Thin Lens Equation and Lensmaker's Equation

Slip on a pair of glasses, raise a magnifying glass to a postage stamp, or press the shutter on a camera, and you are trusting the same two curved pieces of glass to do the same job: gather a fan of rays leaving a point and bend them back toward a point somewhere else. A lens looks like a single object, but optically it is two refracting surfaces working in series — light bends once on entering the glass and once on leaving it. When the glass is thin enough that the gap between those two bends is negligible next to the focal length, the whole element collapses to a single plane carrying a single number: its power. That idealisation is the thin lens, and it carries almost the entire practical load of imaging optics. Two formulas run the show. The lensmaker's equation builds a lens: it turns the two surface curvatures and the glass index into a focal length. The thin-lens equation then uses the lens: given where the object sits, it says where the image lands, and whether that image is real or virtual, upright or inverted, magnified or shrunk. Both drop straight out of one earlier result — refraction at a single spherical surface — applied twice and added. The price of that power is discipline. Every distance, every radius, every focal length carries a sign, and one flipped sign turns a converging lens into a diverging one or a real image into a virtual one. We fix a single convention and hold it for every symbol, because the most common mistake in all of geometric optics is not a wrong formula but a wrong sign.

Key takeaways

A thin lens is two spherical refracting surfaces in the limit of negligible separation; light effectively bends at one plane and adds a fixed convergence, the power $P=1/f$ .
The lensmaker's equation $1/f=(n_l-1)(1/R_1-1/R_2)$ gives the focal length from the two radii and the glass index — the design formula.
The thin-lens equation $1/s_o+1/s_i=1/f$ locates the image; the transverse magnification $M_T=-s_i/s_o$ fixes its size and orientation — the imaging formulas.
Signs are everything: with the Hecht convention $s_o>0$ is a real object, $s_i>0$ a real image, $R>0$ when the centre of curvature lies on the outgoing side, and $f>0$ for a converging lens.
For a converging lens, a real object forms a real, inverted image when $s_o>f$ ; when $s_o<f$ , the image is virtual, erect, and magnified --- the regime used by a magnifying glass.
Key idea: applying the spherical-refraction formula at each surface of a thin lens and then taking the negligible-thickness limit gives $\frac{1}{f}=(n_l-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ , while paraxial ray geometry gives $\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f}$ . The first determines the focal length from the lens geometry; the second determines the image position.

The core insight

A thin lens is nothing more than two spherical refractions stacked at one plane. Compose them in the paraxial, zero-thickness limit and the intermediate image cancels, leaving the lensmaker's equation for the focal length and the thin-lens equation for imaging. Read with a single, rigidly held sign convention, those two relations locate every image a simple lens can form and label it real or virtual, erect or inverted, enlarged or reduced.

Where this lives today

Autofocus relies on the same thin-lens imaging picture. A phone camera moves its lens by small amounts while an autofocus system evaluates image sharpness or phase information, adjusting the lens position until the subject is focused on the sensor. The thin-lens equation provides the underlying geometrical relation between object distance, image distance, and focal length. Eyeglass and contact-lens prescriptions are written in dioptres — $P=1/f$ — precisely because thin lenses in contact add their powers, so the optician's correction simply sums with the eye's own $\approx 60\,\text{D}$ . In the optics lab, placing a fibre tip or pinhole near the front focal plane of a converging lens can produce an approximately collimated beam, while two suitable lenses separated by approximately $f_1+f_2$ can form a Keplerian beam expander with angular-magnification magnitude $f_2/f_1$ . Where the thin-lens idealisation runs out, its successors carry Gauss's programme forward: finite thickness is absorbed into principal planes, the wavelength dependence $f(\lambda)$ that blurs colour is cancelled by cementing a crown lens to a flint lens into an achromatic doublet (powers adding, dispersions opposing), and the residual paraxial errors become the Seidel aberrations that lens-design software tames by ray-tracing. The thin-lens equation and lensmaker's equation remain the starting point for many such first-order optical designs, before more detailed corrections are included.

