Lens formula and combinations

A camera lens, a magnifying glass and the eyepiece of a telescope are all doing the same piece of algebra: bending light at two curved surfaces to place an image exactly where you want it. Once you have the single-surface refraction formula from the previous topic, a thin lens is nothing more than that formula applied twice in a row — once at each face — with the messy intermediate image distance cancelling out neatly. What survives the cancellation are three compact tools: the lens maker's formula (built from the lens's own shape and material), the thin-lens formula (built from where the object actually sits), and a simple addition rule for combining lenses in contact. This article rebuilds all three, works two ISC-style numericals from scratch, and flags the sign-convention traps — especially the lens formula's minus sign — that decide marks in the exam.

The core insight

Applying the single-surface refraction relation at both faces of a thin lens and eliminating the intermediate image distance yields $1/f=(n-1)(1/R_1-1/R_2)$ and $1/v-1/u=1/f$ , and lenses placed in contact simply add their powers, $1/F=1/f_1+1/f_2$ .

The core idea

A thin lens is two refracting surfaces back to back. Light from the object first crosses surface 1 (going from medium $n_1$ , usually air, into the lens material $n$ ) and forms an intermediate image; that intermediate image then acts as the object for surface 2 (going from $n$ back out into $n_1$ ). Applying $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ at each face and adding the two equations makes the intermediate image distance cancel, leaving the lens maker's formula:

\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

where: $f$ is the focal length, $n$ is the refractive index of the lens material relative to the surrounding medium, and $R_1$ , $R_2$ are the signed radii of curvature of the first and second surfaces the light meets (positive if the centre of curvature lies on the outgoing-light side, by the usual convention). Once $f$ is fixed by the lens's shape, the thin-lens formula relates object distance $u$ and image distance $v$ for any object:

\frac1v-\frac1u=\frac1f

Notice the minus sign — the mirror formula uses a plus sign, and mixing the two up is the single most common slip in this chapter. Magnification for a thin lens is $m=h_2/h_1=v/u$ (no leading minus sign, unlike the mirror's $m=-v/u$ ). Power is the reciprocal of focal length in metres, unit dioptre (D): $P=1/f$ , positive for a converging (convex) lens and negative for a diverging (concave) one. For two thin lenses of focal lengths $f_1$ and $f_2$ held in contact (negligible gap between them), the pair behaves as one thin lens of focal length $F$ :

\frac1F=\frac{1}{f_1}+\frac{1}{f_2}\qquad\text{equivalently}\qquad P=P_1+P_2

Powers of lenses in contact simply add — which is exactly why opticians quote prescriptions in dioptres: two lenses worn together (or cemented) combine by plain addition.

Figure

Two rays from an object point crossing a thin convex lens surface and a diverging beam crossing a concave lens surface converging or spreading toward the principal axis

Derivation AFocal length of a biconvex lens from the lens maker's formula

The goal is to compute the focal length and power of a real biconvex lens from its refractive index and two surface radii, applying the lens maker's formula step by step with correct sign convention so the arithmetic can be checked term by term rather than trusted blindly.

Step 1. State the data with signs. The lens is glass of refractive index $n=1.5$ , biconvex, with the first surface (light meets it first) curving with radius $20\ \text{cm}$ toward the outgoing light, so $R_1=+20\ \text{cm}$ ; the second surface curves the opposite way for a biconvex shape, so $R_2=-30\ \text{cm}$ .

Step 2. Substitute into the lens maker's formula, $\dfrac1f=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)=(1.5-1)\left(\dfrac{1}{20}-\dfrac{1}{-30}\right)$ .

Step 3. Simplify the bracket carefully, since subtracting a negative number flips to addition: $\dfrac{1}{20}-\dfrac{1}{-30}=\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{3}{60}+\dfrac{2}{60}=\dfrac{5}{60}=\dfrac{1}{12}$ .

Step 4. Multiply by $(n-1)=0.5$ : $\dfrac1f=0.5\times\dfrac{1}{12}=\dfrac{1}{24}\ \text{cm}^{-1}$ , so $f=24\ \text{cm}$ .

Step 5. Convert to power: $P=1/f$ with $f$ in metres, $f=0.24\ \text{m}$ , so $P=1/0.24\approx4.17\ \text{D}$ . Both $f$ and $P$ are positive, confirming this biconvex lens is converging, as expected for a lens thicker at the centre made of a denser material than its surroundings.

Worked example

An object 4 cm tall stands 30 cm in front of a convex lens of focal length 15 cm. Find the position, nature and size of the image; then place a second thin lens of focal length $-10$ cm (diverging) in contact with the first and find the new combined focal length.

