
Physics III course sequence · Interactive lesson
Spherical-surface refraction & the lensmaker's equation
This is where the lens stops being a black box with a focal length stamped on it. Work out what one curved boundary between two indices does to a paraxial ray, apply the result twice, and f falls out of two radii, the glass, and the part everyone forgets: the index of whatever the lens is sitting in.
UPIII-04.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Build the model
One surface, two indices, one radius. A thin lens is that result applied twice.
The key idea
Paraxial refraction at a spherical surface of radius R between indices n₁ and n₂ obeys n₁/s + n₂/s′ = (n₂ − n₁)/R; applied to both faces of a thin lens it gives 1/f = (nL/nₘ − 1)(1/R₁ − 1/R₂).
Example. An object 30 cm from the convex end (R = 5.0 cm) of a glass rod, n = 1.50, is imaged inside the glass at s′ = 1.50/(0.50/5.0 − 1/30) = 22.5 cm, real and inverted, with m = −22.5/(1.50 × 30) = −0.50.
The complete physical model and assumptions
A spherical boundary between indices n₁ and n₂ bends every ray by Snell's law, and nothing about Snell's law is linear. What turns the surface into an imaging device is the paraxial limit: replace each sine by its angle, write each angle as a height over a distance, and the ray height cancels from the refraction condition, leaving n₁/s + n₂/s′ = (n₂ − n₁)/R. Because h has gone, every paraxial ray from one object point reaches the same image point, which is what the word image means, and it holds only to the accuracy of sin θ ≈ θ. The right-hand side is the surface's power: set by the index step and the curvature together, and its sign, not the shape alone, decides converging or diverging. A lens is two such surfaces back to back; drop the thickness and the first surface's image is the second surface's object at the same distance, so the powers add and 1/f = (nL/nₘ − 1)(1/R₁ − 1/R₂). The textbook's (n − 1) is this with nₘ = 1. Put the same glass in water and the prefactor falls from 0.52 to 0.14, which is why an unaided eye cannot focus underwater. The model fails on three fronts, each one costable: rays at finite height cross the axis short of the paraxial focus by an amount growing as h²; a thickness comparable to the radii adds a term (n − 1)d/(nR₁R₂); and because n depends on wavelength, so does f.
1Set up one surface and declare the signs
Put an object point O on the axis a distance s in front of the vertex V of a spherical surface of radius R that separates index n₁, where the light starts, from n₂. A ray leaves O at angle α to the axis, strikes the surface at height h, refracts, and crosses the axis at the image point I a distance s′ beyond V, arriving at angle β. The normal at the strike point P passes through the centre of curvature C and makes angle φ with the axis. Fix the convention now and keep it for the lens: s > 0 for a real object on the incoming side, s′ > 0 for a real image on the outgoing side, and R > 0 when C lies on the outgoing side, so a face that bulges toward the incoming light has R > 0. Exterior-angle bookkeeping on the two triangles OPC and PCI then gives the angle of incidence θ₁ = α + φ and the angle of refraction θ₂ = φ − β. Nothing has been approximated yet; the geometry is exact.
2Linearise Snell's law and watch the ray height cancel
Snell's law, n₁ sin θ₁ = n₂ sin θ₂, is not linear, and a non-linear law cannot map points to points. The paraxial limit makes it linear: for small angles sin θ ≈ tan θ ≈ θ, and each angle is a height over a distance, α ≈ h/s, β ≈ h/s′, φ ≈ h/R, because P sits so close to V that the axial distances are s, s′ and R to first order. Substituting, n₁(h/s + h/R) = n₂(h/R − h/s′). Every term carries one factor of h, so h cancels and n₁/s + n₂/s′ = (n₂ − n₁)/R. That cancellation is the whole content of the word image: a ray leaving O at any small angle reaches the same I, and it does so only because the sines were replaced by angles. Numbers: a glass rod (n₂ = 1.50) with a convex end of R = 5.0 cm and an object 30 cm away in air gives 1/30 + 1.50/s′ = 0.50/5.0, so 1.50/s′ = 0.1000 − 0.0333 = 0.0667 cm⁻¹ and s′ = 22.5 cm inside the glass, a real image. The ray through the vertex obeys Snell's law with the same linearisation and no geometry, and it gives the magnification m = −n₁s′/(n₂s) = −22.5/(1.50 × 30) = −0.50.