The physical picture

Picture a lens as a device that adds a fixed dose of convergence to whatever bundle of rays passes through. Rays arriving parallel to the axis leave aimed at a single point — the focal point $F'$ a distance $f$ behind the lens. Rays diverging from a nearby object are bent less steeply toward the axis, so they reconverge farther away; the closer the object, the farther the image. That single trade-off is the reciprocal relation $1/s_i = 1/f - 1/s_o$ , and the whole geometry of imaging is contained in it. The cleanest way to see where an image forms is to trace three special rays from the tip of the object. One leaves parallel to the axis and, by definition of the focal point, bends through $F'$ . One passes straight through the centre of a thin lens undeviated. One heads through the front focal point $F$ and emerges parallel to the axis. All three must cross at the image tip — three routes to the same conclusion, which is why the construction is so reliable.

Figure

A converging lens forms a real inverted image — the parallel ray bends through F′, the central ray runs straight, the focal ray leaves parallel, and all three cross at the image tip

Now change one thing — slide the object inside the focal length. The rays from it are diverging too steeply for the lens to bring them back to a crossing on the far side; instead they leave still diverging, and their backward extensions meet on the same side as the object. There is no light there, so the image is virtual: you cannot catch it on a screen, but your eye, looking through the lens, sees an enlarged upright version of the object. That is a magnifying glass. The single reciprocal curve below captures both behaviours at once — and the boundary between them is the sharpest feature on the plot.

Figure

Thin-lens curve

s_i

versus

s_o

for

f = 10\,\text{cm}

— real images to the right of

s_o = f

, the shaded magnifier band (object nearer than

f

) giving virtual erect images, and the crossover where

s_i \to \infty

s_o = f

The blue branch is the familiar world of the projector and camera: for a real object beyond the focal length, the lens forms a real image on the screen or sensor; the image distance increases from $f$ (object at infinity) toward infinity as the object approaches $f$ . Cross into the amber band — where the real object lies closer than the focal length — and the curve moves below the axis: $s_i$ becomes negative, indicating a virtual, erect image whose magnification exceeds one. The vertical divide at $s_o=f$ , where $s_i$ runs off to infinity, is the real-to-virtual crossover, drawn in centimetres. One lens, one equation, two different imaging regimes on either side of a single point.

The core equations

Because this is an imaging article, we commit to one sign convention and never leave it. Hecht convention: light travels left to right; the object distance $s_o$ is positive for a real object (to the left of the lens); the image distance $s_i$ is positive for a real image (to the right, where the light actually goes); a radius $R$ is positive when the surface's centre of curvature lies on the outgoing side; and the focal length $f$ is positive for a converging lens. Every formula below is written in these signs. The foundation is a result we recall from Refraction at a Spherical Surface: a single spherical interface of radius $R$ separating indices $n_1$ and $n_2$ images an axial point by

\frac{n_1}{s_o}+\frac{n_2}{s_i}=\frac{n_2-n_1}{R} \qquad (1)

where: $n_1$ — index on the incoming side; $n_2$ — index on the outgoing side; $R$ — radius of the surface, positive if its centre lies downstream. This is the atom; a lens is two of them. Apply it once at each refracting face of a thin lens of index $n_l$ in air, neglect the thickness, and the intermediate image cancels to give the lensmaker's equation, which connects the surface geometry and refractive index to the focal length:

\frac{1}{f}=(n_l-1)\!\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \qquad (2)

where: $n_l$ — refractive index of the lens material relative to the surrounding medium; $R_1$ — radius of the first (front) surface; $R_2$ — radius of the second (back) surface, each signed by the centre-of-curvature rule. Read it physically: power grows with how strongly the material bends light (through the factor $n_l-1$ ) and how tightly the faces curve. A flat face ( $R\to\infty$ ) adds no power; if the two surface curvatures cancel, the net power can vanish. Once $f$ is known, the thin-lens equation locates the image of an axial object:

