention (Cartesian):

  • Distances measured from the optical centre O.
  • Incident light travels left → right (along +\hat{x}).
  • Distances in the direction of incident light are positive; against it are negative.
  • Heights above the principal axis are positive; below are negative.

Consequence for f: convex lens f > 0; concave lens f < 0.

Lens f Image for real object Everyday use in India
Convex (converging) + real/inverted if \lvert u\rvert > f; virtual/erect/enlarged if \lvert u\rvert < f reading glasses for hypermetropia, camera lens, magnifier, projector
Concave (diverging) - always virtual, erect, diminished spectacles for myopia, peepholes, Galilean telescope eyepiece

The two standard rays used in every ray diagram:

  1. A ray parallel to the principal axis refracts through the second focus F_2 (convex) or appears to diverge from the first focus F_1 (concave).
  2. A ray through the optical centre O passes straight through, undeviated.

Walk into any Bausch+Lomb or Lenskart store and watch what the optician does. They sit you at a slit lamp, flip a dozen small discs of glass in front of your eyes, and each disc says something like "+1.50 D" or "-2.75 D" scratched on its rim. Those discs are thin lenses, and the number is the lens's power. Pick the one that makes the letter chart sharp, and your prescription is done. The entire business of spectacles is arithmetic on the lens formula you are about to derive.

A thin lens is nothing more than a piece of glass with two spherical surfaces. Light enters through one, refracts; travels a negligible distance inside; refracts at the second surface; and emerges. Do the refraction twice, in the limit where the glass is infinitely thin, and the two-surface problem collapses into a single formula relating object distance, image distance, and focal length. That formula — the thin lens equation — is the working tool of all of geometrical optics from here on. Mirrors had 1/v + 1/u = 1/f; lenses have 1/v - 1/u = 1/f, and the sign flip comes from one simple fact: light goes through a lens but bounces off a mirror.

Anatomy of a thin lens

Every thin lens has six points on its principal axis.

Anatomy of a thin convex lens A biconvex lens symbol drawn as a vertical double-arrowed lens shape. Principal axis runs horizontally. Optical centre O at the middle of the lens, first focus F1 on the left at distance f, second focus F2 on the right at distance f. Label for each point. $O$ (optical centre) $F_1$ $F_2$ $f$ $f$ principal axis convex lens
A biconvex thin lens. The optical centre $O$ is the midpoint. The two foci $F_1$ and $F_2$ sit symmetrically at distance $f$ on either side. For a thin lens in air, the focal length is the same on both sides.

Why the symmetry? A ray starting from F_1 and passing into the lens emerges parallel to the axis — by the reversibility of light paths, a parallel ray entering from the other side must converge at F_1. In a single medium, the two focal lengths coincide in magnitude, and we write just one symbol f for both. This stops being true when the lens sits between two different media (an underwater camera, for example), but for every Class 12 and JEE problem, air-on-both-sides is the default.

Convex vs concave. A convex lens is thicker at the middle than at the edges. Parallel rays bend inward and meet at a real focus on the far side — f > 0. A concave lens is thinner at the middle; parallel rays bend outward and appear to diverge from a virtual focus on the near side — f < 0. The sign of f tells you which type of lens you have, without any further labelling.

The sign convention for lenses

The Cartesian convention you learnt for spherical mirrors carries over with one change: light travels left to right through a lens (rather than right to left onto a mirror). Everything else follows the same rule — measure distances from the pole (here, the optical centre), distances in the direction of incident light are positive, and heights above the axis are positive.

Rules for lenses.

  1. The optical centre O is the origin. The principal axis is the x-axis.
  2. Incident light travels from left to right — along +\hat{x}.
  3. Distances measured to the right of O (in the direction of travel) are positive. Distances to the left of O (against the direction of travel) are negative.
  4. Heights above the principal axis are positive; below are negative.
Cartesian sign convention for thin lenses A vertical lens symbol in the middle. Left side labelled "incident light" with distances there being negative. Right side labelled "emergent side" with distances positive. Heights above axis positive, below negative. $O$ incident light $\rightarrow$ distances here are **negative** emergent side distances here are **positive** heights $+$ heights $-$
Cartesian sign convention for lenses: optical centre at origin, incident light travels rightward. Distances left of $O$ are negative; right of $O$ are positive. A real object sits to the left, so $u < 0$. A real image forms to the right, so $v > 0$.

Consequences for standard quantities.