Step 1. Assign signs and apply the thin-lens formula. Taking distances from the lens with the usual sign convention, $f=+15\ \text{cm}$ and $u=-30\ \text{cm}$ (object on the incoming side). From $\dfrac1v=\dfrac1f+\dfrac1u=\dfrac{1}{15}+\dfrac{1}{-30}=\dfrac{2}{30}-\dfrac{1}{30}=\dfrac{1}{30}$ , so $v=+30\ \text{cm}$ .

Step 2. Find magnification and image height. $m=\dfrac{v}{u}=\dfrac{30}{-30}=-1$ , so image height $=m\times h_1=-1\times4=-4\ \text{cm}$ . The negative sign means the image is inverted; the magnitude equal to 1 means it is the same size as the object. Since $v$ is positive (real side), the image is real.

Step 3. So the single lens alone forms a real, inverted image 30 cm behind the lens, the same size (4 cm) as the object — this matches the Pass-1 result exactly.

Step 4. Combine with the diverging lens. Placing $f_2=-10\ \text{cm}$ in contact with $f_1=+15\ \text{cm}$ : $\dfrac1F=\dfrac{1}{f_1}+\dfrac{1}{f_2}=\dfrac{1}{15}+\dfrac{1}{-10}=\dfrac{2}{30}-\dfrac{3}{30}=-\dfrac{1}{30}$ , giving $F=-30\ \text{cm}$ .

Final answer: the convex lens alone forms a real, inverted, same-size image 30 cm away; combined with the $-10$ cm diverging lens, the two-lens system has an equivalent focal length of $-30\ \text{cm}$ — the pair becomes net diverging.

Competitive stretch

Check the combination using power instead of focal length, as an exam shortcut: $P_1=1/0.15=6.67\ \text{D}$ , $P_2=1/(-0.10)=-10\ \text{D}$ , so $P=P_1+P_2=-3.33\ \text{D}$ , and $F=1/P=1/(-3.33)=-0.30\ \text{m}=-30\ \text{cm}$ — the same answer, obtained without ever inverting a fraction of centimetres, which is faster when a problem already gives powers in dioptres.

Common traps

Swapping the mirror and lens signs. Mirror formula uses a plus sign, $\dfrac1v+\dfrac1u=\dfrac1f$ ; lens formula uses a minus sign, $\dfrac1v-\dfrac1u=\dfrac1f$ . Under exam pressure students default to whichever they used last.
Wrong magnification sign convention. For a thin lens $m=v/u$ has no leading minus sign; for a mirror it does. Copying the mirror's rule onto a lens problem gives the wrong nature of image.
Careless $R_1$ , $R_2$ signs in the lens maker's formula. Assign signs from the actual direction light travels through that surface, not from a memorised "convex is always positive" shortcut — a plano-convex or concavo-convex lens will trip that shortcut.
Adding focal lengths directly for a combination. For lenses in contact it is reciprocals (or powers) that add, $1/F=1/f_1+1/f_2$ , never $F=f_1+f_2$ .

Ask the tutor: want a step-by-step walkthrough of the two-surface derivation that produces the lens maker's formula from scratch, or a practice problem using a plano-convex lens where one radius is infinite?

Why does the lens formula have a minus sign but the mirror formula has a plus sign?

The two formulas come from different sign-convention setups even though both use the same Cartesian rule that distances are measured from the optical element. For a lens, $v$ and $u$ are measured on opposite sides in the "normal" case (object and image on opposite faces), while for a mirror both are typically on the same side, which flips the sign that appears when the single-surface refraction relation is manipulated into the final compact formula. Rather than re-deriving the sign each time, just memorise the two forms as fixed results: mirror plus, lens minus, and always double-check with a known case such as an object at infinity.

How do I know whether to use R1 or R2 first when a lens has one flat side?

Label the surfaces by which one light meets first as it travels through the lens, regardless of which looks more "important." A flat surface has infinite radius of curvature, so $1/R=0$ for that surface; if it's the second surface met, the lens maker's formula collapses to $\dfrac1f=(n-1)\dfrac{1}{R_1}$ . The labelling ( $R_1$ first, $R_2$ second) is about the order light crosses the surfaces, not about which surface is curved.

If I put two lenses close but not exactly touching, can I still just add 1/f1 and 1/f2?

Only if the separation between them is genuinely negligible compared to their focal lengths — that is the assumption baked into $1/F=1/f_1+1/f_2$ . If the lenses are separated by a distance you're told to account for, that simple addition is no longer valid; a full two-step calculation (image from lens 1 becomes the object for lens 2, using the actual gap) is needed instead. ISC problems on lens combination almost always specify "in contact" precisely so this shortcut applies.

Lens formula and combinations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads