Lensmaker's Formula and the Imaging Equation

Derivation

Lensmaker's Formula and the Imaging Equation

D-200 Home PU-206 Threads light Depends on Paraxial Ray-Transfer (ABCD) Matrices, Laws of Reflection and Refraction from Fermat

Statement

For a lens of refractive index \(n\) bounded by two spherical surfaces of radii \(R_1\) and \(R_2\), immersed in a medium of index \(n_0\), the focal length \(f\) obeys the lensmaker's equation, and any pair of conjugate object/image distances \(s,s'\) obeys the thin-lens imaging relation. We derive both by composing the ray-transfer (ABCD) matrices for refraction at surface 1, translation through the glass, and refraction at surface 2, then reading the imaging condition off the system matrix.

Why it matters

The lensmaker's equation is the bridge from a lens's geometry and material (two radii, one index) to its single optical parameter, the focal length. Every camera, microscope, telescope, and pair of spectacles is designed by fixing \(f\) through this relation and then chaining lenses.

Deriving it through the ABCD formalism, rather than by tracing individual rays, shows that imaging is a statement about a single matrix element vanishing. The same machinery then generalizes without effort to thick lenses, multi-element systems, and Gaussian-beam propagation, so the thin lens is revealed as one limit of a unified linear theory.

Assumptions

Paraxial (small-angle) rays.If dropped, \(\sin\theta \neq \theta\) and spherical aberration appears: rays through the rim focus nearer than paraxial rays, so a single \(f\) no longer describes the lens.

Rotationally symmetric spherical surfaces.If dropped (cylindrical or toroidal surfaces, or tilt/decenter), the matrix becomes different in the two transverse planes and the lens is astigmatic, needing separate \(f_x, f_y\).

Monochromatic light, single index \(n\).If dropped, \(n(\lambda)\) makes \(f(\lambda)\) wavelength-dependent — chromatic aberration — so white light does not share one focus.

Thin lens: vertex separation \(d \to 0\) relative to \(R_1,R_2,f\).If dropped, the \(d\)-dependent terms survive and one gets the thick-lens equation with distinct principal planes; \(s,s'\) must then be measured from those planes, not from a single vertex.

Homogeneous, isotropic, non-absorbing media.If dropped, index gradients bend rays continuously (GRIN optics) and the piecewise-surface refraction model no longer applies; birefringence would split each ray in two.

Derivation

We use the ray vector \(\begin{pmatrix} y \\ \theta \end{pmatrix}\), where \(y\) is transverse height and \(\theta\) the ray angle to the axis, and the convention that light travels in the \(+z\) direction with the reduced-angle (Snell-invariant) form of the matrices from ray-transfer-abcd-matrices. Sign convention: \(R>0\) when the surface's center of curvature lies to the right of the vertex; \(s\) is negative for a real object to the left.

\[ M_{\text{surf}} = \begin{pmatrix} 1 & 0 \\[2pt] \dfrac{n_1-n_2}{n_2 R} & \dfrac{n_1}{n_2} \end{pmatrix} \]

Refraction at a single spherical surface between indices \(n_1\to n_2\), obtained by linearizing Snell's law \(n_1\sin\theta_1 = n_2\sin\theta_2\) at the surface; this is the prior result snell-law-and-reflection-from-fermat written in matrix form. A

\[ M_1 = \begin{pmatrix} 1 & 0 \\[2pt] \dfrac{n_0-n}{n R_1} & \dfrac{n_0}{n} \end{pmatrix}, \qquad M_2 = \begin{pmatrix} 1 & 0 \\[2pt] \dfrac{n-n_0}{n_0 R_2} & \dfrac{n}{n_0} \end{pmatrix} \]

Apply Step 1 to surface 1 (\(n_0\to n\), radius \(R_1\)) and surface 2 (\(n\to n_0\), radius \(R_2\)). The internal medium is the glass \(n\); the outside is \(n_0\). A

\[ T = \begin{pmatrix} 1 & d \\[2pt] 0 & 1 \end{pmatrix} \]

Free propagation a distance \(d\) (the axial glass thickness) between the two vertices; height grows as \(y\to y+d\theta\), angle unchanged. A

\[ M = M_2\, T\, M_1 \]