Notice the contrast with mirrors: for a mirror, light goes right-to-left, so a real object had u < 0 and a real image also had v < 0. For a lens, light goes left-to-right, so a real object has u < 0 but a real image has v > 0. This sign flip is exactly why the lens formula has a minus sign where the mirror formula has a plus.

Deriving the thin lens formula — ray tracing from first principles

Place a convex lens with optical centre O on the principal axis. Put an object AB of height h at distance \lvert u\rvert to the left of O (so u < 0), farther from the lens than F_1. Trace two rays from the tip B and find where they cross after refracting through the lens — that intersection is the image tip B'.

Ray 1: From B parallel to the axis → refracts and passes through F_2 on the far side. Ray 2: From B through the optical centre O → emerges undeviated (straight line).

The image tip B' lies at the intersection. The image base A' sits directly below B' on the axis. The image distance OA' = v (positive, to the right of O).

Deriving the thin lens formula for a convex lens An object AB of height h on the left, a convex lens at O in the middle, image A'B' formed on the right. Ray 1 goes from B parallel to the axis then bends through F2. Ray 2 goes from B through the optical centre O undeviated. The two rays meet at B'. $O$ $F_1$ $F_2$ $B$ $A$ $h$ ray 1 (parallel $\to F_2$) ray 2 (through $O$) $B'$ $A'$ $h'$
Object $AB$ (height $h$, distance $u$ left of $O$) images at $A'B'$ (height $h'$, distance $v$ right of $O$). Ray 1 traces parallel then through $F_2$. Ray 2 runs straight through the optical centre. Two similar-triangle relations give the thin lens formula.

Step 1. Triangles ABO and A'B'O are similar (both right-angled on the principal axis at A/A', and they share the vertical angle at O).

\frac{A'B'}{AB} \;=\; \frac{OA'}{OA} \;\Rightarrow\; \frac{\lvert h'\rvert}{h} \;=\; \frac{v}{\lvert u\rvert}

Why: ray 2 is a straight line through O. The two triangles formed on either side of O are similar by AA — both right angles, both share the angle the ray makes with the axis. Ratios of corresponding sides are equal.

With the Cartesian convention: h > 0, h' < 0 (inverted real image), u < 0, v > 0. So h'/h is negative and v/u is negative — the signs match on both sides:

\frac{h'}{h} \;=\; \frac{v}{u} \tag{a}

Step 2. Triangles OCF_2 and A'B'F_2 are similar, where C is the top of the lens at height h (ray 1 passes through C before bending). Both triangles share the angle at F_2, and both have a vertical side.

\frac{A'B'}{OC} \;=\; \frac{F_2 A'}{F_2 O}

Since OC = AB = h (ray 1 is horizontal before hitting the lens, so it reaches the lens at the same height as the object):

\frac{\lvert h'\rvert}{h} \;=\; \frac{v - f}{f}

With signs (h' < 0, v - f > 0 for a real image past F_2, f > 0):

\frac{h'}{h} \;=\; -\,\frac{v - f}{f} \tag{b}

Why: the "minus" in (b) is because both triangles on opposite sides of F_2 are mirror images of each other — ray 1 goes down on the right of the lens and the object height points up on the left. The magnitudes are related by the simple ratio, and the sign bookkeeping flips one of them.

Step 3. Equate (a) and (b):

\frac{v}{u} \;=\; -\,\frac{v - f}{f}

Why: both expressions are h'/h, so they must be equal to each other. From here it is straight algebra.

Step 4. Cross-multiply and expand.

v f \;=\; -\,u(v - f) \;=\; -uv + uf

Step 5. Divide through by u v f (all non-zero when a real image forms):

\frac{v f}{u v f} \;=\; \frac{-uv}{uvf} + \frac{uf}{uvf}
\frac{1}{u} \;=\; -\,\frac{1}{f} + \frac{1}{v}

Step 6. Rearrange to the standard form.

\boxed{\;\frac{1}{v} - \frac{1}{u} \;=\; \frac{1}{f}\;}

Why: this is the thin lens formula. It holds for convex and concave lenses, for real and virtual objects and images, so long as the Cartesian sign convention is applied consistently. The minus sign (contrast with the mirror's plus) reflects that light transmits through the lens rather than bouncing off it.

