Refraction by lenses

Hold a magnifying glass over a printed page and the letters swell; tilt a pair of spectacles the wrong way and the same page blurs into a smear. Both are the same physics wearing different faces — light bending as it crosses a curved, transparent boundary, twice in a row, on its way through a thin lens. A camera lens, a microscope objective, the lens in your own eye: every one of them is, at heart, two curved refracting surfaces glued together, and every one of them obeys exactly the same two equations you are about to meet. The previous node built refraction at a single spherical surface — one boundary, one bend. A lens simply repeats that trick twice: light refracts on entry, travels a short distance inside the glass, then refracts again on exit. Because the two bends can be chosen to cooperate (both bending rays toward the axis) or oppose each other in shape, a lens can be built to either converge parallel rays to a real point of focus or diverge them so they appear to spread from a virtual point behind the lens. That single design choice — converging or diverging — is what separates a convex lens from a concave one, and it is fixed entirely by the sign of one number: the focal length $f$ . This section compresses that double refraction into two formulas you will use for the rest of the chapter: the lens maker's formula, which builds $f$ out of a lens's material and curvature, and the thin lens formula, which turns $f$ into predictions about where an image forms and how big it is. Get comfortable here, because every optical instrument later in this chapter — microscope, telescope, the eye itself — is just one or two thin lenses used according to these same two rules.

Key takeaways

A convex (converging) lens bends parallel rays inward to a real focus; a concave (diverging) lens spreads them so they appear to come from a virtual focus on the incoming side.
The lens maker's formula builds focal length from geometry and material: $\dfrac1f=(n-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ , using the same Cartesian sign convention as single-surface refraction.
By convention a convex lens has positive $f$ and a concave lens has negative $f$ — this is fixed once you know the lens shape, before doing any calculation.
The thin lens formula $\dfrac1v-\dfrac1u=\dfrac1f$ locates the image; note the minus sign, unlike the mirror formula's plus sign.
Lens magnification is $m=v/u$ (no minus sign, unlike mirrors); positive $m$ means an erect image, negative means inverted.
Lenses placed in contact combine by adding reciprocals of focal length, $\dfrac1f=\dfrac1{f_1}+\dfrac1{f_2}$ , and their powers simply add, $P=P_1+P_2$ , with $P=1/f$ measured in dioptres when $f$ is in metres.
Key idea: a thin lens is two refracting surfaces in series, so its focal length is fixed by shape and material through $\frac1f=(n-1)(\frac1{R_1}-\frac1{R_2})$ , and once $f$ is known, image position and size follow from $\frac1v-\frac1u=\frac1f$ and $m=v/u$ , with convex lenses positive and concave lenses negative throughout.

The core insight

A thin lens refracts light at two curved surfaces in succession, so its focal length is set by material and curvature through the lens maker's formula, and — once $f$ is known and signed by convention (convex positive, concave negative) — the thin lens formula $\frac1v-\frac1u=\frac1f$ and magnification $m=v/u$ predict image position, nature, and size exactly as the mirror equations do for mirrors, but with the signs of both central formulas flipped.

Where it fits in the syllabus

Refraction by lenses sits in Unit IX, "Ray Optics and Optical Instruments," directly after refraction at a single spherical surface and directly before power of a lens and lens combinations. It is fully within the prescribed 2025-26 CBSE syllabus: the thin lens formula, lens maker's formula, lens magnification, and combination of thin lenses in contact are all explicitly listed as theory content and are fair game for both board and competitive numericals. Two things are deliberately kept out of this section because they are out of the current prescription: aberrations (spherical and chromatic) get at most a passing mention, and the combination of lenses separated by a distance (which needs an equivalent-focal-length formula beyond "in contact") is not board-syllabus content and is left for a later, non-CBSE context if you meet it at all.

The physical picture

A lens is a transparent medium — glass or plastic of refractive index $n$ relative to its surroundings — bounded by two spherical surfaces (or one spherical and one flat, which is just a sphere of infinite radius). Trace a ray of light through it: it refracts once entering the first surface, travels a short distance through the glass, and refracts a second time leaving the second surface. Because the lens is "thin" — its thickness is negligible compared with the object and image distances and the radii of curvature — we can treat both refractions as happening effectively at a single plane, the optical centre, and add up their combined bending effect in one formula. The shape of the two surfaces decides everything. In a biconvex lens both surfaces bulge outward, and a bundle of rays parallel to the principal axis is bent inward at each surface, converging to a single point beyond the lens — the principal focus $F$ , at a distance $f$ from the optical centre, called the focal length. In a biconcave lens both surfaces curve inward, and parallel rays are bent outward at each surface so they never actually meet; extended backward, they appear to diverge from a virtual focus on the same side as the incoming light.

\frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right)

where: $f$ is the focal length of the lens, $n$ is the refractive index of the lens material relative to the surrounding medium (usually air), and $R_1$ , $R_2$ are the radii of curvature of the first and second surfaces the light meets, each signed by the same Cartesian convention used for a single refracting surface — measured from the optical centre, positive if the centre of curvature lies on the outgoing (transmitted-light) side, negative if it lies on the incoming side. Reading it physically: a lens made of a more strongly bending material (larger $n$ , so a larger $n-1$ ) or shaped with more sharply curved surfaces (smaller $|R_1|$ , $|R_2|$ ) has a shorter focal length and bends light more strongly — exactly the everyday sense in which a "fatter," more strongly curved lens is more powerful. Once $f$ is known, the thin lens formula locates the image of an object at distance $u$ from the optical centre:

\frac1v-\frac1u=\frac1f

where: $u$ is the object distance (negative, since the object is always measured against the direction of incident light, on the incoming side), $v$ is the image distance (positive if the image forms on the outgoing side — a real image — negative if it forms on the same side as the object — a virtual image), and $f$ is the signed focal length from the lens maker's formula: positive for a convex (converging) lens, negative for a concave (diverging) lens. This one sign flip — plus instead of minus, relative to the mirror formula $\frac1v+\frac1u=\frac1f$ — is the single most important thing to get automatic in this chapter, because the two formulas are used side by side constantly and are easy to confuse under exam pressure. Magnification for a lens compares image height $h'$ to object height $h$ :

m=\frac{h'}{h}=\frac{v}{u}

where: unlike the mirror magnification $m=-v/u$ , the lens formula carries no minus sign in front of $v/u$ itself — the sign of $m$ comes purely from the signs of $v$ and $u$ that you compute. A positive $m$ (both $v$ and $u$ having the same sign, which for a real object means $v>0$ ) corresponds to an erect image; a negative $m$ (real object, $v<0$ ... more precisely, $v$ and $u$ of opposite sign) corresponds to an inverted image. In practice, for a real object in front of a lens ( $u$ always negative), a positive $v$ (real image) always gives negative $m$ — inverted — while a negative $v$ (virtual image) gives positive $m$ — erect.

Figure

A biconvex lens on the left bending three parallel rays inward to a single real focus on the right, and a biconcave lens on the right bending three parallel rays outward so their backward extensions meet at a virtual focus on the incoming side

Combining two thin lenses in contact — pressed together so both share essentially the same optical centre — adds their bending effects directly:

\frac1f=\frac1{f_1}+\frac1{f_2}

Because power $P=1/f$ (dioptre, when $f$ is in metres), this is the same statement as $P=P_1+P_2$ : powers simply add, which is exactly why opticians describe corrective lenses in dioptres — you can add or subtract prescriptions arithmetically. A converging lens in contact with a diverging lens of the same magnitude cancels to zero net power, a fact used to check corrective-lens numericals quickly.

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

The goal is to find the numerical focal length of a real, symmetric biconvex lens — glass of refractive index $1.5$ , both surfaces of radius of curvature $20$ cm — using only the lens maker's formula and a careful, deliberate application of the sign convention to a genuinely confusing symmetric case, since assigning $R_1$ and $R_2$ correctly here is the step students most often get backwards.

Step 1. Set up the sign convention. Light travels left to right, so the incoming side is to the left of the lens and the outgoing side is to the right; distances are measured from the optical centre, positive to the right (outgoing side) and negative to the left (incoming side), exactly as for a single refracting surface.

Step 2. Assign $R_1$ , the radius of the first surface the light meets. For this biconvex lens, the first (left) surface bulges toward the incoming light, so its centre of curvature lies on the outgoing (right) side — that makes $R_1=+20$ cm.

Step 3. Assign $R_2$ , the radius of the second surface. The second (right) surface, curving the same physical way to complete the symmetric bulge, has its centre of curvature on the incoming (left) side when judged from that surface — that makes $R_2=-20$ cm. This is the step that trips people up: both surfaces "look convex" from outside the lens, but the sign convention judges each one independently, from its own position, and the two signs come out opposite.