\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f} \qquad (3)

where: $s_o$ — object distance ( $>0$ real object); $s_i$ — image distance ( $>0$ real image on the far side, $<0$ virtual image on the near side); $f$ — focal length ( $>0$ converging, $<0$ diverging). The sign of $s_i$ that falls out is the answer to "real or virtual". To finish the description — size and orientation — the transverse magnification compares image height $y'$ to object height $y$ :

M_T=\frac{y'}{y}=-\frac{s_i}{s_o} \qquad (4)

where: $M_T$ — transverse magnification; a negative value means inverted, a positive value erect, and $|M_T|>1$ means enlarged. Finally, opticians and lab optics quote a lens not by focal length but by power in dioptres, which is simply the reciprocal focal length in metres:

P=\frac{1}{f}\quad[\text{dioptres, D}],\qquad 1\,\text{D}=1\,\text{m}^{-1} \qquad (5)

where: $P$ — optical power; $f$ — focal length in metres. Power is the natural currency because thin lenses in contact simply add their dioptres — the reason an optometrist can stack a correction onto the eye's own $\approx 60\,\text{D}$ by addition.

Derivation ABuilding the lensmaker's equation from two surfaces

We want to prove — not assert — that a lens's focal length comes entirely from its two surfaces, by feeding the single-surface refraction law through both faces of the glass and taking the thin limit. The plan: image the object through surface 1 (air into glass, radius $R_1$ ) to get an intermediate image; feed that intermediate image as the object for surface 2 (glass into air, radius $R_2$ ); collapse the two surfaces to one plane; and show the intermediate-image term cancels, leaving a focal length that depends only on $n_l$ , $R_1$ and $R_2$ .

Step 1. Refract at the first surface with $n_1=1$ , $n_2=n_l$ : equation $(1)$ gives $\dfrac{1}{s_o}+\dfrac{n_l}{s_{i1}}=\dfrac{n_l-1}{R_1}$ , where $s_{i1}$ is where surface 1 alone would form the image. Why: the glass front is just one spherical interface, so the surface law applies verbatim with air outside and glass in.

Step 2. Refract at the second surface with $n_1=n_l,n_2=1$ . In the thin-lens limit, the two surfaces are effectively in the same plane. The intermediate image formed by surface 1 lies on the outgoing side of surface 2, so it acts as a virtual object for the second refraction and has object distance $-s_{i1}$ . Equation (1) then reads $-n_l/s_{i1}+1/s_i=-(n_l-1)/R_2$ . This negative sign is what makes the intermediate-image terms cancel when the two surface equations are added.

Step 3. Add the two equations. The $+n_l/s_{i1}$ term from Step 1 and the $-n_l/s_{i1}$ term from Step 2 cancel, giving $1/s_o+1/s_i=(n_l-1)(1/R_1-1/R_2)$ . The intermediate image disappears from the final relation because it is only an internal step in the two-surface refraction.

Step 4. Send the object to infinity, $1/s_o\to 0$ , so the image forms at the back focal point, $s_i=f$ . Comparing with Step 3 identifies $\dfrac{1}{f}=(n_l-1)\!\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ , the lensmaker's equation, and simultaneously delivers $1/s_o+1/s_i=1/f$ . Why: "focal length" is defined as where collimated light converges, so this limit is what pins the abstract $f$ to the physical lens. *Adding the two surface equations confirms that $s_{i1}$ drops out exactly, and the result matches $1/s_o+1/s_i=(n_l-1)(1/R_1-1/R_2)$ term for term.*

Go deeper · Derivation B · 2 minThe thin-lens equation and the sign of the magnification

Derivation BThe thin-lens equation and the sign of the magnification

Having reduced the lens to a single plane of power $1/f$ , we now extract the imaging relation and the magnification, and — crucially for an imaging article — check that the signs behave for both a real and a virtual image. The plan: rearrange the thin-lens equation to solve for the image; use the undeviated central ray and similar triangles to get $M_T$ ; then evaluate a converging lens at an object outside its focus and again at an object inside its focus, reading the sign of $s_i$ and of $M_T$ each time.