Compose in the order the light meets the elements — refract at surface 1, translate, refract at surface 2. Matrices act right-to-left on the incoming ray vector. A

\[ M = \begin{pmatrix} 1 & 0 \\ P_2 & \tfrac{n}{n_0} \end{pmatrix}\!\! \begin{pmatrix} 1 & d \\ 0 & 1 \end{pmatrix}\!\! \begin{pmatrix} 1 & 0 \\ P_1 & \tfrac{n_0}{n} \end{pmatrix}, \quad P_1=\frac{n_0-n}{nR_1},\ P_2=\frac{n-n_0}{n_0R_2} \]

Name the two lower-left entries \(P_1,P_2\) to keep the algebra readable; these are the (reduced) surface powers. B

\[ M = \begin{pmatrix} 1 + dP_1 & \dfrac{n_0}{n}\,d \\[6pt] P_2 + \dfrac{n}{n_0}P_1 + dP_1P_2 & \dfrac{n}{n_0}\cdot\dfrac{n_0}{n} + dP_2\cdot\dfrac{n_0}{n} \end{pmatrix} \]

Multiply the three matrices out. Every entry is exact so far — no thin-lens limit has been taken. B

\[ M \xrightarrow{\,d\to 0\,} \begin{pmatrix} 1 & 0 \\[4pt] P_2 + \dfrac{n}{n_0}P_1 & 1 \end{pmatrix} \]

Thin-lens limit: set \(d\to 0\), which kills every term carrying a factor \(d\). The system reduces to a single refraction-like matrix with unit diagonal. B

\[ C = P_2 + \frac{n}{n_0}P_1 = \frac{n-n_0}{n_0 R_2} + \frac{n}{n_0}\cdot\frac{n_0-n}{nR_1} = \frac{n-n_0}{n_0}\!\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \]

Substitute \(P_1,P_2\) and simplify: the \(n/n_0\) factor cancels the \(n\) inside \(P_1\), and \((n_0-n)/R_1 = -(n-n_0)/R_1\) combines with the \(R_2\) term. This lower-left entry \(C\) is minus the system power. B

\[ \frac{1}{f} \equiv -C = \frac{n-n_0}{n_0}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \]

Define the focal length by \(C=-1/f\), the standard identification of the \(C\)-element of a thin system with its negative power. For a lens in air, \(n_0=1\). A

To obtain the imaging relation we now bracket the thin lens between object- and image-space translations and demand that the output height be independent of the input angle (all rays from one object point reconverge to one image point).

\[ M_{\text{sys}} = \begin{pmatrix} 1 & s' \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -\tfrac{1}{f} & 1 \end{pmatrix} \begin{pmatrix} 1 & -s \\ 0 & 1 \end{pmatrix} \]

Translate from object (at \(z=s<0\)) to lens, refract, translate to image plane (at \(z=s'>0\)). Distances enter as \(-s\) and \(s'\) with the sign convention above. A

\[ M_{\text{sys}} = \begin{pmatrix} 1 - \dfrac{s'}{f} & -s + s' + \dfrac{ss'}{f} \\[6pt] -\dfrac{1}{f} & 1 + \dfrac{s}{f} \end{pmatrix} \]

Carry out the product. The upper-right (B) element is the one that must vanish for imaging: it is the coefficient of the input angle \(\theta\) in the output height. B

\[ B = -s + s' + \frac{ss'}{f} \overset{!}{=} 0 \]

Imaging condition: output height \(y' = A\,y + B\,\theta\) must be the same for every incoming angle \(\theta\), so \(B=0\). This is the defining requirement that a point maps to a point. A

\[ s' - s + \frac{ss'}{f} = 0 \ \Longrightarrow\ \frac{1}{s} - \frac{1}{s'} = -\frac{1}{f} \ \Longrightarrow\ \frac{1}{s'} - \frac{1}{s} = \frac{1}{f} \]

Divide through by \(ss'\) and rearrange. Dividing \(s'-s+ss'/f=0\) by \(ss'\) gives \(1/s - 1/s' + 1/f = 0\). A

Result

\[ \boxed{\ \frac{1}{f} = \frac{n-n_0}{n_0}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \qquad\text{and}\qquad \frac{1}{s'}-\frac{1}{s} = \frac{1}{f}\ } \]

Reading. The focal length is set entirely by the index contrast \((n-n_0)/n_0\) and the curvature difference \(1/R_1 - 1/R_2\); a lens in a medium of matched index (\(n=n_0\)) has \(1/f=0\) and does nothing. The imaging relation then says that once \(f\) is fixed, object and image distances are reciprocally linked: pushing the object toward the front focal point (\(s\to -f\)) sends the image to infinity, and vice versa.