Magnification. From Step 1, the signed ratio of heights is

\boxed{\;m \;=\; \frac{h'}{h} \;=\; \frac{v}{u}\;}

A positive m means erect; negative means inverted. \lvert m\rvert > 1 means enlarged; \lvert m\rvert < 1 means diminished. Because for a real object u < 0, a positive v (real image) gives negative m — real images of real objects are always inverted through a single thin lens. A negative v (virtual image) gives positive m — virtual images are erect. That is the whole classification.

Explore: drag the object, watch the image form

Below is a convex lens of focal length f = 15 cm. Drag the red point along the principal axis to change the object distance. The curves and markers below it show the image distance v and the magnification m computed from the lens formula. Two regimes fall out immediately:

Interactive: image distance vs object distance for a convex lens Two curves on a coordinate plane: image distance v (red) and magnification m (dark) as functions of object distance u. A vertical dashed line at u=-15 marks the focus, where v diverges. Drag the red point on the u-axis to change the object distance and read off v and m. object distance $u$ (cm) image distance $v$ (cm) −60 −45 −30 −15 0 60 30 −30 −60 $u = -f$ $v(u)$ drag the red point along the axis
Drag the red point along the horizontal axis to change the object distance. The image distance $v$ is computed from $1/v - 1/u = 1/f$ with $f = 15$ cm. Watch $v$ diverge (goes off the chart) as $u \to -f$: a candle flame exactly one focal length from the lens produces parallel rays, no image. Move the object inside $F_1$ and $v$ flips negative — that is the virtual, erect, magnified image you see through a hand lens.

Image formation — the six standard cases for a convex lens

The lens formula handles every case, but the qualitative behaviour falls into six regimes depending on where the object sits relative to F_1 and 2F_1 (the point at distance 2f on the incident side). Memorising these is wasteful; deriving them from the formula takes ten seconds each.

Object position Image position Nature Size Where you see it
At infinity At F_2 real, inverted point sunlight focused by a magnifier
Beyond 2F_1 Between F_2 and 2F_2 real, inverted diminished camera photographing a distant friend
At 2F_1 At 2F_2 real, inverted same size the "unit magnification" setup in a physics lab
Between F_1 and 2F_1 Beyond 2F_2 real, inverted enlarged cinema projector throwing an image on a screen
At F_1 At infinity — — headlamp bulb at the focal point of a convex-lens headlight (parallel beam)
Between O and F_1 On the same side as object virtual, erect enlarged hand magnifier, simple microscope

For a concave lens, the story is simpler — there is only one regime. No matter where the real object sits, the image is virtual, erect, diminished, and on the same side as the object. That is why concave lenses are used as corrective lenses for myopia (short-sightedness): they take a real object at any finite distance and produce a virtual image closer to the eye, which the short-sighted retina can then focus on.

Ray diagram for a concave (diverging) lens A concave lens symbol in the middle. Object arrow on the left. Ray 1 parallel to axis, diverging after the lens as if it came from F1 on the near side. Ray 2 straight through the optical centre. The two diverging rays, when traced back, intersect on the object side to form a virtual erect diminished image. $O$ $F_1$ $F_2$ $B$ $B'$ ray 1 diverges ray 2 (through $O$)
A concave lens always produces a virtual, erect, diminished image. Ray 1 (parallel) diverges after the lens as if coming from $F_1$ on the near side (dashed back-extension). Ray 2 passes through $O$ undeviated. The back-extensions of the two diverging rays cross between $F_1$ and the lens — that crossing is the virtual image $B'$.

Power of a lens — what the optician writes on your prescription

A strong lens bends light sharply; a weak lens bends it gently. Quantitatively, the strength of a lens is measured by its power:

P \;=\; \frac{1}{f} \qquad (f \text{ in metres})

The unit is the dioptre (symbol D). A lens of focal length +0.5 m has power +2 D. A lens of focal length -0.25 m has power -4 D.

Why define P = 1/f rather than f itself? Because powers of lenses in contact add. If you stack two thin lenses of focal lengths f_1 and f_2 against each other, the combination behaves like a single lens of focal length f where

\frac{1}{f} \;=\; \frac{1}{f_1} + \frac{1}{f_2} \qquad\text{so}\qquad P \;=\; P_1 + P_2

(The derivation of this formula is done in Lensmaker's Equation and Combinations.) Addition of powers is the reason opticians use dioptres: fitting you for spectacles is literally adding dioptres until the letters on the chart come into focus. If a patient's myopic eye needs -1.75 D and then develops a slight astigmatism that needs an extra -0.25 D cylindrical correction, the total prescription is -2.00 D sphere with cylinder — two numbers that add.