3Read the right-hand side as a power and test its limits
P = (n₂ − n₁)/R is the surface's power, in dioptres when R is in metres, and it is fixed by the surface alone, whatever the object does. Its sign is the product of two signs. A convex face (R > 0) into a denser medium (n₂ > n₁) has P > 0 and converges; the same R > 0 into a rarer medium, water into an air bubble, has P < 0 and diverges. Send light out of a glass sphere instead and both signs flip together, C now on the incoming side so R < 0 and n₂ < n₁, and P is positive again: a surface converges or diverges regardless of which way the light crosses it. Now run the parameters to their limits. n₂ → n₁: the right side vanishes, s′ = −s, and the image sits on the object, as it must with no surface there. R → ∞: the face is flat, s′ = −(n₂/n₁)s, a virtual image at a reduced distance with m = +1; a coin 1.00 m under water (n₁ = 1.33) seen from air appears at 0.75 m, the apparent-depth result, its size unchanged. s → ∞: the image lands at f₂ = n₂R/(n₂ − n₁), the surface's image-side focal length, while a parallel bundle sent the other way focuses at n₁R/(n₂ − n₁): one surface has two focal lengths, in the ratio n₂/n₁. The cornea shows how strong one surface can be: (1.376 − 1)/0.0078 m ≈ 48 D from a single air-to-tissue boundary of radius 7.8 mm, most of the eye's roughly 60 D.
4Apply it twice: the first image is the second surface's object
A lens is two surfaces in a row, so image sequentially. At the first face the light enters the glass: nₘ/s₁ + nL/s₁′ = (nL − nₘ)/R₁. Its image at s₁′ is the object for the second face, but measured from the second vertex, a distance d further on: s₂ = d − s₁′. If the first image lies beyond the second face, s₂ is negative, a virtual object, and the convention absorbs it with no special case. At the second face the light leaves the glass, so the indices swap roles: nL/s₂ + nₘ/s₂′ = (nₘ − nL)/R₂. Now let d → 0, so s₂ = −s₁′, and add the two equations. The nL/s₁′ terms cancel, leaving nₘ/s₁ + nₘ/s₂′ = (nL − nₘ)(1/R₁ − 1/R₂). Divide by nₘ: 1/s + 1/s′ = (nL/nₘ − 1)(1/R₁ − 1/R₂). The right side is the same for every object distance, so it earns a name, 1/f. For a biconvex lens with R₁ = +20 cm and R₂ = −20 cm (its centre lies on the incoming side) of nL = 1.52 in air, 1/f = 0.52 × (1/20 + 1/20) = 0.052 cm⁻¹ and f = 19.2 cm. Turn the lens round and the two radii swap and change sign together, so f is the same from either side.
5The (n − 1) is a lens-in-air result
The prefactor is nL/nₘ − 1, and (n − 1) is what it becomes at nₘ = 1. Submerge the lens and the glass is unchanged; what changes is the index step at each face, so both surface powers P₁ = (nL − nₘ)/R₁ and P₂ = (nₘ − nL)/R₂ shrink. The 19.2 cm lens under water, nₘ = 1.333: 1.52/1.333 − 1 = 0.140, 1/f = 0.140 × 0.10 = 0.0140 cm⁻¹, f = 71.3 cm, 3.7 times longer. The medium enters twice. The surfaces contribute P₁ + P₂ = (nL − nₘ)(1/R₁ − 1/R₂), and 1/f = (P₁ + P₂)/nₘ, so the lens is weaker both because each step is smaller and because a given power buys less bending in a denser medium. Match the indices, nₘ = nL, and the lens vanishes, which is why a glass rod in oil of the same index cannot be seen. Go past that, nₘ > nL, and the sign flips: a biconvex pocket of air in water has prefactor 1/1.333 − 1 = −0.25 and diverges. The unaided eye under water loses its 48 D cornea to a 5.5 D one, (1.376 − 1.333)/0.0078 m, more than the lens inside the eye can recover: hence goggles, which put air back in front of the cornea.
6Where thin and paraxial stop being true
Two approximations were made and each can be costed. Thickness: keeping d in the sequential trace gives, for a lens in air, 1/f = (n − 1)[1/R₁ − 1/R₂ + (n − 1)d/(nR₁R₂)]. For the 19.2 cm lens with d = 1.0 cm the extra term is 0.52 × 1.0/(1.52 × 20 × (−20)) = −8.6 × 10⁻⁴ cm⁻¹ against the thin part's 0.100 cm⁻¹: f moves to 19.4 cm, under 1%, and is now measured from principal planes inside the glass rather than from the vertices. Paraxiality: keep the sines. A parallel ray at height h meeting a single convex surface has sin φ = h/R exactly, sin θ₂ = (n₁/n₂) sin φ, and the sine rule in triangle CPF puts its axis crossing at VF = R + R sin θ₂/sin(φ − θ₂). Expand for small φ and the leading term is n₂R/(n₂ − n₁), the paraxial f₂; the next term is negative and grows as h², so marginal rays focus short. For a glass surface (n₂ = 1.52) of R = 5.0 cm the paraxial focus is 14.6 cm from V; a ray at h = 2.0 cm crosses at 14.1 cm, 3.6% short. That gap is longitudinal spherical aberration, and it is why a fast lens is never one spherical surface.