Step 4. Substitute into the lens maker's formula. With $n=1.5$ :

\frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right)=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right)=0.5\left(\frac{1}{20}+\frac{1}{20}\right)

\frac1f=0.5\times\frac{2}{20}=0.5\times0.1=0.05\ \text{cm}^{-1}

Step 5. Invert and interpret. $f=\dfrac{1}{0.05}=20$ cm. The positive sign confirms the lens is converging, consistent with a convex lens under this convention, and the power is $P=1/f=1/0.20\ \text{m}=5$ dioptres — a moderately strong reading-glass-strength lens. Sanity check: for a symmetric biconvex lens with $R_1=+R$ and $R_2=-R$ , the formula reduces to $\frac1f=(n-1)\frac{2}{R}$ , so $f=\frac{R}{2(n-1)}=\frac{20}{2(0.5)}=20$ cm — matches Step 5 exactly, confirming the arithmetic.

Worked examples

Board level - focal length of a symmetric biconvex lens

Question: A convex lens is made of glass with refractive index $1.5$ . One surface has a radius of curvature of $20$ cm, convex toward the incoming light, and the other surface curves the opposite way with the same magnitude of radius, $20$ cm, giving a symmetric biconvex lens. Find its focal length and its power.

Step 1. Assign signs by the Cartesian convention: for this biconvex lens, $R_1=+20$ cm (first surface's centre of curvature on the outgoing side) and $R_2=-20$ cm (second surface's centre of curvature on the incoming side).

Step 2. Apply the lens maker's formula. $\dfrac1f=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)=(1.5-1)\left(\dfrac{1}{20}-\dfrac{1}{-20}\right)=0.5\left(\dfrac{1}{20}+\dfrac{1}{20}\right)=0.5\times0.1=0.05\ \text{cm}^{-1}$ .

Step 3. Invert to find $f$ . $f=\dfrac{1}{0.05}=20$ cm $=0.20$ m. The positive value confirms this symmetric biconvex lens is converging.

Step 4. Convert to power. $P=\dfrac1f=\dfrac{1}{0.20\ \text{m}}=5$ D (dioptres).

The focal length is $\boldsymbol{f=+20\ \text{cm}}$ and the power is $\boldsymbol{P=+5\ \text{D}}$ .

Competitive level - image, magnification, and a lens combined in contact

Question: An object is placed $30$ cm in front of the convex lens from the example above ( $f=20$ cm). (a) Find the image position and magnification, and describe the image. (b) A second, thin concave lens of focal length $-40$ cm is now placed in contact with the first. Find the new combined focal length and power of the pair.

Step 1. Locate the image of the single convex lens. Given $u=-30$ cm (object measured against incident light) and $f=+20$ cm (convex lens), the thin lens formula gives $\dfrac1v=\dfrac1f+\dfrac1u=\dfrac{1}{20}+\dfrac{1}{-30}=\dfrac{3-2}{60}=\dfrac{1}{60}$ , so $v=60$ cm.

Step 2. Interpret $v$ . Positive $v=60$ cm means the image forms $60$ cm on the far side of the lens from the object — a real image, since it lies on the outgoing side.

Step 3. Find the magnification. $m=\dfrac{v}{u}=\dfrac{60}{-30}=-2$ . The image is inverted ( $m<0$ ) and twice the object's size — real, inverted, magnified, exactly as expected for an object between $f$ and $2f$ of a convex lens ( $f=20$ cm, $2f=40$ cm, object at $30$ cm lies between them).

Step 4. Combine the two lenses in contact. With $f_1=+20$ cm and $f_2=-40$ cm, $\dfrac{1}{f}=\dfrac{1}{f_1}+\dfrac{1}{f_2}=\dfrac{1}{20}+\dfrac{1}{-40}=\dfrac{2-1}{40}=\dfrac{1}{40}$ , so $f=+40$ cm.

Step 5. Convert to power to sanity-check the addition. $P_1=\dfrac{1}{0.20}=+5$ D, $P_2=\dfrac{1}{-0.40}=-2.5$ D, so $P=P_1+P_2=+2.5$ D, giving $f=1/P=0.40$ m $=40$ cm — matches Step 4.

The single-lens image is real, inverted, $2\times$ magnified at $\boldsymbol{v=60}$ cm; the combined pair has $\boldsymbol{f=+40\ \text{cm}}$ , $\boldsymbol{P=+2.5\ \text{D}}$ , still net converging. Timing note: in a competitive-exam setting, do not re-derive image position for the combined pair from scratch by tracing through both lenses in sequence unless the question demands it — for lenses genuinely in contact, jump straight to $1/f=1/f_1+1/f_2$ (or add powers) and treat the pair as one equivalent lens; this alone typically saves thirty to forty-five seconds per question and avoids a second, unnecessary application of the thin lens formula.

Common traps

Using the mirror equation's $+$ sign instead of the lens equation's $-$ sign, or vice versa — $\dfrac1v+\dfrac1u=\dfrac1f$ is for mirrors, $\dfrac1v-\dfrac1u=\dfrac1f$ is for lenses; this single slip is the most common source of wrong answers in this chapter.
Using $m=-v/u$ (the mirror formula) for a lens instead of the correct $m=v/u$ — this flips the predicted orientation (erect versus inverted) of the image.
Forgetting that convex lenses have positive $f$ and concave lenses have negative $f$ under this convention — the opposite pairing from convex/concave mirrors — and mixing the two families up under time pressure.
Assigning $R_1$ and $R_2$ by "which way the lens bulges" instead of judging each surface independently from its own position, which is exactly where the symmetric biconvex/biconcave case in the derivation above goes wrong for most students on a first attempt.
Applying $1/f=1/f_1+1/f_2$ to lenses that are not in contact — a separated pair needs an equivalent-focal-length formula that also accounts for the gap between them, and that extended case is outside this syllabus; do not invent it under exam pressure.
Forgetting to convert $f$ to metres before computing power in dioptres — $P=1/f$ only gives dioptres directly when $f$ is expressed in metres, not centimetres.

You should be able to…

Distinguish convex and concave lenses by shape, by the sign of their focal length, and by whether they converge or diverge parallel rays.
Apply the lens maker's formula, correctly assigning the signs of $R_1$ and $R_2$ by the Cartesian convention, to find a lens's focal length and power.
Apply the thin lens formula and the lens magnification formula together to find an image's position, nature (real or virtual, erect or inverted), and size.
State the sign convention for convex versus concave lenses and never confuse it with the sign convention for convex versus concave mirrors.
Combine two thin lenses in contact using $1/f=1/f_1+1/f_2$ or equivalently $P=P_1+P_2$ , and convert between focal length in metres and power in dioptres.

Ask the tutor: want to work through a concave-lens numerical, or practice assigning $R_1$ and $R_2$ signs for a biconcave lens step by step?

Why does the lens formula have a minus sign but the mirror formula has a plus sign, if both are describing basically the same idea of focusing light?

It comes down to how the sign convention interacts with each device's geometry. A mirror sends light back the way it came, so image and object distances end up related through addition in the standard convention. A lens sends light through to the far side, and object and image distances are measured with opposite orientation relative to how the formula was set up historically, which turns the relation into a subtraction. There is no way to "see" this from first principles quickly — the safest approach is to simply memorise the two formulas as a pair and always double-check which device the question is about before you write anything down.

If a concave lens always produces a virtual image, why bother with the full sign convention at all — can I not just skip straight to the answer?

A single concave lens with a real object does always give a virtual, erect, diminished image, so in that one specific case you could shortcut it. But the same sign convention has to work for every other situation too — convex lenses, virtual objects formed by a previous lens in a combination, objects at different distances from the focus — and examiners often set questions precisely in those less-obvious cases to test whether you actually understand the convention or have just memorised one outcome. Learning the general method once is faster than memorising a list of special cases.

When two lenses are in contact, does the order in which I write f1 and f2 in the combination formula matter?

No — addition is commutative, so $1/f_1+1/f_2$ gives the same $1/f$ regardless of which lens you call "first" and which you call "second." What does matter is getting the sign of each individual $f$ right before you add: a convex lens contributes a positive term, a concave lens contributes a negative term, and the combined $f$ can come out positive, negative, or (in the special case where the two powers exactly cancel) infinite, meaning the pair behaves like plain glass with no net focusing power at all.

My textbook mentions cutting a lens in half or silvering one face — is that in my board syllabus?

No. Cutting, silvering, and similar constructed variations of lenses are practical or advanced extensions that are not part of the current CBSE Class 12 prescribed theory, so you do not need them for board exams. Stick to the lens maker's formula, the thin lens formula, magnification, and lenses combined in contact — those four cover everything examinable in this specific topic.