Lenses in Contact & Combination

Almost no real optical instrument uses a single lens. A camera lens is a stack of six or eight elements; a microscope eyepiece is two lenses in a tube; your spectacles add a lens to the lens already inside your eye. So the question that closes this chapter is the practical one: put two thin lenses together — how does the pair behave? The answer is astonishingly clean when the lenses are in contact:

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}, \qquad P = P_1 + P_2

The combined lens has a focal length $F$ whose reciprocal is the sum of the reciprocals — which, translated into the dioptres of the last article, says simply that powers add. A $+5$ D lens and a $+3$ D lens in contact make a $+8$ D lens. Two equal converging lenses double their power and halve their focal length. A converging lens paired with a weaker diverging one stays net converging, its power just the signed sum. This single rule is why the dioptre was worth naming: power is the currency that adds. That "powers add" is the whole reason Huygens could design a two-lens eyepiece, why an optician's lens simply supplements your eye's own power, and — looking ahead — why an achromatic doublet can be built by pairing a converging crown-glass lens with a diverging flint one so their focal lengths combine to the value wanted while their colour errors cancel (the cancellation itself is a coming chapter). When the two lenses are separated by a distance $d$ , one extra term appears, $P = P_1 + P_2 - d\,P_1 P_2$ , and the contact rule is just its $d = 0$ case. This page derives the contact rule from the thin-lens formula, works the standard combinations, and shows exactly how signed powers add.

Key takeaways

Two thin lenses in contact combine to a single lens whose power is the sum: $P = P_1 + P_2$ , equivalently $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$ .
Powers add with sign — a converging ( $+$ ) and a diverging ( $-$ ) lens give the signed sum, and the combination can be net converging or diverging depending on which dominates.
Anchors: $f_1 = +20$ , $f_2 = +30$ cm in contact give $\dfrac{1}{F} = \dfrac{1}{12}$ , so $F = +12$ cm and $P = +5 + 3.33 = +8.33$ D. With $f_2 = -30$ : $F = +60$ cm, $P = +5 - 3.33 = +1.67$ D (still converging).
Two equal $+5$ D lenses in contact make $+10$ D, $F = +10$ cm — power doubles, focal length halves.
Separated lenses need one more term: $P = P_1 + P_2 - d\,P_1 P_2$ ; the contact rule is its $d = 0$ limit.
Key idea: lenses in contact add their powers, $P = P_1 + P_2$ — the cleanest reason the dioptre is the natural unit of a lens — and this is the principle behind eyepieces, spectacles and the achromatic doublet.

The core insight

Two thin lenses in contact behave as a single lens whose power is the sum of the individual powers, $P = P_1 + P_2$ (equivalently $1/F = 1/f_1 + 1/f_2$ ), because the image of the first lens is the object of the second at zero separation; powers add with their signs, and separated lenses carry an extra $-d\,P_1 P_2$ term.

Where this lives today

Adding powers is the design principle of every multi-lens instrument. The eyepiece of a microscope or telescope — Huygens' or Ramsden's or a modern wide-field design — is a computed combination of two or more lenses whose powers and spacing set the magnification and correct the field. Camera lenses stack six to a dozen elements, and the first thing a designer does is add their powers to hit a target focal length before optimising the residual aberrations on a computer. Spectacles are the most personal example: your prescription lens adds its power to your eye's own fixed power, so an optometrist prescribes exactly the dioptres needed to bring the sum to a healthy eye's value — the "powers add" rule applied to the eye-plus-glasses pair. And the achromatic doublet that makes sharp, colour-free images possible — the crown-and-flint pairing invented in the eighteenth century — is a lens combination chosen so the powers sum to the wanted value while the colour errors cancel; that colour cancellation is the subject of the next chapter, on dispersion and chromatic aberration, where these combinations become the tool that defeats it. From your glasses to a space telescope, the optics of the built world are lenses added together, one power at a time.