Units check. Radii and distances carry units of length \([\mathrm{m}]\); the index ratio is dimensionless. Thus \(1/f\) has units \([\mathrm{m}^{-1}]\) (dioptres), and \(1/s'-1/s\) is also \([\mathrm{m}^{-1}]\) — both sides match. Multiplying by \(ss'\) restores \([\mathrm{m}]\) throughout the imaging equation.

Limiting cases

Object at infinity (\(s\to-\infty\)): \(1/s'=1/f\), so \(s'=f\) — parallel light focuses at the back focal point, the operational meaning of \(f\).
Symmetric biconvex lens (\(R_1=R=-R_2\)): \(1/f = 2(n-n_0)/(n_0 R)\); the two surfaces contribute equally.
Plano-convex (\(R_2\to\infty\)): \(1/f=(n-n_0)/(n_0 R_1)\); the flat face has zero power.
Index-matched medium (\(n\to n_0\)): \(1/f\to 0\), \(f\to\infty\); the lens is optically invisible.
Lens reversed (\(R_1\leftrightarrow -R_2\)): \(1/f\) is unchanged — thin-lens focal length is independent of orientation, though aberrations are not.
Immersion (\(n_0>1\), e.g. water): the prefactor \((n-n_0)/n_0\) shrinks, so a glass lens is weaker underwater — the origin of blurred vision without goggles.

Breaks when

Rays are non-paraxial. For finite aperture and large angles, \(\sin\theta\to\theta\) fails; the true focus varies with ray height (spherical aberration) and a single \(f\) no longer images sharply. The formula is the leading term of an expansion in \(y^2\).
The lens is not thin. When \(d\) is comparable to \(R_1,R_2,f\), the dropped \(d\)-terms of Step 6 matter; imaging distances must be measured from two separated principal planes, and \(1/f\) gains a \(-(n-n_0)^2 d/(n\, n_0 R_1 R_2)\) correction.
Light is polychromatic. Because \(n=n(\lambda)\), each wavelength has its own \(f\); the single-\(f\) result holds only per wavelength (chromatic aberration).
Surfaces are non-spherical or misaligned. Tilt, decenter, cylindrical or freeform surfaces break rotational symmetry; the \(2\times2\) scalar matrix must be replaced by a \(4\times4\) or plane-dependent treatment (astigmatism, coma).
Diffraction dominates. For apertures only a few wavelengths across, ray optics itself fails; the "focus" is a diffraction spot of finite size and geometric imaging is meaningless.

Failure modes

Sign-convention scrambling. Using \(1/s'+1/s=1/f\) (a different convention) together with radii signed by the \(R>0\)-to-the-right rule; the two conventions must be kept internally consistent or focal lengths come out with the wrong sign.
Forgetting the surrounding index. Writing \(1/f=(n-1)(1/R_1-1/R_2)\) for a lens in water; the correct prefactor is \((n-n_0)/n_0\), not \(n-1\).
Mislabeling \(R_2\). \(R_2\) is signed by the same rule as \(R_1\); for a biconvex lens \(R_2<0\), so \(1/R_1-1/R_2\) is a sum of magnitudes, not a difference. Students often drop the sign and get \(1/f\approx 0\).
Confusing focal length with image distance. \(s'=f\) holds only for an object at infinity; for finite objects \(s'\neq f\).
Applying the thin-lens result to a thick lens or a compound system without shifting to principal planes.
Treating the two surface powers as simply additive with the wrong internal index, i.e. forgetting the \(n/n_0\) weighting that Step 8 shows is exactly what cancels.

Discussion

The derivation makes precise what "a lens images" means: it is the vanishing of a single matrix element, the \(B\) (upper-right) entry of the object-to-image system matrix. That element is the coefficient of the input angle in the output height, so \(B=0\) is exactly the statement that all rays leaving one object point — whatever their angle — arrive at one image point. Everything else about the image (magnification, orientation) is then read from the surviving entries: the diagonal \(A=1-s'/f\) reduces, once \(B=0\), to \(s'/s\), which is precisely the transverse magnification \(m=y'/y\).

The lensmaker's equation separates cleanly into a material factor \((n-n_0)/n_0\) and a geometric factor \((1/R_1-1/R_2)\). This factorization is why an optician can grind a prescription: the geometry (curvature) is the adjustable knob once the glass is chosen. The dioptre, \(1/f\) in \(\mathrm{m}^{-1}\), is additive for thin lenses in contact — a direct corollary of matrix multiplication when \(d\to 0\), since the lower-left entries then simply add.