The sign of P. A convex (converging) lens has f > 0, so P > 0. A concave (diverging) lens has f < 0, so P < 0. When an Indian optician writes "+1.50 D" on a reading-glasses prescription, they mean a convex lens of focal length \frac{1}{1.5} \approx 0.67 m — matched to the working distance for reading a book held at roughly arm's length. A myopia prescription of "-3.50 D" means a concave lens of focal length \frac{1}{-3.5} \approx -0.29 m.

Worked examples

Example 1: Camera lens photographing a distant friend

A compact camera uses a convex lens of focal length 5 cm. Your friend, who is 1.7 m tall, stands 3 m away. Find the position of the image on the camera's sensor, the height of the image, and verify that the image is real and inverted.

Before-and-after diagram for a camera lens imaging a distant person On the left, a tall person-arrow at distance u = −300 cm from a convex lens. On the right, an inverted small image at v ≈ +5.08 cm on the sensor side. $O$ friend $u = -300$ cm $h = +170$ cm $v \approx +5.08$ cm $h' \approx -2.88$ cm convex lens, $f = +5$ cm
Object (friend) at $u = -300$ cm, lens $f = +5$ cm, image formed just past the focus on the sensor side. Diagram not drawn to scale — the image is 60 times smaller than the object.

Step 1. Apply the sign convention.

u = -300 cm (object on the incident side, left of O); f = +5 cm (convex lens); h = +170 cm (friend is upright).

Why: the Cartesian sign convention gives these signs directly. The focal length is positive because the lens is convex.

Step 2. Solve the lens formula for v.

\frac{1}{v} - \frac{1}{u} \;=\; \frac{1}{f} \;\Rightarrow\; \frac{1}{v} \;=\; \frac{1}{f} + \frac{1}{u} \;=\; \frac{1}{5} + \frac{1}{-300}
\frac{1}{v} \;=\; \frac{60}{300} - \frac{1}{300} \;=\; \frac{59}{300}
v \;=\; \frac{300}{59} \;\approx\; 5.08 \text{ cm}

Why: the image distance is positive, so the image sits on the emergent side of the lens. It forms just past the focus — a near-distant object images nearly at F_2, which is exactly where a camera places its sensor when focused on a scene 3 m away.

Step 3. Compute the magnification.

m \;=\; \frac{v}{u} \;=\; \frac{5.08}{-300} \;\approx\; -0.01695

Why: the ratio is negative (image inverted) and small (image much smaller than object) — exactly what a camera does.

Step 4. Compute the image height.

h' \;=\; m \cdot h \;=\; -0.01695 \times 170 \;\approx\; -2.88 \text{ cm}

Why: the image on the sensor is about 2.88 cm tall, inverted. A full-frame sensor (36 mm × 24 mm) can just fit this.

Result: The image forms at v \approx 5.08 cm behind the lens, is inverted (negative sign on h'), and is about 2.88 cm tall. This is why a compact camera's lens-to-sensor distance is a tiny bit more than the focal length.

What this shows: For objects much farther than f, the image distance v is barely different from f itself. That is why a camera's focus-ring travels only a few millimetres between "infinity" and "near subject" — all the image positions are squeezed into the narrow band just outside F_2.

Example 2: Your grandfather's reading glasses

Your grandfather, who has hypermetropia (long-sightedness), wears reading glasses of power +2.5 D. He holds a newspaper at 25 cm — the standard near point for a healthy young eye, which his presbyopic eye can no longer focus on. (His own near point has shifted to about 50 cm.) Find the focal length of the lens, and determine where the virtual image forms so that his eye can see it sharply.

Reading glasses forming a virtual image farther than the newspaper On the left, a newspaper-arrow at distance u=−25 cm from a convex reading-glass lens. The virtual, enlarged, erect image appears farther away on the same side at about v=−50 cm. $O$ (lens) newspaper $u = -25$ cm virtual image $v = -50$ cm convex lens, $P=+2.5$ D, $f=+40$ cm
The newspaper is inside $F_1$ of the reading-glass lens ($f = 40$ cm). The lens forms a virtual, erect, enlarged image at about 50 cm — exactly where grandfather's presbyopic eye can focus.

Step 1. Convert power to focal length.

P = +2.5 \text{ D} \;\Rightarrow\; f \;=\; \frac{1}{P} \;=\; \frac{1}{2.5} \text{ m} \;=\; 0.40 \text{ m} \;=\; +40 \text{ cm}

Why: power in dioptres is defined as 1/f with f in metres. So f = 1/P in metres, which converts to 40 cm.

