Combination of thin lenses in contact

A camera lens that looks like one thick piece of glass is very often not one lens at all — crack open a real photographic lens and you find a stack of thin lenses, convex and concave, cemented together in a row, each one correcting for something the others leave behind. Microscope eyepieces and telescope objectives do the same trick. Nobody wants to grind one impossibly precise piece of glass when four or five ordinary thin lenses, glued together, can be made to behave as a single, better-controlled optical unit. The question this article answers is deceptively small: if you push several thin lenses together until they touch, how do you predict what the combination does to light, given only what each individual lens does on its own? The answer turns out to be one of the cleanest results in the whole ray-optics chapter. Focal lengths themselves do not add — that is the single most common mistake students make on this topic. What adds, beautifully and exactly, is the reciprocal of the focal length, which is precisely the quantity called power. Two lenses in contact, one of focal length $f_1$ and one of $f_2$ , behave as a single lens whose power is $P_1+P_2$ . A weak diverging lens tucked behind a strong converging one does not necessarily wreck the system — it just nudges the net power down a little, and the sign of the final total tells you everything about whether the assembled stack converges or diverges. This article builds that rule from the thin lens formula applied twice in a row, checks it against the two worked numbers you have already seen in the Pass-1 version of this topic (a 20 cm convex lens paired with a 40 cm concave lens, and a three-lens 4 D + 4 D − 3 D stack), and then pushes further into how examiners like to disguise the same addition — mixed units, three or four lenses, a "find the missing lens" twist, and the sign traps that cost marks under time pressure in JEE and NEET-style multiple-choice questions.

Key takeaways

Two or more thin lenses in contact (touching, negligible gap between them) act as one equivalent lens.
Focal lengths do not add directly; their reciprocals add: $\dfrac1f=\dfrac1{f_1}+\dfrac1{f_2}+\cdots$
Equivalently, powers add: $P=P_1+P_2+P_3+\cdots$ , with $P$ in dioptres (D) when $f$ is in metres.
The sign of the net power decides everything: $P>0$ means the combination converges overall; $P<0$ means it diverges overall — regardless of how many individual lenses of each type are present.
Net magnification of the stack is the product of the individual magnifications, $m=m_1\times m_2\times\cdots$
This additive rule holds only for lenses in contact; lenses separated by a finite distance need an extra separation term that is outside this syllabus.
Key idea: for thin lenses in contact, power (the reciprocal of focal length) simply adds, $P=P_1+P_2+\cdots$ , so the whole stack behaves as one lens whose sign and magnitude tell you at a glance whether it converges or diverges.

The core insight

When thin lenses touch, their powers add algebraically ( $P=P_1+P_2+\cdots$ , so $1/f=1/f_1+1/f_2+\cdots$ ), and the sign of that single net power tells you whether the whole stack converges or diverges.

Where it fits in the syllabus

In Unit IX, "Ray Optics and Optical Instruments," this topic follows directly after power of a lens and directly precedes refraction through a prism. The CBSE 2025-26 syllabus explicitly restricts this to lenses placed in contact — touching, with negligible separation between them; combinations of lenses separated by a finite distance, which need an extra "equivalent focal length" formula carrying a separation term, are outside the prescribed scope and should never appear on a board answer. You are expected to derive the combined-focal-length and combined-power formulas from the thin lens formula, and to apply them confidently — with correct signs — to two, three, or more lenses in contact.

The physical picture

Picture two thin lenses, L1 and L2, pushed together so there is essentially no gap between them, both sharing the same principal axis. Parallel light enters L1 first. L1 alone would form a real or virtual image at some point; because L2 sits right where that image would form, the still-travelling rays never actually complete that first image — instead L2 immediately bends them further, using that unformed image as its own "object." Because the lenses are thin and touching, the object distance for L2 is (to excellent approximation) the same as the image distance produced by L1, just carried across the shared plane with a sign flip that the lens formula already handles for us.

\frac1v-\frac1u=\frac1f

where: $u$ is the object distance, $v$ is the image distance, and $f$ is the focal length, all measured from the lens using the Cartesian sign convention (distances measured in the direction of the incident light are positive). Applying this equation once for L1 and once for L2, then eliminating the intermediate image distance because it is common to both, leaves a relation purely between the final image position and the two focal lengths — this is the seed of the combination formula worked out fully below.

Figure

Two thin convex lenses touching on a common axis with parallel rays entering from the left bending at the first lens toward an intermediate image point then bending again at the second lens to a final image closer in, next to a single equivalent lens producing the same final image

Because power $P=1/f$ (with $f$ in metres, $P$ in dioptres, D), the same relation reads as an addition of powers, $P=P_1+P_2+\cdots$ , and this is why opticians habitually specify lenses in dioptres rather than in focal length: combining lens effects becomes ordinary addition instead of the awkward reciprocal-then-add arithmetic that focal lengths force on you.

