Power of a lens, combination of lenses: Visual walkthrough

We will lean on one prerequisite: the Thin Lens Equation. Everything else we build here.

Step 1 — What a single ray and a lens actually do

WHAT. Before combining lenses, we must agree on what one lens does to one ray. Look at the figure: a ray travels parallel to the central axis (the horizontal dotted line) at a height $h$ above it. It strikes the lens and is bent — it changes direction — and afterwards it crosses the axis at a special point called the focus, a distance $f$ away.

WHY. Every claim about "power" is a claim about bending. So we need a clean number for "how much did this ray bend?". The natural number is the bending angle $δ$ — the angle between the ray's original direction and its new direction.

PICTURE.

Figure — Power of a lens, combination of lenses

Read the little right triangle in the figure. Its horizontal side is $f$ (lens to focus), its vertical side is $h$ (the ray's entry height), and $δ$ is the angle at the focus.

Why does the angle $δ$ satisfy $tan δ = h / f$ ? Because $tan$ of an angle in a right triangle is defined as opposite over adjacent — here the side opposite $δ$ is $h$ , and the side adjacent is $f$ . For thin lenses the rays stay close to the axis, so the angles are tiny, and for tiny angles $tan δ \approx δ$ . Hence:

$how much it bent δ \approx height \div focal length \frac{h}{f}$

The bend per unit height is $δ / h = 1/ f$ . That number — deviation produced per unit ray height — is exactly what we christen the power of the lens.

Step 2 — Setting up two lenses in contact

WHAT. Now place a second thin lens right up against the first, with zero gap between them. An object $O$ sits to the left. We want the final image after light passes through both lenses.

WHY. "In contact" is the whole trick. Because there is no space between them, a ray leaving lens 1 enters lens 2 at (essentially) the same height — it has no room to drift sideways. That shared height is what will let the bends add cleanly. (When we later allow a gap, that assumption breaks — see Step 8.)

PICTURE.

The key idea shown in the figure: the intermediate image $I_{1}$ is the object for lens 2. Light doesn't know a lens ended; it just keeps going. Whatever lens 1 "aimed" the light at is where lens 2 thinks the light is coming from.

Step 3 — Lens 1 acting alone

WHAT. Apply the Thin Lens Equation to lens 1 by itself. The object is at $u$ ; the image lens 1 forms is at $v_{1}$ .

WHY. We handle one lens at a time because we already know the law for a single lens. Divide-and-conquer: solve lens 1, then feed its output into lens 2.

PICTURE.

$where lens 1 sends the image \frac{1}{v _{1}} - where the object is \frac{1}{u} = bending power of lens 1 \frac{1}{f _{1}}$

Every term:

$1/ v_{1}$ — reciprocal of the intermediate image distance.
$1/ u$ — reciprocal of the object distance.
$1/ f_{1}$ — the "strength" of lens 1 (this is $P_{1}$ ).

Nothing new here — this is the raw thin-lens law, just labelled with a subscript 1 so we can keep track. The output is $v_{1}$ , the location of $I_{1}$ , drawn as the orange point in the figure.

Step 4 — Lens 2 acting on lens 1's image

WHAT. Now apply the thin-lens equation to lens 2. Its object is $I_{1}$ , sitting at distance $v_{1}$ ; its image is the final image at $v$ .

WHY. Because the lenses touch, the object distance for lens 2 is the same $v_{1}$ that came out of lens 1 — no gap means no change in that distance. This is the moment where "in contact" pays off.

PICTURE.

$final image \frac{1}{v} - object for lens 2 = I_{1} \frac{1}{v _{1}} = bending power of lens 2 \frac{1}{f _{2}}$

Every term:

$1/ v$ — reciprocal of the final image distance (what we ultimately want).
$1/ v_{1}$ — reciprocal of $I_{1}$ 's distance; the same number that appeared in Step 3.
$1/ f_{2}$ — strength of lens 2 (this is $P_{2}$ ).

Notice the plot twist already: $v_{1}$ appears with a minus sign here, but appeared with a plus sign in Step 3. That mismatch is not an accident — it is exactly what we exploit next.

Step 5 — Adding the two equations: the intermediate image vanishes

WHAT. Add the Step 3 and Step 4 equations. The $v_{1}$ terms are $+ \frac{1}{v _{1}}$ and $- \frac{1}{v _{1}}$ — they cancel.

WHY. The intermediate image $v_{1}$ was only internal bookkeeping. A user of the lens pair never measures it — they only see the object $u$ and the final image $v$ . So a correct final law must not contain $v_{1}$ . Adding the equations is the algebraic act that erases it.

PICTURE.

$(\frac{1}{v _{1}} - \frac{1}{u}) + (\frac{1}{v} - \frac{1}{v _{1}}) = \frac{1}{f _{1}} + \frac{1}{f _{2}}$

The two $\frac{1}{v _{1}}$ boxes annihilate (shown crossed out in orange in the figure), leaving:

$object and final image only \frac{1}{v} - \frac{1}{u} = the two strengths, summed \frac{1}{f _{1}} + \frac{1}{f _{2}}$

This already looks like a single thin-lens equation — for one imaginary "equivalent" lens. We name that next.

Step 6 — Naming the equivalent lens: powers add

WHAT. Compare Step 5's result with the thin-lens equation of a single equivalent lens of focal length $F$ : $\frac{1}{v} - \frac{1}{u} = \frac{1}{F} .$ Matching the two gives our result.

WHY. If the pair behaves exactly like one lens, we may as well give that one lens a name $F$ and a power $P = 1/ F$ . Physics doesn't care whether there are one or two pieces of glass — only the net bending matters.

PICTURE.