Key equations
Read the units, assumptions and limits before substituting.
Paraxial refraction at one spherical surface
n₁/s + n₂/s′ = (n₂ − n₁)/R
Linear in 1/s and free of the ray height, so every paraxial ray from one point reaches one image point: that is what makes it an imaging law.
s, s′, R in m; n₁ on the incoming side, n₂ on the outgoing; R > 0 when the centre of curvature is on the outgoing sidePower of a single surface
P = (n₂ − n₁)/R
Convex into the denser medium converges; keep R > 0 but cross into a rarer medium, as light from water entering an air bubble, and P flips sign, so the bubble is a diverging element.
P in dioptres (m⁻¹) with R in m; the sign of P, not the shape alone, says whether the surface convergesImage-side focal length of a surface
f₂ = n₂R/(n₂ − n₁)
One surface has two unequal focal lengths; only a lens with the same medium on both sides has a single f.
f₂ in m: where a parallel incident bundle (s → ∞) crosses the axis; the object-side value is n₁/n₂ times thisLateral magnification at a surface
m = −n₁s′/(n₂s)
The index ratio is real: a flat surface (R → ∞) gives m = +1, so a submerged coin looks its true size but nearer.
m = h′/h, dimensionless; from Snell's law applied to the ray through the vertexLensmaker's equation in any medium
1/f = (nL/nₘ − 1)(1/R₁ − 1/R₂)
Two surface powers added with the thickness dropped; nₘ = 1 gives the textbook (n − 1), and nₘ = nL switches the lens off.
nL lens, nₘ surroundings, both dimensionless; R₁ first face, R₂ second, each signed by the outgoing-side rule; f in mThick-lens correction (lens in air)
1/f = (n − 1)[1/R₁ − 1/R₂ + (n − 1)d/(nR₁R₂)]
Prices the word thin: for a symmetric lens the extra term is a fraction (n − 1)d/(2n|R|) of the thin result, under 1% for d = 1 cm and R = 20 cm.
d = centre thickness in m; f is measured from the principal planes, not the verticesInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Leave both radii at 10 cm and raise the medium index from 1.00 to 1.33: F′ slides from 9.6 cm out to 35 cm while the open air-focus stays put, because the prefactor is 1.52/nₘ − 1, not 1.52 − 1. Then set nₘ back to 1.00 and pull R₂ in to −6 cm: the second face's power climbs from 5.2 D to 8.7 D.
Live results
- FACE 1 POWER P₁
- 5.20 D
- FACE 2 POWER P₂
- 5.20 D
- nₘ/f = P₁ + P₂
- 10.40 D
- FOCAL LENGTH f
- 9.6 cm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. First face radius R₁: 10 cm; Second face radius R₂: −10 cm; Medium index nₘ: 1.00
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Students carry 1/f = (n − 1)(1/R₁ − 1/R₂) into a problem about a lens under water and keep the same f, arguing that the lens has not changed. The lens has not; the equation has, because it was derived with air on both sides. Each face bends light by the index step across it, (nL − nₘ)/R, not by the glass index alone, and 1/f is the sum of the two powers divided by nₘ. For crown glass in water the step at each face falls from 0.52 to 0.19 and the focal length grows 3.7 times; make the surroundings denser than the lens and the same biconvex shape diverges. The companion error is to write R₂ positive because the second face is also convex. The sign belongs to where its centre of curvature sits relative to the outgoing light, so a biconvex lens has radii of opposite sign, and that is precisely why its two faces add rather than cancel.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
The end of a long acrylic rod (n = 1.49) is ground to a convex spherical surface of radius 4.0 cm. A small LED sits on the axis in air, 12.0 cm from the vertex. Locate its image, say whether it is real, and find the magnification.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Declare the signs: light goes from air (n₁ = 1.00) into acrylic (n₂ = 1.49); the centre of curvature lies inside the rod, on the outgoing side, so R = +4.0 cm; the LED is a real object, s = +12.0 cm.
- Surface power: (n₂ − n₁)/R = 0.49/4.0 cm = 0.1225 cm⁻¹, which is 12.25 D. Positive, so this face converges.
- Imaging equation: 1.00/12.0 + 1.49/s′ = 0.1225, so 1.49/s′ = 0.1225 − 0.0833 = 0.0392 cm⁻¹ and s′ = 1.49/0.0392 = 38.0 cm.
- s′ > 0 on the outgoing side: a real image 38 cm inside the rod, provided the rod is longer than that. Magnification m = −n₁s′/(n₂s) = −(1.00 × 38.0)/(1.49 × 12.0) = −2.13.
Answer. s′ = 38 cm inside the acrylic: a real, inverted image 2.1 times the LED's size.
MediumExample 2
A positive meniscus lens of n = 1.55 has faces of radii R₁ = +8.0 cm and R₂ = +12.0 cm, both centres on the outgoing side. Find its focal length in air and in glycerol (n = 1.47), give the two surface powers in each case, and name the medium index at which it would have no power at all.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Geometry factor first, since the medium never touches it: 1/R₁ − 1/R₂ = 1/8.0 − 1/12.0 = 0.1250 − 0.0833 = 0.04167 cm⁻¹. Both radii positive means one face converges and the other diverges; the more curved face wins.
- In air: 1/f = (1.55 − 1)(0.04167) = 0.02292 cm⁻¹, f = 43.6 cm. The surface powers are P₁ = 0.55/0.080 m = 6.88 D and P₂ = −0.55/0.120 m = −4.58 D, summing to 2.29 D = 1/(0.436 m).
- In glycerol: the prefactor is 1.55/1.47 − 1 = 0.0544, so 1/f = 0.0544 × 0.04167 = 0.002268 cm⁻¹ and f = 441 cm, ten times longer.
- Check by powers: P₁ = (1.55 − 1.47)/0.080 m = 1.00 D, P₂ = −0.08/0.120 m = −0.667 D, sum 0.333 D. That sum is nₘ/f, so 1/f = 0.333/1.47 = 0.227 D and f = 4.41 m, the same answer.
- Zero power needs nL/nₘ − 1 = 0, so nₘ = 1.55, where the lens is optically invisible. The geometry factor cannot do it, because 1/R₁ − 1/R₂ ≠ 0.
Answer. f = 43.6 cm in air and 4.41 m in glycerol; P₁, P₂ = +6.88 D, −4.58 D in air and +1.00 D, −0.67 D in glycerol; no power at nₘ = 1.55.
HardExample 3
A biconvex lens of n = 1.60 has R₁ = +10.0 cm, R₂ = −15.0 cm and a centre thickness of 2.0 cm, in air. Find (a) its thin-lens focal length, (b) where parallel light actually focuses, by imaging through the two faces in turn, measured from the back vertex, and (c) the effective focal length from the thick-lens formula and where the second principal plane sits.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- (a) Thin: 1/f = 0.60 × (1/10.0 + 1/15.0) = 0.60 × 0.16667 = 0.1000 cm⁻¹, so f = 10.0 cm.
- (b) Face 1, s₁ → ∞: 1.60/s₁′ = 0.60/10.0 = 0.0600 cm⁻¹, so s₁′ = 26.67 cm beyond the first vertex, where the image would form if the glass continued: well past the back face at 2.0 cm.
- That image is the second face's object, measured from the back vertex 2.0 cm on: s₂ = 2.0 − 26.67 = −24.67 cm, a virtual object, because the light is still converging when it reaches face 2.
- Face 2, glass to air: 1.60/(−24.67) + 1.00/s₂′ = (1.00 − 1.60)/(−15.0) = +0.0400 cm⁻¹, so 1/s₂′ = 0.0400 + 0.0649 = 0.1049 cm⁻¹ and s₂′ = 9.54 cm. Parallel light focuses 9.54 cm behind the back vertex, which is 10.54 cm from the lens centre; the thin model said 10.0 cm.
- (c) Thick formula: 1/f = 0.60[0.16667 + 0.60 × 2.0/(1.60 × 10.0 × (−15.0))] = 0.60[0.16667 − 0.00500] = 0.0970 cm⁻¹, f = 10.31 cm. This f is measured from the second principal plane H′, which must therefore sit 10.31 − 9.54 = 0.77 cm inside the glass from the back vertex, 0.23 cm past the centre.
- Consistency: the standard offset of H′ from the back vertex, −f(n − 1)d/(nR₁) = −10.31 × 0.60 × 2.0/(1.60 × 10.0) = −0.77 cm, agrees. The thin model is 3% low on f and 5% out on where the focus sits, both from the (n − 1)d/(nR₁R₂) term that the word thin discards.
Answer. (a) f = 10.0 cm; (b) the focus is 9.54 cm behind the back vertex, 10.54 cm from the centre; (c) f = 10.3 cm from a principal plane 0.77 cm inside the glass; the thin lens is 3% low on f.
Quick check
Select an answer and explain it before checking.
Review & continue
Paraxial refraction at a spherical surface of radius R between indices n₁ and n₂ obeys n₁/s + n₂/s′ = (n₂ − n₁)/R; applied to both faces of a thin lens it gives 1/f = (nL/nₘ − 1)(1/R₁ − 1/R₂).
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.