Power of a lens, combination of lenses: Worked examples

Before we compute anything, one reminder that we will lean on constantly:

The scenario matrix

Every problem about lens power lives in one of these cells. Below the table, examples E1–E9 each announce which cell they clear.

#	Cell (what makes it distinct)	Key danger	Cleared by
C1	Single convex lens, $f > 0$	forget cm→m	E1
C2	Single concave lens, $f < 0$	drop the minus sign	E2
C3	Two lenses in contact, both converging	none — pure addition	E3
C4	Contact: convex + concave (opposite signs)	sign cancellation	E4
C5	Contact giving net zero power (degenerate)	$F \to \infty$ , undefined $1/ P$	E5
C6	Two lenses separated by $d$	must use $- d P_{1} P_{2}$	E6
C7	Limiting check: let $d \to 0$ in the gap formula	must collapse to C3/C4	E7
C8	Word problem (real world: eyeglasses / corrective)	translate words to signs	E8
C9	Exam twist: given combined $F$ , find the missing lens	reverse the arithmetic	E9

E1 — Single convex lens (cell C1)

E2 — Single concave lens (cell C2)

E3 — Two converging lenses in contact (cell C3)

Figure — Power of a lens, combination of lenses

Worked example Statement

A $+ 5 D$ convex lens is placed in contact with a convex lens of $f_{2} = 20 cm$ . Find the combined power $P$ and focal length $F$ .

Forecast: two converging lenses back-to-back bend light twice the same way — the combo should be stronger (higher power, shorter focal length) than either alone.

Step 1 — Get both powers in the same units. $P_{1} = + 5 D, P_{2} = \frac{1}{0.20} = + 5 D$ Why this step? We can only add powers, never focal lengths — so convert $f_{2}$ to $P_{2}$ first.

Step 2 — Add (they are in contact). $P = P_{1} + P_{2} = 5 + 5 = + 10 D$ Why this step? In contact means $d = 0$ : the ray bent by lens 1 hits lens 2 at the same height, so the bends add cleanly (look at the two stacked lenses in the figure — no gap, no spreading).

Step 3 — Back to focal length. $F = \frac{1}{P} = \frac{1}{10} = 0.10 m = 10 cm$ Why this step? To report a focal length we invert the power once more.

Verify: Combined $F = 10 cm$ is shorter than either $f_{1} = 20 cm$ or $f_{2} = 20 cm$ ✓ — two lenses converge harder than one, exactly as forecast. This is the Resistors in Parallel pattern: two equal "resistances" halve.

E4 — Contact: convex + concave (cell C4)

Worked example Statement

A convex lens of $P_{1} = + 5 D$ is in contact with a concave lens of $f_{2} = - 50 cm$ . Find $P$ and $F$ .

Forecast: the concave lens subtracts bending power. So the net should be positive but weaker than $+ 5 D$ — the combo still converges, just lazily.

Step 1 — Convert the concave lens. $P_{2} = \frac{1}{- 0.50} = - 2 D$ Why this step? Concave $\Rightarrow$ negative power; carry the sign.

Step 2 — Add the signed powers. $P = P_{1} + P_{2} = 5 + (- 2) = + 3 D$ Why this step? Contact $\Rightarrow$ add; the minus sign of $P_{2}$ does the "cancelling" for us automatically.

Step 3 — Focal length. $F = \frac{1}{3} \approx + 0.333 m = + 33.3 cm$ Why this step? Positive $F$ confirms a converging combination.

Verify: $+ 3 D < + 5 D$ ✓ (weakened by the concave partner) and $F = 33.3 cm > 20 cm$ ✓ (weaker lens = longer focal length). Everything points the same way.

E5 — Degenerate case: net zero power (cell C5)

E6 — Separated lenses (cell C6)

Worked example Statement

Two convex lenses, $f_{1} = 20 cm$ and $f_{2} = 30 cm$ , are separated by $d = 10 cm$ . Find the combined power $P$ and focal length $F$ .

Forecast: they no longer touch, so the clean sum $P_{1} + P_{2}$ is spoiled by the gap term. Expect the combined power to be a bit less than $P_{1} + P_{2}$ .

Step 1 — Everything in metres, everything as power. $P_{1} = \frac{1}{0.20} = 5 D, P_{2} = \frac{1}{0.30} = 3.33 D, d = 0.10 m$ Why this step? The gap formula multiplies $d$ by two powers, so all three must live in SI units together.

Step 2 — Use the separation formula. $P = P_{1} + P_{2} - d P_{1} P_{2}$ Why this step? When $d \neq = 0$ the ray strikes lens 2 at a different height (see the figure: the red ray from lens 1 drifts over the gap before hitting lens 2). That drift is exactly what $- d P_{1} P_{2}$ corrects for.