Step 1. Write equation $(3)$ as $\dfrac{1}{s_i}=\dfrac{1}{f}-\dfrac{1}{s_o}$ . A positive result means $s_i>0$ : a real image on the outgoing side. A negative result means $s_i<0$ : a virtual image on the incoming side. Why: the algebra itself decides real versus virtual — there is no separate rule to remember, only the sign that emerges.

Step 2. The ray through the centre of a thin lens is undeviated, so object and image tips subtend equal and opposite angles at the centre: $y'/y=-s_i/s_o$ , which is $M_T=-s_i/s_o$ . A real image ( $s_i>0$ ) with a real object ( $s_o>0$ ) gives $M_T<0$ — inverted; a virtual image ( $s_i<0$ ) gives $M_T>0$ — erect. Why: the minus sign is geometric, born of the ray crossing the axis at the centre, and it is what couples "which side" to "which way up".

Step 3 (real). For a converging lens, $f=+10\,\text{cm}$ , with the real object beyond the focal point at $s_o=15\,\text{cm}$ : $1/s_i=1/10-1/15=1/30$ , so $s_i=+30\,\text{cm}$ and $M_T=-30/15=-2$ — a real, inverted image twice the object size. The lens converts the diverging rays from the object into a converging bundle that crosses on the far side of the lens.

Step 4 (virtual). Now slide the object inside the focus, $s_o=5\,\text{cm}<f$ : $1/s_i=1/10-1/5=-1/10$ , so $s_i=-10\,\text{cm}$ and $M_T=-(-10)/5=+2$ — a virtual, erect image twice the size. As $s_o\to f$ from either side, $1/s_i\to 0$ and $s_i\to\pm\infty$ : that divergence is the real-to-virtual crossover. Why: inside the focus the lens can no longer force a real crossing, so it hands back a magnified upright image — the magnifying glass, living entirely in the $s_o<f$ band. *Both cases reproduce $s_i=+30$ ( $M=-2$ , real inverted) and $s_i=-10$ ( $M=+2$ , virtual erect), with the signs confirmed explicitly.*

Limiting cases

Case	Result	Meaning
Object at infinity ( $s_o\to\infty$ )	$s_i=f$	collimated input focuses at the back focal point $F'$ — this defines $f$
Object at the focus ( $s_o=f$ )	$s_i\to\infty$	output collimated; the real $\leftrightarrow$ virtual crossover
Object inside the focus ( $s_o<f$ )	$s_i<0,\ M_T>1$	virtual, erect, magnified — the simple magnifier / eyepiece
Symmetric conjugates ( $s_o=2f$ )	$s_i=2f,\ M_T=-1$	unit-magnification inverted real image — the relay
Diverging lens ( $f<0$ ), real object	$s_i<0,\ 0<M_T<1$	always virtual, erect, minified — no real image possible
Flat or equal radii ( $R_1=R_2$ )	$1/f=0$	zero power — a window, not a lens

Common trap“A magnifying glass magnifies, so a converging lens always makes things bigger”

It does not. A converging lens produces a magnified image when the object lies between the lens and twice the focal length: for $s_o<f$ the image is virtual, erect and $M_T>1$ — the magnifying-glass regime — while for $f<s_o<2f$ it is real, inverted and also magnified, with $|M_T|>1$ . At $s_o=2f$ the image is the same size as the object, and for $s_o>2f$ it is real, inverted and reduced ( $|M_T|<1$ ) — which is precisely how a camera fits a whole landscape onto a tiny sensor. The trap has a sign at its heart: a virtual image returns $s_i<0$ , and it is that negative $s_i$ that flips $M_T=-s_i/s_o$ positive (erect) and can push it above one. Whenever you quote a magnification, check the sign of $s_i$ first: positive means real and inverted, negative means virtual and erect. And beware the mirror-image trap in the lensmaker's equation — for a biconcave (diverging) lens the front radius is negative and the back radius positive, the reverse of a biconvex lens, so mixing the signs turns a diverging lens into a converging one on paper.

