Power of a lens, combination of lenses: Question bank

Before we start, one reminder of the only three facts every answer leans on:

Picture the four facts first

Every trap on this page is really about one of four pictures. Anchor them now, then the reveals become obvious.

Picture 1 — Why $P = 1/ f$ (the triangle behind the definition). A ray comes in parallel to the axis at height $h$ . After the thin lens it heads down to the focus, a distance $f$ away. Look at the red triangle: the ray dropped a height $h$ over a horizontal run $f$ . The bending angle $δ$ is the angle at the lens.

Figure — Power of a lens, combination of lenses

Picture 2 — What $R_{1}$ and $R_{2}$ mean (sign convention for the lensmaker formula). The lensmaker formula $P = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ hides two signed radii. Here $R_{1}$ is the radius of the first surface light hits and $R_{2}$ the second. The convention (light travelling left→right): a surface that bulges toward the incoming light is convex and its radius is positive; a surface that curves away is negative.

Picture 3 — Contact lenses add like parallel resistors (the conductance analogy). Resistors in parallel add their reciprocals (conductances): $\frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$ . Lenses in contact add their reciprocals (powers): $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ . Same shape of equation — see the side-by-side.

Picture 4 — Where the $- d P_{1} P_{2}$ term comes from. Follow a ray parallel to the axis at height $h_{1}$ into lens 1. Lens 1 bends it by $δ_{1} = P_{1} h_{1}$ . Over the gap $d$ it drops to a smaller height $h_{2} = h_{1} - d δ_{1}$ before hitting lens 2. Lens 2 bends that smaller height: $δ_{2} = P_{2} h_{2}$ .

True or false — justify

A lens with a larger focal length is a more powerful lens.

False. Power is

P = 1/ f

, so power is inversely related to

f

— a longer focal length means a weaker lens that bends rays gently (Picture 1).

Power is always a positive quantity because it measures "strength".

False. Power carries the sign of the focal length: convex

f > 0

gives

P > 0

, concave

f < 0

gives

P < 0

. The sign encodes direction of bending (converging vs diverging), not just magnitude.

Two lenses in contact always give a stronger (higher-magnitude) combination than either alone.

False. Powers add with sign: a

+ 5 D

and a

- 2 D

give only

+ 3 D

, weaker than the

+ 5 D

lens. A concave lens partly cancels a convex one.

f

is doubled, the power is halved.

True. Since

P = 1/ f

, doubling

f

multiplies

P

1/2

. This is the whole meaning of "reciprocal relationship".

For lenses in contact, focal lengths add just like the resistances of resistors in series.

False. It is the powers (reciprocals) that add, so lenses in contact mimic Resistors in Parallel, where reciprocals add:

\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}

(Picture 3).

A lens of power

+ 5 D

has focal length

5 m

False.

f = 1/ P = 1/5 = 0.2 m = 20 cm

, not

5 m

. Power is the reciprocal of focal length in metres.

The separation formula

P = P_{1} + P_{2} - d P_{1} P_{2}

can never give a power smaller than

P_{1} + P_{2}

False (for two converging lenses). With

P_{1}, P_{2} > 0

the term

- d P_{1} P_{2}

is negative, so separating two convex lenses reduces the combined power below

P_{1} + P_{2}

(Picture 4).

The magnifications of a lens combination add, just like the powers.

False. Powers add (

P = P_{1} + P_{2}

) but magnifications multiply:

m = m_{1} \times m_{2} \times \dots

, because each stage scales the previous image, seen in Magnification of Lenses.

The unit "dioptre" is the same as "per centimetre".

False.

1 D = 1 m^{- 1}

, defined for focal length in metres. A focal length in centimetres must be converted to metres before taking the reciprocal.

Spot the error

f = 20 cm

, so

P = 1/20 = 0.05 D

The error is using centimetres. Convert first:

20 cm = 0.2 m

, so

P = 1/0.2 = 5 D

. The dioptre is defined only for metres.

"A concave lens of

f = - 40 cm

has

P = 1/40 = 2.5 D

The sign was dropped.

f = - 0.4 m

, so

P = 1/ (- 0.4) = - 2.5 D

. Concave lenses have negative power — the minus is what lets them cancel converging power.

"Lenses in contact:

P_{1} = + 4 D

P_{2} = - 1 D

, so

F = f_{1} + f_{2}

You cannot add focal lengths. Add powers:

P = 4 + (- 1) = + 3 D

, then

F = 1/3 m \approx + 33.3 cm

. Focal-length addition is only correct for resistor-series problems, not lenses.

"Two lenses separated by

d

P = P_{1} + P_{2} - d P_{1} P_{2}

with

d = 10

Here

d

must be in metres to match the dioptre. Use

d = 0.10 m

; plugging

d = 10

makes the correction term absurdly huge.

"For lenses in contact I used

P = P_{1} + P_{2} - d P_{1} P_{2}

with

d = 0

, and I'm worried the answer changed."

No error, actually — at

d = 0

the term

- d P_{1} P_{2}

vanishes and the formula collapses to

P_{1} + P_{2}

. That is the built-in sanity check that the two formulas agree.

"For a biconvex lens I set

R_{1} < 0

and

R_{2} > 0

in the lensmaker formula."

Signs reversed. For light entering left→right on a biconvex lens, the first surface bulges toward the light so

R_{1} > 0

, and the second curves away so

R_{2} < 0

; both terms then add (Picture 2).

"Since

P = (n - 1) (1/ R_{1} - 1/ R_{2})

, using a lens in water (where the surrounding

n

rises) keeps its power unchanged."

The factor is really

(n_{lens} / n_{medium} - 1)

; in water this shrinks toward zero, so the lens becomes much weaker. Power depends on the relative refractive index, from Lensmaker's Equation.

Why questions

Why is power defined as

1/ f

rather than just using

f

directly?

A ray at height

h

makes the red triangle of Picture 1 with sides

h

and

f

, so

tan δ = h / f

; for small angles

tan δ \approx δ

, giving bending-per-height

δ / h = 1/ f

. That deviation strength is what "power" names, and strengths of stacked lenses add.

Why do the powers of two lenses in contact add, but not their focal lengths?

Adding the two thin-lens equations cancels the intermediate image distance

v_{1}

, leaving

1/ F = 1/ f_{1} + 1/ f_{2}

— a sum of reciprocals, i.e. of powers, not of focal lengths (Picture 3).

Why does separating two convex lenses reduce their combined power?

A ray bent by lens 1 drops to a smaller height over the gap

d

, so lens 2 bends a smaller height and contributes less; the correction

- d P_{1} P_{2}

(negative for two positives) subtracts from the total (Picture 4).

Why does a denser glass or a more sharply curved surface give more power?

P = (n - 1) (1/ R_{1} - 1/ R_{2})

, larger

n

makes

(n - 1)

bigger and smaller

∣ R ∣

makes the curvature term bigger — both increase the bending, hence the power.

Why do opticians prescribe spectacles in dioptres rather than centimetres?

Dioptres add directly (eye power + spectacle power), so correcting a defect is simple addition, whereas focal lengths would require awkward reciprocal arithmetic — see Defects of Vision.

Why does a microscope or telescope quote powers for objective and eyepiece separately?

Because their behaviour combines through power/magnification rules; the design is built from these combination laws, as used in Microscope and Telescope.

Why must

d

be in metres in the separation formula, but appears "just as a distance"?

The term

d P_{1} P_{2}

has units of

m \times D \times D = m \cdot m^{- 2} = m^{- 1} = D

; only metres make the units come out as power to add to

P_{1} + P_{2}

Edge cases

What is the power of a flat glass plate (both surfaces flat,

R_{1}, R_{2} \to \infty

1/ R_{1} - 1/ R_{2} \to 0

, so

P = 0

and

f \to \infty

. It bends nothing — a "lens" of zero power, which is why windows don't focus light.

What happens to the combined power if a convex lens (

+ P

) is placed in contact with a concave lens of equal magnitude (

- P

P_{net} = P + (- P) = 0

, giving an effective flat plate: the pair converges and diverges by equal amounts and does no net bending.

In the separation formula, what value of

d

makes the combined power zero for two convex lenses?

Set

P_{1} + P_{2} - d P_{1} P_{2} = 0

, giving

d = (P_{1} + P_{2}) / (P_{1} P_{2}) = f_{1} + f_{2}

. At that spacing the pair acts as a zero-power (telescopic) system.

What is the power of a lens as its focal length approaches zero (an ideal ultra-strong lens)?

P = 1/ f \to \infty

. Physically no real lens reaches this, but the formula shows an arbitrarily short focal length means arbitrarily large power.

If a lens has power exactly

0 D

, is it convex or concave?

Neither — it is effectively a plane plate. Zero power sits at the boundary between positive (convex) and negative (concave), bending no rays.

For lenses in contact, does the order (which lens first) change the net power?

No.

P_{1} + P_{2} = P_{2} + P_{1}

— addition is commutative, so swapping the lenses gives the same combined power (and same

F

Two lenses separated by

d

equal to

f_{1} + f_{2}

: what is special about this arrangement?

The combined power is zero (from the edge case above), so parallel rays enter and leave parallel — this is exactly the afocal, telescope-like configuration used in Microscope and Telescope.

Recall One-line self-test

Cover every answer above; a "trap" you can't justify in one sentence is a concept to revisit before an exam.

Connections

Thin Lens Equation — the equation we add to derive combined power.
Lensmaker's Equation — source of the $(n - 1) (1/ R_{1} - 1/ R_{2})$ traps.
Magnification of Lenses — magnifications multiply, powers add.
Microscope and Telescope — afocal and combination arrangements.
Resistors in Parallel — same reciprocal-addition mathematics.
Defects of Vision — why prescriptions use dioptres.