P=P_1+P_2+P_3+\cdots

where: each $P_i=1/f_i$ carries its own sign — positive for a converging (convex) lens, negative for a diverging (concave) lens — and $P$ is the net power of the whole stack. A combination can end up converging overall even with a diverging lens inside it, provided the converging lens's power magnitude wins; the sign of $P$ , not the mere presence of a concave element, decides the outcome. The overall linear magnification of the stack is the product of the individual magnifications, $m=m_1\times m_2\times\cdots$ , because each lens in turn magnifies (or demagnifies) the image handed to it by the one before.

Derivation ACombined focal length of two thin lenses in contact

Starting from the thin lens formula applied separately to each of two thin lenses sharing a common axis and touching so closely that the image formed by the first lens becomes the object for the second before any light travels a measurable distance between them, the goal is to show that the pair behaves exactly like one single lens, and to obtain the formula $1/f=1/f_1+1/f_2$ (and hence $P=P_1+P_2$ ) that lets that single equivalent focal length be computed directly from the two individual ones without tracing the ray through both surfaces by hand every time.

Step 1. Let an object be placed at distance $u$ from the first lens L1 (focal length $f_1$ ). If L1 acted alone, it would form an image at distance $v_1$ , related by the thin len formula: $\dfrac{1}{v_1}-\dfrac{1}{u}=\dfrac{1}{f_1}$ .

Step 2. Because L2 is placed in contact with L1, that same image (whether or not it actually forms, since L2 intercepts the rays first) serves as the object for L2. With the lenses touching, the object distance for L2 is $v_1$ , the very same value found for L1's image. Applying the thin lens formula to L2, with final image distance $v$ : $\dfrac{1}{v}-\dfrac{1}{v_1}=\dfrac{1}{f_2}$ .

Step 3. Add the two equations from Step 1 and Step 2. The intermediate term $v_1$ appears as $-1/v_1$ in Step 1 (rearranged) and $-1/v_1$ in Step 2, so adding the left-hand sides directly cancels it: $\left(\dfrac{1}{v_1}-\dfrac{1}{u}\right)+\left(\dfrac{1}{v}-\dfrac{1}{v_1}\right)=\dfrac{1}{f_1}+\dfrac{1}{f_2}$ , which simplifies to $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$ .

Step 4. Compare this to the thin lens formula for a single equivalent lens producing the same final image from the same original object: $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ . Matching the right-hand sides of Step 3 and Step 4 term by term gives the combination formula directly: $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$ .

Step 5. Generalise and reinterpret. The same cancellation trick extends to any number of thin lenses in contact by chaining the algebra one more lens at a time, giving $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}+\dfrac{1}{f_3}+\cdots$ Since power is defined as $P=1/f$ , multiplying through by nothing more than a change of name turns this into $P=P_1+P_2+P_3+\cdots$ — the additive-power statement used throughout this article, now traced back to two straightforward applications of the thin lens formula.

Worked examples

Board level - convex and concave lens in contact

Question: A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 40 cm. Find the focal length and power of the combination, and state whether it is converging or diverging overall.

Step 1. Assign signs using the Cartesian convention: the convex lens is $f_1=+20\ \text{cm}=+0.20\ \text{m}$ , the concave lens is $f_2=-40\ \text{cm}=-0.40\ \text{m}$ .

Step 2. Convert to powers, since adding powers is the more direct route: $P_1=\dfrac{1}{0.20}=5\ \text{D}$ and $P_2=\dfrac{1}{-0.40}=-2.5\ \text{D}$ .

Step 3. Add the powers: $P=P_1+P_2=5+(-2.5)=2.5\ \text{D}$ .

Step 4. Read off the physical meaning and convert back to focal length: since $P>0$ , the combination is converging overall, with equivalent focal length $f=\dfrac{1}{P}=\dfrac{1}{2.5}=0.4\ \text{m}=40\ \text{cm}$ .

The combination behaves as a single convex lens of focal length 40 cm and power 2.5 D — the stronger convex lens (5 D) dominates the weaker concave lens (−2.5 D), but only partially, so the net power is smaller than either lens taken alone.

Competitive level - three-lens stack with a missing power

Question: Two thin convex lenses, each of power $+4$ D, are placed in contact. A third thin concave lens of power $-3$ D is then also placed in contact with them. Find the net power and equivalent focal length of the full three-lens system, and find the power of a fourth lens that, added to the stack, would make the whole system exactly afocal (net power zero).

Step 1. Add the three given powers directly, since power addition works for any number of lenses in contact: $P=P_1+P_2+P_3=4+4+(-3)=5\ \text{D}$ .

Step 2. Convert to the equivalent focal length: $f=\dfrac{1}{P}=\dfrac{1}{5}=0.2\ \text{m}=20\ \text{cm}$ , a converging system since $P>0$ .

Step 3. For the fourth lens to bring the total power to zero, its power $P_4$ must exactly cancel the existing net power: $P+P_4=0 \Rightarrow P_4=-P=-5\ \text{D}$ . This means a diverging lens of focal length $f_4=1/P_4=1/(-5)=-0.20\ \text{m}=-20\ \text{cm}$ , i.e. a concave lens of focal length 20 cm.