Step 2. Apply the sign convention.

u = -25 cm (newspaper on incident side); f = +40 cm (convex lens).

Note that \lvert u\rvert < f — the object sits between O and F_1. This is the magnifying-glass regime: expect a virtual, erect, enlarged image.

Step 3. Solve the lens formula for v.

\frac{1}{v} \;=\; \frac{1}{f} + \frac{1}{u} \;=\; \frac{1}{40} + \frac{1}{-25}

Common denominator 200:

\frac{1}{v} \;=\; \frac{5}{200} - \frac{8}{200} \;=\; -\,\frac{3}{200}
v \;=\; -\,\frac{200}{3} \;\approx\; -66.7 \text{ cm}

Why: the negative v confirms a virtual image — it lies on the same (incident) side as the object. Your grandfather's eye receives diverging rays that appear to come from a point 66.7 cm away, which is comfortably beyond his 50 cm near point.

(Check: if the image were to sit exactly at his 50 cm near point, the required 1/f = 1/v - 1/u = -1/50 + 1/25 = 1/50, giving f = 50 cm = 0.50 m and P = 1/0.50 = +2.0 D. Grandfather's actual +2.5 D pushes the image a bit farther — comfortable, not strained.)

Step 4. Compute the magnification.

m \;=\; \frac{v}{u} \;=\; \frac{-66.7}{-25} \;\approx\; +2.67

Why: m is positive, so the image is erect. \lvert m\rvert > 1, so it is enlarged — the newsprint appears nearly 2.7 times larger than it actually is.

Result: Reading glasses of +2.5 D (convex, f = +40 cm) form a virtual, erect image of the newspaper at about 66.7 cm from the lens, magnified 2.67 times. Grandfather sees the print sharply.

What this shows: Every pair of reading glasses is a weak convex lens used inside its focal length — the magnifying-glass case. A prescription of higher +P is needed for worse presbyopia because a shorter focal length moves the virtual image closer to where a stiffened eye can still focus.

Example 3: Correcting a classmate's myopia

Your classmate reads the blackboard from the first row with ease but cannot read it from the back row. An optometrist tests her and finds that her far point — the farthest point at which her eye can form a sharp image on its retina without effort — is only 2.5 m. She is myopic. Prescribe a concave spectacle lens that, when placed immediately in front of her eye, brings distant objects (treated as at infinity) to a sharp virtual image exactly at her 2.5 m far point.

Concave lens correcting myopia by imaging infinity at the far point Parallel rays from infinity enter a concave lens. They diverge after refraction as if coming from a virtual image at v=−2.5 m, the far point of the myopic eye. $O$ (concave lens) virtual image at $v = -2.5$ m parallel rays from $\infty$
A concave lens takes parallel rays from infinity and bends them so that they appear to diverge from a virtual image at the myopic eye's far point. The eye then relaxes and focuses this virtual image on the retina.

Step 1. State the requirement in lens-formula language.

Object at infinity: u \to -\infty, so 1/u \to 0. Virtual image at the far point, on the incident side: v = -2.5 m = -250 cm.

Why: the spectacle lens must take parallel rays (from a distant blackboard) and produce an image exactly at the farthest point the classmate's naked eye can still focus on. Any closer would still be sharp; farther would be blurry.

Step 2. Apply the lens formula.

\frac{1}{v} - \frac{1}{u} \;=\; \frac{1}{f}
\frac{1}{-2.5} - 0 \;=\; \frac{1}{f}
f \;=\; -2.5 \text{ m}

Step 3. Convert to power.

P \;=\; \frac{1}{f} \;=\; \frac{1}{-2.5} \text{ m}^{-1} \;=\; -0.4 \text{ D}

Why: the negative power (and negative focal length) tells you this is a concave (diverging) lens — which is always the answer for myopia correction. A real, opaque classmate corrected to -0.40 D is a very mild case; serious myopes wear -3 D to -8 D.

Result: A concave spectacle lens of power P = -0.4 D, focal length f = -2.5 m, brings distant blackboard writing to a sharp virtual image at her far point. She can now read the back-row view.

What this shows: The power of a corrective lens for simple myopia is the reciprocal of the (negative of the) far-point distance in metres. A far point of 50 cm (-0.5 m) needs P = -2 D. A far point of 1 m needs P = -1 D. Opticians compute this in five seconds and hand you a frame with the right lens.