Worked examples

1 · One lens, both images — building and using a biconvex lens

A biconvex lens is ground from crown glass, $n_l=1.5$ , with front radius $R_1=+10\,\text{cm}$ and back radius $R_2=-10\,\text{cm}$ . (a) Find its focal length and power. (b) An object stands $15\,\text{cm}$ in front of it — locate and describe the image. (c) The object is moved to $5\,\text{cm}$ — locate and describe the image.

Step 1. Build the lens with the lensmaker's equation. With the centre of the front surface downstream, $R_1=+10\,\text{cm}$ ; the back surface bulges the other way, so $R_2=-10\,\text{cm}$ . Then $1/f=(1.5-1)\left(\tfrac{1}{10}-\tfrac{1}{-10}\right)=0.5\times\tfrac{2}{10}=0.1\,\text{cm}^{-1}$ , giving $\boxed{f=+10\,\text{cm}}$ and power $P=1/(0.10\,\text{m})=\boxed{+10\,\text{D}}$ . The positive $f$ confirms a converging lens.

Step 2 (b, real). With $s_o=15\,\text{cm}>f$ : $1/s_i=1/f-1/s_o=1/10-1/15=1/30$ , so $\boxed{s_i=+30\,\text{cm}}$ — positive, hence a real image $30\,\text{cm}$ behind the lens. Its magnification is $M_T=-s_i/s_o=-30/15=\boxed{-2}$ : inverted and twice life-size. Put a screen there and the image lands on it.

Step 3 (c, virtual). With $s_o=5\,\text{cm}<f$ : $1/s_i=1/10-1/5=-1/10$ , so $\boxed{s_i=-10\,\text{cm}}$ — negative, hence a virtual image $10\,\text{cm}$ in front of the lens, on the object's side. Its magnification is $M_T=-(-10)/5=\boxed{+2}$ : erect and twice life-size. No screen can catch it; the eye looking through the lens sees the enlarged upright image. This is the magnifying glass.

Step 4. The two regimes are two branches separated by the focal point: crossing $s_o=f=10$ cm changes the image from real and inverted to virtual and erect, while the image distance grows without bound as $s_o$ approaches $f$ from either side. This is why a magnifier is not useful when the object is placed extremely close to the focal point.

2 · A diverging lens and its dioptre prescription

A biconcave lens has $n_l=1.5$ , $R_1=-10\,\text{cm}$ and $R_2=+10\,\text{cm}$ . Find its power, and locate the image of an object $15\,\text{cm}$ away.

Step 1. The signs reverse for a biconcave lens: the front surface's centre lies upstream, so $R_1=-10\,\text{cm}$ , and $R_2=+10\,\text{cm}$ . The lensmaker's equation gives $1/f=(1.5-1)\left(\tfrac{1}{-10}-\tfrac{1}{10}\right)=0.5\times(-\tfrac{2}{10})=-0.1\,\text{cm}^{-1}$ , so $\boxed{f=-10\,\text{cm}}$ and $P=1/(-0.10\,\text{m})=\boxed{-10\,\text{D}}$ — a strong diverging lens, the kind that corrects severe short-sight.

Step 2. Image the object with $s_o=15\,\text{cm}$ : $1/s_i=1/f-1/s_o=-1/10-1/15=-\tfrac{3}{30}-\tfrac{2}{30}=-\tfrac{5}{30}=-\tfrac{1}{6}$ , so $\boxed{s_i=-6\,\text{cm}}$ . The negative sign marks a virtual image, $6\,\text{cm}$ in front of the lens on the object's side.