The physical picture

Combining lenses sounds as though it should be complicated — two focal lengths, two sign conventions, an image handed from one lens to the next. In contact, it collapses to a single idea. Light from the object is focused by the first lens toward an image; but before it gets there it meets the second lens, for which that not-yet-formed image is simply the object. Because the two lenses touch — their separation is zero — the object handed to the second lens is at exactly the image distance of the first, measured from the same place. When you write the thin-lens formula for each lens and add them, the intermediate image distance cancels, and what remains is one thin-lens formula for the pair, with $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$ . The tidy way to read that result is in powers. Dividing through (with focal lengths in metres) turns it into $P = P_1 + P_2$ : the combination's strength is just the sum of the strengths. This is why the dioptre earned its own name back in the last article — focal lengths combine through an awkward reciprocal sum, but powers simply add, like resistances in series or masses on a scale. And they add with sign: a diverging lens contributes a negative power, subtracting from a converging partner. That signed addition is the whole behaviour of every lens stack in this book, and it is the same relation that lets an optician add a corrective lens to the fixed power of your eye.

Figure

Two thin lenses in contact act as one lens whose power is the sum: a converging lens of power P1 and a second lens of power P2, touching, behave exactly as a single lens of power P equal to P1 plus P2, bringing parallel light to a combined focus F. For f1 equal to 20 cm and f2 equal to 30 cm the combination is 8.33 dioptres with F equal to 12 cm.

The combination rule

For two thin lenses placed in contact on a common axis, the pair acts as a single thin lens whose focal length $F$ obeys

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \qquad (1)

where: $f_1$ and $f_2$ are the focal lengths of the two lenses in New Cartesian signs, and $F$ is the focal length of the equivalent single lens. Every focal length keeps its own sign — converging $+$ , diverging $-$ — so the reciprocals add algebraically. Note that this is a sum of reciprocals, which is the same shape as the thin-lens formula itself: the combination is, in every respect, one lens of focal length $F$ , and you feed it objects with the ordinary $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{F}$ of the previous article — the minus-sign lens equation, not the mirror's plus (see The Mirror Formula for that contrast). The rule is far more memorable in power form. Multiply through by the appropriate metres and $1/f$ becomes $P$ :

P = P_1 + P_2 \qquad (2)

where: $P_1 = 1/f_1$ and $P_2 = 1/f_2$ are the individual powers in dioptres, and $P$ is the combined power. This is the punchline of the whole chapter on power: powers add. It is why opticians think in dioptres, why a lens stack is designed by summing a column of powers, and why the combination of a strong converging lens and a weaker diverging one is trivially the signed sum. Two lenses, one addition. When the lenses are not in contact but separated by a distance $d$ along the axis, one extra term enters:

P = P_1 + P_2 - d\,P_1 P_2 \qquad (3)

where: $d$ is the separation in metres. Setting $d = 0$ recovers equation (2), so the contact rule is the special case of a general result — and the separation term is exactly what lets a Huygens eyepiece, whose two lenses are held apart, be designed for a particular combined power.

Derivation APowers add for lenses in contact

The goal is to prove $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$ , hence $P = P_1 + P_2$ , for two thin lenses in contact, using only the thin-lens formula of the previous article applied to each lens in turn. The engine is the idea that the image of the first lens is the object of the second — and because the lenses touch, that hand-off is exact, with the intermediate image distance cancelling cleanly.

Step 1. Refract at the first lens. An object at $u$ is imaged by lens 1 (focal length $f_1$ ) to an intermediate image at $v_1$ : by the thin-lens formula, $\dfrac{1}{v_1} - \dfrac{1}{u} = \dfrac{1}{f_1}$ . This $v_1$ is where the light heads after the first lens.

Step 2. Refract at the second lens. The intermediate image at $v_1$ is the object for lens 2 (focal length $f_2$ ). Because the two lenses are in contact — separation zero — this object is at distance $v_1$ from lens 2 as well, measured from the common plane. So $\dfrac{1}{v} - \dfrac{1}{v_1} = \dfrac{1}{f_2}$ , where $v$ is the final image distance.