Common confusions

If you came here to use the lens formula, compute powers, and correct a friend's spectacles, you have what you need. What follows is for readers heading into JEE Advanced — the Newtonian form of the lens formula, the silvered lens, and why the lens formula fails when the object or image is not small.

The Newtonian form of the lens equation

Measure object and image distances from the foci rather than from the optical centre:

x_1 \;=\; \lvert u\rvert - f \qquad (\text{object distance from } F_1)
x_2 \;=\; v - f \qquad (\text{image distance from } F_2)

Substitute into 1/v - 1/u = 1/f (with u = -\lvert u\rvert):

\frac{1}{x_2 + f} + \frac{1}{x_1 + f} \;=\; \frac{1}{f}

Clear the fractions and simplify:

f(x_1 + f) + f(x_2 + f) \;=\; (x_1 + f)(x_2 + f)
fx_1 + f^2 + fx_2 + f^2 \;=\; x_1 x_2 + fx_1 + fx_2 + f^2
f^2 \;=\; x_1 x_2

Why: after cancelling the cross terms, a remarkable identity survives. The product of distances from the foci equals f^2. This is Newton's form of the lens equation.

\boxed{\;x_1 \, x_2 \;=\; f^2\;}

Newton's form is what you use when you measure distances from the foci (as in a photographic setup on an optical bench, where you mark F_1 and F_2 directly on the rail). It is also the form that makes projective geometry of lens systems clean: lens systems are, at heart, maps that preserve the product x_1 x_2 = f^2.

The silvered lens — a lens that doubles as a mirror

Silver the back surface of a convex lens. Light entering from the front now refracts at surface 1, reflects off the silvered back, and refracts again at surface 1 on the way out. The combination behaves like a single spherical mirror of effective focal length f_{\text{eff}}, where the power adds:

P_{\text{eff}} \;=\; 2 P_{\text{lens}} + P_{\text{mirror}}

For a plano-convex lens (f_{\text{lens}} = 20 cm, so P_{\text{lens}} = +5 D) with its flat side silvered (flat mirror, P_{\text{mirror}} = 0), the silvered lens has P_{\text{eff}} = +10 D, so f_{\text{eff}} = 10 cm. This is the source of a favourite JEE problem: given a silvered lens, find its effective focal length. The answer is always the addition-of-powers formula, counting the lens twice (light passes through it twice) and the mirror once.

When does the thin-lens approximation break?

The derivation above used two crucial simplifications.

  1. The lens is thin. The distance between the two refracting surfaces is negligible compared to \lvert u\rvert, \lvert v\rvert, and \lvert f\rvert. For a real lens of thickness t comparable to f (a hand magnifier 1 cm thick with f = 5 cm is borderline), a thick-lens correction is needed — you must treat each surface separately and compose them with matrix optics.

  2. Paraxial rays. All rays stay close to the principal axis. This is the regime where \sin\theta \approx \tan\theta \approx \theta. Outside it, different rays focus at different points — spherical aberration. It is why the camera lenses in your phone are not single thin lenses but stacks of carefully shaped elements that cancel each other's aberrations.

Also: different wavelengths of light refract differently through glass (dispersion), so red and blue images form at slightly different positions — chromatic aberration. Even a perfect paraxial lens suffers from chromatic aberration. The fix is an achromatic doublet: a convex crown-glass lens cemented to a concave flint-glass lens, designed so the two lenses' chromatic aberrations cancel. Every decent telescope objective, from the dome at the Indian Institute of Astrophysics at Kavalur to the binoculars used in an Indian Army regiment, is an achromatic doublet.

The full object-image map is a Möbius transformation

Writing 1/v - 1/u = 1/f in terms of dimensionless variables \tilde{u} = u/f, \tilde{v} = v/f:

\tilde{v} \;=\; \frac{\tilde{u}}{\tilde{u} + 1}

This is a Möbius transformation — a rational function of the form (a\tilde{u} + b)/(c\tilde{u} + d) with a = 1, b = 0, c = 1, d = 1. Möbius transformations are exactly the maps that preserve circles and lines on the Riemann sphere; the physics of lenses inherits this geometric cleanness, which is why combinations of lenses always give another lens (another Möbius transformation), and why matrix optics with 2 \times 2 ray-transfer matrices is the right formalism for multi-lens systems. A product of Möbius transformations is again a Möbius transformation; a product of lenses is again a lens.

Where this leads next