Step 3. Magnification: $M_T=-s_i/s_o=-(-6)/15=\boxed{+0.4}$ — erect and reduced to 40%. A diverging lens does this for every real object: the image is always virtual, erect and minified, sitting between the lens and its front focal point. That is why a peephole or a strong myopia lens shows a shrunken, upright, wide field.

Step 4. Two examples, one lesson: the sign of $f$ distinguishes converging from diverging lenses, while the sign of $s_i$ identifies real and virtual images under the chosen convention. For these paraxial thin-lens cases, the arithmetic then captures the essential imaging geometry.

Physics of the portrait

Lenses were ground and used for centuries before anyone reduced a whole optical system to a handful of numbers. The person who did — who showed that any train of lenses, however thick and however many, behaves exactly like a single equivalent lens once you know its focal points and principal planes — turned optics into an algebra.

Origins · 1 minCarl Friedrich Gauss and how this was discovered

Carl Friedrich Gauss

1777–1855 · Göttingen

Gauss is remembered for number theory, the method of least squares, and terrestrial magnetism, but late in life he also refounded optics. In his 1841 Dioptrische Untersuchungen (Dioptrical Investigations) he showed that, in the paraxial regime where angles are small, an arbitrarily complicated centred lens system is completely characterised by a set of cardinal points — two focal points and two principal planes (with nodal points for systems in different media). Once these are known, one locates images and computes magnifications with the same simple formulas that govern a single thin lens, treating the entire instrument as one black box of known focal length. This is why the paraxial, first-order regime of imaging is called Gaussian optics to this day: it is the framework in which the thin-lens and lensmaker's equations live, and the foundation on which every camera, microscope and telescope is first designed before aberrations are addressed.

History

1611

— Kepler's Dioptrice analyses the converging lens and lays out the theory of the astronomical (Keplerian) telescope.

1637

— Descartes's La Dioptrique treats refraction at lens surfaces and the aspheric "Cartesian ovals" that image perfectly.

1690

— Huygens's Traité de la Lumière develops the wavefront construction and applies it to lenses and compound eyepieces.

1841

— Gauss's Dioptrische Untersuchungen founds paraxial imaging: focal lengths, principal planes, and the cardinal points of any centred system.

Going further

Texts. Hecht, Optics, 5th ed., Ch. 5 (Geometrical Optics — the thin lens, the lensmaker's equation, and the sign convention adopted here) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 1 (Ray Optics) · Jenkins & White, Fundamentals of Optics, Ch. 3–5 (lenses and the Gaussian formulation) · Born & Wolf, Principles of Optics, Ch. 4 (paraxial imaging and cardinal points). Both formulas descend from single-surface refraction $n_1/s_o+n_2/s_i=(n_2-n_1)/R$ applied at each face and added in the thin limit. Key numerical anchors: the biconvex case $n_l=1.5,\,R_1=+10,\,R_2=-10\,\text{cm}\Rightarrow f=+10\,\text{cm}$ , power $+10\,\text{D}$ ; the thin-lens imaging pair $s_o=15\,\text{cm}\Rightarrow s_i=+30\,\text{cm},\,M=-2$ (real) and $s_o=5\,\text{cm}\Rightarrow s_i=-10\,\text{cm},\,M=+2$ (virtual); the diverging lens $f=-10\,\text{cm}$ , power $-10\,\text{D}$ , with $s_o=15\,\text{cm}\Rightarrow s_i=-6\,\text{cm},\,M=+0.4$ . The relaxed-eye power $\approx 60\,\text{D}$ and $n_{\text{water}}\approx 1.33$ are standard figures.

Prerequisites. Refraction at a Spherical Surface · Snell's Law of Refraction · Ray Model of Light · Refractive Index and Optical Path Length. Leads to. Image Formation and Magnification · Thick Lenses and Principal Planes · Stops, Pupils, and Field of View · Dispersion and Prisms.

Why is a real image inverted but a virtual image erect?