Step 3. Add the two equations. The intermediate term $1/v_1$ appears as $+1/v_1$ in Step 1 and $-1/v_1$ in Step 2, so adding cancels it: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$ . The pair therefore obeys a single thin-lens formula with an equivalent focal length $F$ given by $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$ — the combination behaves as one lens.

Step 4. Translate to powers. Since $P = 1/f$ (focal lengths in metres), dividing through gives $P = P_1 + P_2$ : powers add. Test on the anchor: $f_1 = +20$ , $f_2 = +30$ cm give $\dfrac{1}{F} = \dfrac{1}{20} + \dfrac{1}{30} = \dfrac{1}{12}$ , $F = +12$ cm, and $P = +5 + 3.33 = +8.33$ D — the two forms agree, as $1/8.33 = 0.12$ m confirms. The signs carry through, so a diverging lens subtracts its power, exactly as the worked examples show. Verified: the combined $F = +12$ cm, $P = +8.33$ D, and the signed cases are recomputed in verify_hs2-lens-combinations.py.

Go deeper · Derivation B · 1 minSigned powers: converging plus diverging

Derivation BSigned powers: converging plus diverging

The goal is to show explicitly that powers add with their signs, so that a converging lens paired with a diverging one gives the algebraic sum — and to see why the combination in the second anchor stays converging while being much weaker. This is the case that trips students who add magnitudes and forget the minus.

Step 1. Assign the signed powers. The converging lens $f_1 = +20$ cm has $P_1 = +5$ D; the diverging lens $f_2 = -30$ cm has $P_2 = 1/(-0.30) = -3.33$ D. The diverging lens's power is genuinely negative — that is what "diverging" means in dioptres.

Step 2. Add them with signs. $P = P_1 + P_2 = 5 + (-3.33) = +1.67$ D. The result is positive, so the pair is still a converging lens, but at $+1.67$ D it is far weaker than the $+5$ D lens alone — the diverging partner has cancelled two-thirds of its power.

Step 3. Cross-check through focal lengths, a different road. $\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} = \dfrac{1}{20} + \dfrac{1}{-30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$ , so $F = +60$ cm, and $P = 1/0.60 = +1.67$ D — the same answer. The positive $F$ confirms "converging"; the large $F$ confirms "weak".

Step 4. Read the design lesson. Whether a converging-plus-diverging pair ends up converging or diverging is decided by which power is larger in magnitude — here the converging one wins. Pairing lenses of opposite sign but nearly equal magnitude gives a very weak combination of either sign, which is exactly the freedom a designer needs to cancel a colour error (the achromatic doublet) while keeping a chosen net power — a payoff developed in the coming chapter on chromatic aberration, not claimed here. Verified: the signed sum $P = +1.67$ D and the focal-length cross-check $F = +60$ cm are recomputed in verify_hs2-lens-combinations.py.

Figure

A converging lens of plus 5 dioptres in contact with a diverging lens of minus 3.33 dioptres. The diverging lens cancels most of the converging one, so parallel light is bent only gently to a distant focus. The net power is the signed sum plus 1.67 dioptres, a weak converging lens of focal length 60 cm.

Limiting cases

Case	Result	Meaning
two convergers in contact	$P = P_1 + P_2$ , larger	the pair is stronger than either lens
equal lenses $P_1 = P_2$	$P = 2P_1$ , $F = f_1/2$	power doubles, focal length halves
converging $+$ diverging	$P$ = signed sum	net converging or diverging by which dominates
$f_2 \to \infty$ (plano/flat)	$F = f_1$	a zero-power lens leaves the other unchanged
$P_2 = -P_1$ (equal, opposite)	$P = 0$ , $F \to \infty$	the pair neither converges nor diverges
separated by $d$	$P = P_1 + P_2 - d P_1 P_2$	contact rule is the $d = 0$ case

Common trapAdd powers with sign, and only for lenses in contact

Two disciplines keep this rule honest. First, add powers, not focal lengths — never write $F = f_1 + f_2$ . Focal lengths combine through the reciprocal sum $1/F = 1/f_1 + 1/f_2$ ; only their reciprocals, the powers, add directly. Second, carry the signs: a diverging lens has negative power and subtracts. A $+5$ D and a $-3.33$ D lens make $+1.67$ D, not $+8.33$ D — adding the magnitudes is the classic wrong answer, and it turns a weak converging pair into a strong one on paper. Third, the plain "powers add" rule holds only in contact. The moment the lenses are separated by a distance $d$ , the extra term $-d\,P_1 P_2$ appears, and $d$ must be in metres to match the dioptre — a separation quoted in centimetres, dropped straight into the formula, inflates the correction a hundredfold. Check whether the lenses touch before you decide which formula to use.

Worked examples

Board level — three combinations in contact (CBSE and NEET pattern)

Find the combined focal length and power when two thin lenses are placed in contact: (a) $f_1 = +20$ cm and $f_2 = +30$ cm; (b) $f_1 = +20$ cm and $f_2 = -30$ cm; (c) two identical lenses of power $+5$ D each.

Step 1. Part (a), two converging lenses. $\dfrac{1}{F} = \dfrac{1}{20} + \dfrac{1}{30} = \dfrac{3 + 2}{60} = \dfrac{5}{60} = \dfrac{1}{12}$ , so $F = \boxed{+12\ \mathrm{cm}}$ . In powers: $P_1 = 1/0.20 = +5$ D, $P_2 = 1/0.30 = +3.33$ D, so $P = P_1 + P_2 = \boxed{+8.33\ \mathrm{D}}$ — check: $1/P = 1/8.33 = 0.12$ m $= 12$ cm, consistent. The pair is stronger than either lens alone, as adding two converging lenses must be.

Step 2. Part (b), converging plus diverging. $\dfrac{1}{F} = \dfrac{1}{20} + \dfrac{1}{-30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$ , so $F = \boxed{+60\ \mathrm{cm}}$ . In powers: $P = +5 + (-3.33) = \boxed{+1.67\ \mathrm{D}}$ . The combination is still converging (positive power), but much weaker — the diverging lens has cancelled most of the converging one's strength. Powers subtracted with their signs.

Step 3. Part (c), two equal $+5$ D lenses. $P = 5 + 5 = \boxed{+10\ \mathrm{D}}$ , so $F = 1/10 = 0.10$ m $= \boxed{+10\ \mathrm{cm}}$ . Doubling the lenses doubled the power and halved the focal length — the cleanest illustration that powers add.

Step 4. Read the pattern across the three. Two convergers strengthen (part a), a diverger weakens its partner (part b), and equal lenses double (part c). Every case is just $P = P_1 + P_2$ with the signs kept — no new physics beyond the last article's dioptre, only its additivity.

Competitive level — the same pair, now held apart (JEE pattern)

Take the $+20$ cm and $+30$ cm lenses of the first example, but now separate them by $d = 10$ cm along the axis. Find the combined power and equivalent focal length, and compare with the in-contact result.

Step 1. Convert everything to consistent units. $P_1 = 1/0.20 = +5$ D, $P_2 = 1/0.30 = +3.33$ D, and the separation $d = 10$ cm $= 0.10$ m — the separation must be in metres because it multiplies powers, which are per-metre.

Step 2. Apply equation (3): $P = P_1 + P_2 - d\,P_1 P_2 = 5 + 3.33 - 0.10 \times 5 \times 3.33 = 8.33 - 1.67 = \boxed{+6.67\ \mathrm{D}}$ . The separation term $d P_1 P_2 = 1.67$ D has reduced the combined power from the in-contact $8.33$ D.

Step 3. Equivalent focal length: $F = 1/P = 1/6.67 = 0.15$ m $= \boxed{+15\ \mathrm{cm}}$ , longer than the $12$ cm the lenses gave in contact. Separating two converging lenses weakens their combination — a fact every telescope and eyepiece designer uses to tune the effective focal length by adjusting a spacing.

Step 4. Check the limit. Set $d = 0$ in equation (3) and the extra term vanishes, returning $P = 8.33$ D and $F = 12$ cm — the in-contact answer of the first example. The two results are the same formula at two separations, which is the sanity check that you have used equation (3) correctly. Timing note: about 75 seconds. The one competitive trap is units — the separation $d$ must be in metres to match the dioptre, so $10$ cm is $0.10$ m, not $10$ . Put $d$ in metres and the arithmetic is a single line; put it in centimetres and the correction term is a hundred times too big.

Exam traps

Adding focal lengths. The combination rule adds powers, or equivalently reciprocals of focal length: $1/F = 1/f_1 + 1/f_2$ . Writing $F = f_1 + f_2$ is simply wrong.

Dropping the sign of a diverging lens. A diverging lens has negative power and subtracts. Add magnitudes instead of signed powers and a converging-plus-diverging pair comes out far too strong.

Using the contact rule for separated lenses. Powers add plainly only when the lenses touch. Separated by $d$ , use $P = P_1 + P_2 - d\,P_1 P_2$ , and put $d$ in metres.

Metres for power and separation. Both $P = 1/f$ and the $d\,P_1 P_2$ term demand metres: a $20$ cm lens is $+5$ D and a $10$ cm gap is $0.10$ m. Centimetres poison both.

Board scope. Lenses-in-contact combination, $1/F = 1/f_1 + 1/f_2$ and $P = P_1 + P_2$ , is examined by CBSE, ICSE, JEE Main, JEE Advanced and NEET. It is not part of any mainstream A-Level, AP or IB core — those courses stop at a single lens. So this is an Indian-syllabus result; treat it as the natural finish to the power article if you sit those boards, and as enrichment otherwise. The separated-lens formula with $-d P_1 P_2$ is a JEE-Advanced extension.

Origins · 1 minChristiaan Huygens and how this was discovered

Christiaan Huygens

1629–1695 · The Hague & Paris

Huygens was a maker as much as a theorist, and the lens combination is his in the most literal sense. A Dutch polymath who discovered Saturn's rings and its moon Titan and wrote an early wave theory of light, he ground his own lenses with his brother Constantijn, reaching a quality no workshop could match. To tame the colour fringing of his telescopes he designed the Huygens eyepiece — two plano-convex lenses held a calculated distance apart — which is precisely a lens combination, governed by the addition of powers. He was the first to treat an eyepiece as a system of lenses to be computed, not a single glass to be polished. His wavefront principle belongs to the coming wave-optics chapter; here he is the craftsman who first combined lenses by design.

History

1611

— Johannes Kepler describes the two-lens astronomical telescope, the first deliberate lens combination.

1655

— Christiaan Huygens grinds his own lenses and discovers Titan and Saturn's rings with a telescope of his own making.

1690

— Huygens publishes his wave theory of light and, in his telescope work, the two-lens compound eyepiece that bears his name.

1729

— Chester Moore Hall invents the achromatic doublet, a converging crown lens paired with a diverging flint to combine powers while cancelling colour.

1758

— John Dollond patents and popularises the achromatic doublet, making combined-lens optics a commercial craft.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), §9.6 "Combination of Thin Lenses in Contact" · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" · D. C. Pandey, Optics and Modern Physics, lens-combination problems.

Provenance. Standard paraxial optics. Recompute: in contact $f_1 = +20$ , $f_2 = +30$ cm $\to 1/F = 1/12$ , $F = +12$ cm, $P = +5 + 3.33 = +8.33$ D; $f_1 = +20$ , $f_2 = -30 \to F = +60$ cm, $P = +5 - 3.33 = +1.67$ D; two $+5$ D $\to +10$ D, $F = +10$ cm. Separated $d = 10$ cm $= 0.10$ m: $P = 5 + 3.33 - 0.10 \times 5 \times 3.33 = 6.67$ D, $F = +15$ cm, reducing to $8.33$ D at $d = 0$ . Equal-and-opposite $P = 0$ .

Prerequisites. Lens Power & Magnification — the dioptre and $P = 1/f$ this rule adds; The Thin Lens Formula — the single-lens equation the derivation applies twice. Leads to. Optical Instruments (university track), where combinations become telescopes and microscopes.

Go deeper (UG). How real instruments combine lenses — the eyepiece designs, the objective, and the matrix method for a system of separated lenses — is developed in Optical Instruments in the university optics track. It is enrichment; the contact rule here follows from the thin-lens formula applied twice.

Why do powers add but focal lengths do not?

Because power is defined as $1/f$ , and it is the reciprocals that add in the combination rule $1/F = 1/f_1 + 1/f_2$ . Rewrite each $1/f$ as a power $P$ and the rule becomes $P = P_1 + P_2$ — a plain sum. Focal lengths themselves combine through the clumsier reciprocal relation, so $F$ is not $f_1 + f_2$ . This is exactly why the dioptre was worth defining as a unit: it turns the awkward reciprocal-sum of focal lengths into the easy addition of powers, the same way conductance turns the reciprocal-sum of series resistances into a sum. Nature made the reciprocal quantity the additive one, so we named it and work in it.

How can a converging and a diverging lens together still converge?

By having the converging lens win. Powers add with their signs, so a $+5$ D converging lens and a $-3.33$ D diverging lens combine to $+5 - 3.33 = +1.67$ D — positive, hence still converging, but much weaker than the converging lens alone. If instead the diverging lens were the stronger, say $-6$ D, the sum would be $-1$ D and the pair would diverge. The net behaviour is decided entirely by which power is larger in magnitude. This freedom — to set the net power anywhere by choosing the two lenses — is precisely what makes combinations useful: you can build a weak lens of either sign, or cancel one lens's power almost entirely, just by pairing.

Does the order of the two lenses matter?

For thin lenses in contact, no — the combined power $P_1 + P_2$ is a plain sum, and addition does not care about order, so a $+5$ D lens then a $+3$ D lens gives the same $+8$ D as the reverse. The image location for a given object is also the same either way. Order and orientation begin to matter only when the lenses are thick, or separated, or when you track aberrations and where exactly the light bends — considerations beyond a thin-lens combination. For the school-level contact rule, swap the two lenses freely; the combined focal length and power are unchanged.

What happens if the two powers are equal and opposite?

They cancel: $P = P_1 + P_2 = P_1 + (-P_1) = 0$ . A zero-power combination has infinite focal length, meaning it does not converge or diverge at all — parallel light comes out parallel, as through a flat window. A converging lens and a diverging lens of exactly equal strength, in contact, form such a "null" pair. This is more than a curiosity: it is the starting point for a zoom lens, where moving one element slightly away from the cancelling position gives a small, continuously adjustable net power. It is also why a diverging lens can exactly neutralise a converging one for testing or calibration.

Why does separating the lenses change the power?

Because once there is a gap, the light spreads or converges as it travels from the first lens to the second, so the second lens receives it at a different height and bends it by a different amount than it would in contact. The correction term $-d\,P_1 P_2$ captures exactly this: for two converging lenses (both powers positive) it is negative, so separating them weakens the combination, which is why pulling the elements of an eyepiece apart lengthens its effective focal length. The term vanishes when $d = 0$ , recovering the contact rule, and grows with the separation. Remember to put $d$ in metres, since it multiplies powers measured per metre.

Is this the same rule as resistors or capacitors?

It rhymes with them, which is a useful memory aid. Lens powers in contact add like series resistances (both are plain sums of the "reciprocal" quantity — resistance is $1/\text{conductance}$ , power is $1/f$ ), and correspondingly focal lengths combine like parallel resistances, through a reciprocal sum $1/F = 1/f_1 + 1/f_2$ . So if you can add resistors in parallel, you can combine focal lengths, and if you can add resistors in series, you can add powers. The analogy is not a physical identity — light and current are different things — but the algebra is genuinely the same shape, and many students find the combination rule easier to remember once they see it as "focal lengths in parallel, powers in series".