Step 3 — Plug in. $P = 5 + 3.33 - (0.10) (5) (3.33) = 8.33 - 1.67 = 6.67 D$ Why this step? The gap term $- 1.67 D$ reduces the naive sum $8.33 D$ .

Step 4 — Focal length. $F = \frac{1}{6.67} \approx 0.15 m = 15 cm$

Verify: $6.67 D < 8.33 D$ ✓ (the gap always steals power for two convex lenses). Exact form: $P = 5 + \frac{10}{3} - \frac{1}{10} \cdot 5 \cdot \frac{10}{3} = \frac{20}{3} D$ , and $\frac{20}{3} = 6.67$ ✓. This principle is the heart of every Microscope and Telescope, where objective and eyepiece sit a fixed distance apart.

E7 — Limiting check: let the gap vanish (cell C7)

E8 — Real-world word problem: reading glasses (cell C8)

Worked example Statement

An optician prescribes reading glasses of power $+ 2.5 D$ for a far-sighted person. (a) What focal length is that, and what type of lens? (b) If a mild $- 0.5 D$ scratch-coating lens is cemented in contact with it, what is the net prescription?

Forecast: positive prescription $\Rightarrow$ converging (convex) lens, since far-sighted eyes need help converging. The $- 0.5 D$ add-on should shave the power down slightly.

Step 1 — Translate power to focal length. $f = \frac{1}{P} = \frac{1}{2.5} = 0.40 m = 40 cm$ Why this step? Opticians speak in dioptres; to picture the lens we invert to focal length.

Step 2 — Classify. Positive power $\Rightarrow$ convex/converging. Far-sight (hypermetropia) is corrected with a converging lens, so this checks out physically (see Defects of Vision).

Step 3 — Combine in contact. $P_{net} = 2.5 + (- 0.5) = + 2.0 D$ Why this step? Cemented = in contact = powers add, signs included.

Step 4 — New focal length. $F = \frac{1}{2.0} = 0.50 m = 50 cm$

Verify: $+ 2.0 D < + 2.5 D$ ✓ (the negative coating weakened it) and $50 cm > 40 cm$ ✓ (weaker $\Rightarrow$ longer focal length). Still positive, still converging — the glasses still work, just a touch milder.

E9 — Exam twist: find the missing lens (cell C9)

Worked example Statement

Two thin lenses in contact form a combination of focal length $F = + 15 cm$ . One of them is a concave lens of power $- 4 D$ . Find the power and focal length of the other lens.

Forecast: the combination is converging ( $F > 0$ ), yet one member is diverging ( $- 4 D$ ). So the unknown lens must be a strong convex one — strong enough to beat the $- 4 D$ and still net positive.

Step 1 — Combined power first. $P = \frac{1}{F} = \frac{1}{0.15} = 6.67 D$ Why this step? We know the result; to reverse-engineer a part, express the whole as a power.

Step 2 — Reverse the addition rule. Contact means $P = P_{1} + P_{2}$ , so with $P_{2} = - 4 D$ : $P_{1} = P - P_{2} = 6.67 - (- 4) = 6.67 + 4 = 10.67 D$ Why this step? Solving $P_{1} + P_{2} = P$ for the unknown is plain subtraction — this is why powers (not focal lengths) are the friendly quantity.

Step 3 — Its focal length. $f_{1} = \frac{1}{10.67} \approx 0.0938 m \approx 9.4 cm$ Why this step? Convert the found power back to a focal length for a physical answer.

Verify: Add them back: $10.67 + (- 4) = 6.67 D$ , giving $F = 1/6.67 = 0.15 m = 15 cm$ ✓. Positive, strongly convex, exactly as forecast — it out-muscles the concave lens.

Recall Which cell is each example?

E1 :: C1 single convex E2 :: C2 single concave E3 :: C3 two convex in contact E4 :: C4 convex + concave in contact E5 :: C5 net-zero degenerate E6 :: C6 separated by d E7 :: C7 limiting d→0 check E8 :: C8 real-world eyeglasses E9 :: C9 exam reverse-engineering

Connections

Thin Lens Equation — the equation we add together for combinations.
Lensmaker's Equation — the other route to $P$ .
Magnification of Lenses — net $m = m_{1} m_{2} \dots$ for these combos.
Microscope and Telescope — the separated-lens case (E6) in action.
Resistors in Parallel — the same reciprocal arithmetic as contact lenses.
Defects of Vision — where E8's dioptre prescriptions come from.