It comes straight from the magnification sign, $M_T=-s_i/s_o$ . For a real object $s_o>0$ always, so the sign of $M_T$ is set entirely by the sign of $s_i$ . A real image forms on the far side with $s_i>0$ , making $M_T<0$ — inverted. A virtual image forms on the near side with $s_i<0$ , making $M_T>0$ — erect. Geometrically, the central ray crosses the axis at the lens centre only for a real image, and that crossing is what flips the picture upside down. No crossing, no inversion.

What does a negative image distance actually mean physically?

A negative $s_i$ means the image lies on the same side as the incoming object, where no real light converges. The emerging rays are diverging, and only their backward extensions meet — so there is no glowing spot in space you could land on a screen. Your eye, however, interprets those diverging rays as though they came from that point, and sees an image there. That is a virtual image: perfectly visible, entirely real to your eye, but not projectable. Magnifying glasses, diverging lenses, and flat mirrors all form them.

Why does the same lens have a longer focal length underwater?

Because power depends on the index contrast between lens and surroundings, not the lens index alone. In air the bending factor is $(n_l-1)\approx 0.5$ ; immerse the lens in water and it becomes $(n_l/n_{\text{water}}-1)\approx(1.5/1.33-1)\approx 0.13$ , nearly four times smaller. The radii are unchanged, so $1/f$ shrinks and $f$ grows — the lens is much weaker. This is exactly why your eye cannot focus underwater (its cornea loses most of its power) and why swim goggles, by restoring an air gap in front of the eye, bring the world back into focus.

Why do the two radii subtract in the lensmaker's equation instead of adding?

Because the two surfaces contribute with opposite algebraic sense once the sign convention is applied. The combination $(1/R_1-1/R_2)$ is what falls out of adding the two single-surface equations, where the second surface enters with $R_2$ and the index step reversed. For a biconvex lens $R_1>0$ and $R_2<0$ , so $-1/R_2$ is positive and the two terms reinforce — both surfaces converge. Writing it as a subtraction with signed radii is what lets one formula handle biconvex, biconcave, plano-convex and meniscus lenses without special cases.

Is the thin-lens equation exact, or an approximation?

It is a double idealisation. First, it is paraxial: it assumes rays make small angles so that $\sin\theta\approx\theta$ , which is what makes all points of the object image to sharp points; real wide-angle rays miss slightly, producing spherical aberration. Second, it is thin: it ignores the lens thickness, so the object and image distances are both measured from one plane. Real lenses need the thick-lens correction, which restores the thickness through a term $(n_l-1)d/(n_l R_1 R_2)$ and splits the single plane into two principal planes. For thin lenses and modest apertures, though, the simple equation is accurate to a fraction of a percent.

What exactly is a dioptre, and why do opticians use it instead of focal length?

A dioptre is one inverse metre: a lens of focal length $f$ (in metres) has power $P=1/f$ dioptres, so a $+2\,\text{D}$ lens has $f=0.5\,\text{m}$ and a $-4\,\text{D}$ lens has $f=-0.25\,\text{m}$ . Opticians prefer power because thin lenses in contact add their powers — combining the eye's optics with a corrective lens is then simple addition, and a prescription can be read off directly. Power also grows smoothly through zero (a flat window is $0\,\text{D}$ ), whereas focal length blows up to infinity there, which is awkward to tabulate.

Can a converging lens ever make an object look smaller?

Yes — that is its everyday job in a camera. Once the object sits beyond twice the focal length ( $s_o>2f$ ), the real image it forms has $|M_T|<1$ : inverted and reduced. That is how a camera lens squeezes a whole scene onto a small sensor, and how your own eye images a distant building onto the few millimetres of your retina. The enlarging behaviour people associate with a "lens" only appears in the narrow window $s_o<f$ , where the image turns virtual and erect. The same piece of glass shrinks or magnifies depending entirely on where the object sits relative to $f$ and $2f$ .