With the three original lenses, the system has net power +5 D and equivalent focal length +20 cm (converging); adding a fourth lens of power −5 D (a concave lens of focal length 20 cm) would make the whole four-lens stack afocal, with zero net power. Timing note: in an MCQ setting, do not waste time converting every power to a focal length and back — add the powers directly in dioptres (whole numbers here: 4, 4, −3), get the net power in one line, and only take the reciprocal once, right at the end, to report $f$ if the question explicitly asks for a length rather than a power.

Common traps

Adding focal lengths directly, $f=f_1+f_2$ , instead of adding their reciprocals — focal lengths never add on their own; only $1/f$ , or equivalently power, adds.
Dropping or mishandling signs when a convex and a concave lens are combined — a careless magnitude-only addition can flip a genuinely diverging system into an apparently converging one, or vice versa, and this single sign slip is one of the most common mark-losing errors in this topic.
Mixing units mid-calculation — using centimetres for one lens's focal length and metres for another before converting to power, which silently corrupts the dioptre value; always convert every focal length to metres before computing $P=1/f$ .
Assuming this additive rule applies to lenses separated by any distance apart — the simple rule $1/f=1/f_1+1/f_2$ (and $P=P_1+P_2$ ) holds only for lenses in contact; a finite separation needs an extra term that is outside this syllabus and must never be used on a CBSE answer.
Believing a stack containing a diverging lens must itself diverge, or that a stack of only convex lenses must have a huge combined power — the net sign and magnitude depend on the actual numeric powers involved, not on a quick glance at how many lenses of each type are present.

You should be able to…

State and apply $\dfrac1f=\dfrac1{f_1}+\dfrac1{f_2}+\cdots$ and $P=P_1+P_2+\cdots$ for any number of thin lenses in contact.
Derive the two-lens combination formula from the thin lens formula applied twice, cancelling the intermediate image distance.
Compute the equivalent focal length and power of a stack of lenses with correct signs, and read off whether the result converges or diverges from the sign of $P$ alone.
Convert consistently between focal length (in metres or centimetres) and power (in dioptres) without mixing units mid-calculation.
Recognise that this additive rule applies only to lenses in contact, not to lenses separated by a finite distance.
Combine individual magnifications as a product, $m=m_1\times m_2\times\cdots$ , for a multi-lens stack.

Ask the tutor: want to see how this two-lens derivation extends step by step to three or more lenses, or to try a numerical where you must find one unknown lens's power from a stated combination result?

Why does power add but focal length does not, when they are basically the same information

Power and focal length carry the same physical information, but they are related by a reciprocal, $P=1/f$ , and reciprocals of a sum are not the sum of reciprocals. The derivation above shows exactly where the addition comes from: applying the thin lens formula twice in a row and cancelling the intermediate image distance naturally produces $1/f=1/f_1+1/f_2$ , not $f=f_1+f_2$ . Physically, power measures how strongly a lens bends (converges or diverges) parallel rays per unit length, and bending abilities of successive elements really do add up directly — that additive quantity happens to be $1/f$ , not $f$ itself.

If the two lenses in contact have equal and opposite powers, does the light just pass straight through undeviated

Almost, but this needs care. If $P_1+P_2=0$ , the combination is called afocal — its net power is zero and its equivalent focal length is infinite, meaning parallel rays entering stay parallel on leaving (no net converging or diverging effect on a distant object). This is not quite the same as the light "just passing straight through" as if no lenses were there at all — each individual lens still bends the ray path locally, and for a nearby object (not at infinity) the two lenses in an afocal pair can still produce magnification effects, which is exactly how some afocal beam-expander and telescope-relay systems are built. For CBSE numericals, though, treating an afocal pair as behaving like plane glass for parallel light is the expected level of understanding.

Does it matter which lens I put first when combining them in contact

For thin lenses in contact and for the final image position and equivalent power, no — the order does not change $1/f=1/f_1+1/f_2$ or $P=P_1+P_2$ , since addition does not care about order. Physically this makes sense: whichever lens the light meets first, the same total bending is eventually applied by the pair together. In real, non-idealised optics, putting a strongly curved surface first versus second can change how much certain image defects show up, but that refinement belongs to aberration theory, which is outside this syllabus — for CBSE purposes, combination order is irrelevant to the formulas you are tested on.

My textbook problem gives three lenses with one focal length in centimetres and the powers of the other two in dioptres directly. How do I combine them

Convert everything to the same quantity before adding — the safest habit is to convert every given focal length to metres and then take its reciprocal to get power in dioptres, so all terms in your sum are honestly dioptres. For example, a lens quoted as $f=25\ \text{cm}=0.25\ \text{m}$ has $P=1/0.25=4\ \text{D}$ ; add this to any other powers given directly in dioptres exactly as they are. Never try to add a power in dioptres to a focal length in centimetres term by term — that mixes two different physical quantities and will silently give a meaningless number, even though the arithmetic looks like it "worked."