Physically, the imaging relation is the paraxial shadow of Fermat's principle: among all paths from object point to image point, the optical path lengths are made equal by the lens's extra glass thickness at the center, so all rays arrive in phase. The ABCD route hides this stationary-path content inside the surface matrix (itself derived from Snell's law, which is itself Fermat), but it is worth remembering that the reason a lens focuses at all is the equal-optical-path condition, not any accident of the algebra.

At the deepest level, the \(2\times2\) ray matrices are elements of \(SL(2,\mathbb{R})\) acting on the projective ray coordinate \(y/\theta\), and imaging is a fixed-point/collineation statement in that group — the same structure that governs Gaussian-beam \(q\)-parameter propagation via the identical ABCD law \(q' = (Aq+B)/(Cq+D)\). The determinant of every matrix here equals the ratio of initial to final index (unity for a lens in a single medium), which is nothing but the conservation of the optical étendue, or in quantum language the Liouville invariance of phase-space area \(n\,y\,\theta\). The thin lens is thus a shear in ray phase space; the imaging condition is the demand that a vertical line (a point source, all angles) maps to another vertical line (a point image).

Common misconceptions. A lens does not "focus because it is curved" in isolation — a curved slab of index-matched material has \(1/f=0\). It focuses because of the index contrast across a curved interface. Likewise, focal length is a property of the lens-plus-surrounding-medium, not of the glass alone; the same spectacle lens is measurably weaker in water.

Worked examples

Example 1 — Design a biconvex lens for a target focal length.

\[ \frac{1}{f} = \frac{n-n_0}{n_0}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \]

Start from the lensmaker's equation. Given: crown glass \(n=1.52\), in air \(n_0=1\), equal-curvature biconvex so \(R_1=+R\), \(R_2=-R\); target \(f=+100\ \mathrm{mm}\). A

\[ \frac{1}{f} = (n-1)\left(\frac{1}{R}+\frac{1}{R}\right) = \frac{2(n-1)}{R} \]

Insert \(R_1=R,\ R_2=-R\): \(1/R_1-1/R_2 = 1/R-(-1/R)=2/R\). Symbols rearranged before numbers. A

\[ R = 2(n-1)f = 2(1.52-1)(100\ \mathrm{mm}) = 2(0.52)(100\ \mathrm{mm}) = 104\ \mathrm{mm} \]

Solve for \(R\) and substitute the numbers with units. A

\[ R_1 = +104\ \mathrm{mm}, \quad R_2 = -104\ \mathrm{mm} \]

Reading. Each face is ground to a 104 mm radius; together they give a 100 mm (10-dioptre) lens. If instead \(n_0=1.33\) (water), \(1/f=2(n-n_0)/(n_0 R)\) gives \(f\approx 296\ \mathrm{mm}\) for the same glass — nearly three times weaker.

Units check. \((n-1)\) dimensionless \(\times\) length \(=\) length; \(R\) comes out in mm as required.

Example 2 — Locate the image and its magnification.

\[ \frac{1}{s'} - \frac{1}{s} = \frac{1}{f} \ \Longrightarrow\ \frac{1}{s'} = \frac{1}{f} + \frac{1}{s} \]

Thin-lens imaging relation, solved for the image distance. Given: the \(f=+100\ \mathrm{mm}\) lens above, real object 150 mm in front so \(s=-150\ \mathrm{mm}\). A

\[ \frac{1}{s'} = \frac{1}{100\ \mathrm{mm}} + \frac{1}{-150\ \mathrm{mm}} = \frac{3-2}{300\ \mathrm{mm}} = \frac{1}{300\ \mathrm{mm}} \]

Insert numbers over a common denominator of 300 mm. A

\[ s' = +300\ \mathrm{mm}, \qquad m = \frac{s'}{s} = \frac{300}{-150} = -2 \]

Read the image distance; the transverse magnification for this convention is \(m=s'/s\), the surviving diagonal element \(A\) of the system matrix. B

\[ s' = +300\ \mathrm{mm}, \quad m = -2 \]

Reading. The image is real (\(s'>0\), on the far side), 300 mm behind the lens, inverted (\(m<0\)), and twice life size — the arrangement of a slide projector. Moving the object to \(s=-100\ \mathrm{mm}=-f\) would send \(s'\to\infty\): the projector is out of range.