$\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}} ⟺ P = P_{1} + P_{2}$

$F$ — focal length of the single lens that could replace the pair.
$P = 1/ F$ — its power. Because $P_{1} = 1/ f_{1}$ and $P_{2} = 1/ f_{2}$ , the reciprocal relation for focal lengths turns into plain addition of powers. That is the whole reason power is such a beloved quantity: awkward reciprocals of $f$ become simple sums.

Step 7 — Every sign: convex, concave, and the cancelling case

WHAT. The addition $P = P_{1} + P_{2}$ carries signs. We now walk all cases so no scenario surprises you.

WHY. A concave lens has negative power, and the sum can therefore shrink, vanish, or flip sign. Skipping signs is the single most common error, so we make each case its own picture.

PICTURE.

Three cases, all drawn:

Both convex ( $+ 5 D$ and $+ 3 D$ ): $P = + 8 D$ . Stronger converging — shorter combined focal length. (Teal panel.)
Convex + weak concave ( $+ 5 D$ and $- 2 D$ ): $P = + 3 D$ . The concave lens subtracts, weakening the convergence. (Orange panel.)
Exactly cancelling ( $+ 4 D$ and $- 4 D$ ): $P = 0 \Rightarrow F = \infty$ . The pair bends nothing — parallel rays leave parallel. This is the degenerate case: the "equivalent lens" is effectively a flat window. (Plum panel.)

Step 8 — The gap case: WHERE the drift term $d P_{1} P_{2}$ comes from

WHAT. Pull the lenses apart by a distance $d$ . The rule gains a correction term: $P = P_{1} + P_{2} - d P_{1} P_{2} .$ We now derive that term, not just quote it.

WHY. Our contact derivation relied on the ray hitting lens 2 at the same height it left lens 1 (Step 2). With a gap, the ray — already bent by lens 1 — drifts to a new height before reaching lens 2. Lens 2's bend depends on height, so the total bend changes. Let us track that height with the small-angle bending law from Step 1.

PICTURE.

The drift, in three short lines. Take a ray arriving parallel to the axis at height $h_{1}$ on lens 1 (so its incoming slope is $0$ ).

Lens 1 bends it. From Step 1, the bend angle is $δ_{1} = h_{1} / f_{1} = h_{1} P_{1}$ . So after lens 1 the ray slopes downward toward the axis with slope $- h_{1} P_{1}$ .
It drifts across the gap $d$ . Travelling a horizontal distance $d$ with that slope, its height changes. On reaching lens 2 the new height is $h_{2} = h_{1} - d (h_{1} P_{1}) = h_{1} (1 - d P_{1}) .$ The term $d P_{1}$ is exactly the fractional drop in height caused by the gap — no gap ( $d = 0$ ) means $h_{2} = h_{1}$ , recovering Step 2.
Lens 2 bends again. Lens 2 adds bend $δ_{2} = h_{2} P_{2}$ . The total downward bend of the ray is $δ_{1} + δ_{2} = h_{1} P_{1} + h_{2} P_{2} = h_{1} P_{1} + h_{1} (1 - d P_{1}) P_{2} .$

Now divide the total bend by the entry height $h_{1}$ to read off the effective power (bend per unit height, our Step-1 definition of $P$ ): $P = \frac{δ _{1} + δ _{2}}{h _{1}} = P_{1} + (1 - d P_{1}) P_{2} = clean sum P_{1} + P_{2} - drift correction d P_{1} P_{2} .$

$- d P_{1} P_{2}$ is precisely the height-drop factor $d P_{1}$ acting on the second lens's power $P_{2}$ .
Set $d = 0$ and it vanishes, restoring $P = P_{1} + P_{2}$ . That sanity check is built right in. ✓

The one-picture summary

One figure, the whole story: object → lens 1 (bend $P_{1}$ , makes $I_{1}$ ) → lens 2 (bend $P_{2}$ , uses $I_{1}$ as object) → final image. The intermediate $v_{1}$ cancels when the two thin-lens equations are added, leaving a single equivalent lens with $P = P_{1} + P_{2}$ — and, once a gap opens, the ray drifts and we subtract $d P_{1} P_{2}$ .

Recall Feynman retelling — the walkthrough in plain words

Picture a single light ray coming in flat and high above the middle line. A lens grabs it and tips it down toward one spot — the focus. How hard it tips is the lens's power: strong lens, sharp tip, close focus; weak lens, gentle tip, far focus. Power is just $1/ f$ with $f$ in metres.

Now stack a second lens right against the first. The first lens aims the light at some in-between spot. Before the light can even travel anywhere, the second lens grabs it and tips it again. Because the two lenses are touching, the light is at the same height for both — so the two tips just pile on top of each other. Two tips = one bigger tip = the two powers added. When we write the neat lens-law for each lens and add the two lines, that in-between spot literally cancels out on paper, proving it was never part of the real answer.

Signs matter: a convex lens tips light inward (plus power), a concave lens tips it outward (minus power). Add a plus and a minus and they fight — a big enough minus cancels the plus completely, giving a window that bends nothing. Finally, if you slide the lenses apart by a gap $d$ , the already-tipped light drifts to a slightly lower height before reaching lens 2. Lens 2 tips a shorter ray, so its contribution shrinks by the fraction $d P_{1}$ . Multiply that shrink by lens 2's power and you get exactly the correction we subtract, $d P_{1} P_{2}$ — a correction that politely disappears the moment the lenses touch again.

Connections

Thin Lens Equation — the single law we applied twice and added.
Lensmaker's Equation — where each individual power comes from.
Magnification of Lenses — net magnification multiplies while powers add.
Resistors in Parallel — focal lengths mimic parallel resistances; powers mimic conductances.
Microscope and Telescope — real instruments built from separated lens pairs.
Defects of Vision — corrective lens powers combine with the eye's own power.