Units check. Both \(1/s'\) terms are \([\mathrm{mm}^{-1}]\); \(s'\) returns in mm. \(m\) is a ratio of lengths, dimensionless, as a magnification must be.

Problems

A plano-convex lens has a convex first face of radius \(R_1=+50\ \mathrm{mm}\) and a flat second face (\(R_2\to\infty\)), made of glass \(n=1.60\) in air. Find its focal length.
Solution
\(\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)=(0.60)\left(\dfrac{1}{50\ \mathrm{mm}}-0\right)=\dfrac{0.60}{50\ \mathrm{mm}}=0.012\ \mathrm{mm}^{-1}\). Thus \(f=83.3\ \mathrm{mm}\) (a converging lens; \(+12\) dioptres if we use metres: \(1/f=12\ \mathrm{m}^{-1}\)).
The same lens from Problem 1 is submerged in water (\(n_0=1.33\)). What is its new focal length, and by what factor has its power dropped?
Solution
\(\dfrac{1}{f}=\dfrac{n-n_0}{n_0}\left(\dfrac{1}{R_1}\right)=\dfrac{1.60-1.33}{1.33}\cdot\dfrac{1}{50\ \mathrm{mm}}=\dfrac{0.27}{1.33}\cdot\dfrac{1}{50\ \mathrm{mm}}=0.2030\cdot 0.02\ \mathrm{mm}^{-1}=4.06\times10^{-3}\ \mathrm{mm}^{-1}\). So \(f=246\ \mathrm{mm}\). The power drops by the factor \(\dfrac{(n-1)}{(n-n_0)/n_0}=\dfrac{0.60}{0.203}\approx 2.96\) — about a factor of 3 weaker.
An object is placed 300 mm in front of a converging lens of focal length \(f=200\ \mathrm{mm}\). Find the image distance, the magnification, and state whether the image is real or virtual, erect or inverted.
Solution
\(\dfrac{1}{s'}=\dfrac{1}{f}+\dfrac{1}{s}=\dfrac{1}{200}+\dfrac{1}{-300}=\dfrac{3-2}{600}=\dfrac{1}{600}\ \mathrm{mm}^{-1}\), so \(s'=+600\ \mathrm{mm}\). Magnification \(m=s'/s=600/(-300)=-2\). Image is real (\(s'>0\)), inverted (\(m<0\)), twice size.
A diverging lens has \(f=-150\ \mathrm{mm}\). A real object sits 100 mm in front of it. Locate the image and give its magnification.
Solution
\(\dfrac{1}{s'}=\dfrac{1}{f}+\dfrac{1}{s}=\dfrac{1}{-150}+\dfrac{1}{-100}=\dfrac{-2-3}{300}=\dfrac{-5}{300}=-\dfrac{1}{60}\ \mathrm{mm}^{-1}\). Hence \(s'=-60\ \mathrm{mm}\): the image is virtual (same side as object), \(m=s'/s=(-60)/(-100)=+0.6\) — erect and reduced, as a diverging lens always gives for a real object.
Derive, from the exact system matrix (Step 6, before the thin-lens limit), the correction to \(1/f\) that survives for a lens of finite thickness \(d\). Assume \(n_0=1\). [Hint: the focal length is read from the \(C\) element of the full matrix.]
Solution
The \(C\) element (lower-left) of \(M=M_2 T M_1\) with \(n_0=1\) is \(C=P_2+\dfrac{n}{1}P_1+dP_1P_2\) where \(P_1=\dfrac{1-n}{nR_1}\), \(P_2=\dfrac{n-1}{R_2}\). The first two terms give the thin-lens \(-1/f_{\text{thin}}=-(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)\). The extra term is \(dP_1P_2=d\cdot\dfrac{1-n}{nR_1}\cdot\dfrac{n-1}{R_2}=-\dfrac{d(n-1)^2}{n R_1 R_2}\). Therefore \(\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)+\dfrac{(n-1)^2\,d}{n\,R_1 R_2}\), the standard thick-lens lensmaker's equation. The correction vanishes as \(d\to 0\), confirming the thin-lens limit, and it is positive for a biconvex lens (\(R_1>0,R_2<0\Rightarrow R_1R_2<0\) makes the term reduce the power), so a thick biconvex lens is slightly weaker than its thin idealization predicts.