ARCHETYPE: Lens & Combination Numericals + Ray Diagrams

Somewhere in every CBSE, ICSE, JEE and NEET paper that reaches ray optics there is a lens sitting in wait — a projector, a magnifying glass, a pair of lenses glued together, an object and a screen a metre apart. It is worth three to five marks, it is almost entirely mechanical, and it is one of the most dependable places on the whole paper to bank those marks or to throw them away on a single misplaced sign. Students rarely fail lens numericals because the physics is deep. They fail because they improvise the signs under the clock, forget that a lens uses $m = v/u$ and not the mirror's $-v/u$ , or add focal lengths when they should be adding powers. This is the problem gym for the thin lens. The refraction spine earlier in this chapter built the thin-lens formula and proved it; here the setup is the whole point. Every lens numerical, however it is dressed, is one of a small family: locate the image of a single lens, classify a magnifier, combine two lenses in contact or separated, or find a focal length by the displacement trick with an object and a screen. Learn the four-step recipe until it is a reflex and the entire family collapses into fill-in-the-blanks. We run the drill on the board staples — a converging lens making a real image, and the same lens turned into a magnifying glass — then climb to the two questions that separate top scorers: lenses combined by adding their powers in dioptres, and the displacement (Bessel) method that finds a focal length without ever knowing the object distance. Threading it all is the craft the technique article makes a habit of: draw the two easy rays, sign every distance, and read the nature off the answer. By the end you should see a lens problem and know its shape before you write a line.

Key takeaways

Every single-lens problem is one drill: sign the data (real object $u<0$ ; converging $f>0$ , diverging $f<0$ ), solve $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$ , read $m = v/u$ , and state the nature.
For a lens the magnification is $m = v/u$ (with a $+$ sign) — not the mirror's $m = -v/u$ . Muddling the two is the classic lost mark.
Board anchors ( $f = +20$ cm): object at $u = -30 \to v = +60$ cm, $m = -2$ (real, inverted); object inside the focus at $u = -15 \to v = -60$ cm, $m = +4$ (the magnifying glass).
Lenses in contact add powers: $+5$ D and $+2$ D give $+7$ D, so $F = 100/7 = 14.29$ cm — add dioptres, never focal lengths.
The displacement method: an object and screen $D = 100$ cm apart with two sharp-image positions $d = 40$ cm apart give $f = (D^2 - d^2)/4D = 21$ cm — a focal length with no $u$ or $v$ needed.
A diverging lens with a real object is boringly reliable: $f = -20$ , $u = -30 \to v = -12$ cm, $m = +0.4$ — always virtual, erect, diminished.
Key idea: solve every lens numerical with one fixed procedure — sign the data, apply $1/v = 1/f + 1/u$ , read $m = v/u$ , state the nature — and combine lenses by adding powers in dioptres.

The core insight

There is only one lens numerical, solved by one procedure: attach New Cartesian signs (real object $u<0$ , converging $f>0$ ), apply the thin-lens formula $1/v - 1/u = 1/f$ to locate the image, read the magnification $m = v/u$ to size and orient it, and state whether it is real or virtual, erect or inverted; combinations are handled by adding powers, $P = P_1 + P_2$ in contact.

Where this lives today

Lens numericals are the design arithmetic of every instrument that forms an image. A camera or phone lens places the sensor at the image distance $v$ that equation (1) predicts for a subject at $u$ ; autofocus is a motor solving $1/v = 1/f + 1/u$ many times a second. A projector sits its slide just outside the focus so $v \gg |u|$ and the image is thrown large onto a distant screen — Example 1 in hardware. A magnifying glass, a jeweller's loupe and the reading segment of varifocal spectacles are Example 2: the object held inside the focus for an enlarged upright virtual image. Opticians prescribe in dioptres precisely because lenses in contact add powers, equation (3), so a $+2.5$ D reading add on a $-1.0$ D distance lens is simply $+1.5$ D. The separated-lens formula, equation (4), sizes camera zoom groups and telescope eyepieces where the spacing is a design variable. Even the humble laboratory optical bench measuring a focal length by the displacement method, equation (5), survives in modern metrology because it needs no knowledge of the object or image distance — just two ruler readings. One formula, signed and summed, quietly places every lens image in the built world.

The setup: one recipe for every lens

In an archetype the bookkeeping is the physics, so learn the bookkeeping cold. The master relation, quoted from the thin-lens formula established earlier in this chapter and applied here with the New Cartesian sign convention, is

\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \qquad (1)

where: $u$ is the object distance, $v$ the image distance and $f$ the focal length, all measured from the optical centre with New Cartesian signs — incident light travels in the positive direction, so a distance measured against it (a real object, on the incoming side) is negative. Rearranged for the usual unknown, $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$ , and once you have $u$ and $f$ the image distance drops straight out; the sign of $v$ then tells you the side — a positive $v$ means a real image on the far side of the lens, a negative $v$ a virtual image on the same side as the object. The companion for size and orientation is

m = \frac{h'}{h} = \frac{v}{u} \qquad (2)

where: $h$ is the object height, $h'$ the image height, and $m$ the magnification. Note the sign carefully: for a lens the magnification is $+v/u$ , with no leading minus. Because a real object has $u<0$ and its real image has $v>0$ , the ratio $v/u$ comes out negative — correctly flagging a real image as inverted; a virtual image has $v<0$ too, so $v/u$ is positive, correctly flagging erect. The mirror's formula wore a minus sign, $m = -v/u$ ; the lens does not, and swapping the two is the single most common orientation error on the paper. The four-step recipe that turns any word problem into a number never changes:

Sign the data. A real object is $u<0$ . A converging (convex) lens has $f>0$ ; a diverging (concave) lens has $f<0$ ; a lens power quoted in dioptres is $P = 1/f(\text{m})$ , positive for converging and negative for diverging. Write the sign before the digit.

Solve for $v$ . Use $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$ , work with a common denominator, and keep the sign of $v$ .

Read the magnification. $m = v/u$ , and if a height is given, $h' = m\,h$ .

State the nature. $v>0$ real (far side), $v<0$ virtual (same side); $m<0$ inverted, $m>0$ erect; $|m|>1$ magnified, $|m|<1$ diminished. Boards mark the nature explicitly — never leave the answer as a bare number.

Figure

Two-ray construction for a converging lens: for a real object beyond the focus the parallel ray bends through the far focus and the ray through the optical centre goes straight, meeting to form a real inverted image, while for an object inside the focus the two rays diverge and their backward dashed extensions meet to form a magnified virtual image

Reading the two easy rays

Behind every lens numerical is a ray diagram, and you only ever need two rays to pin the image. A ray leaving the object parallel to the axis bends at the lens and passes through the far focus $F'$ . A ray heading straight through the optical centre of a thin lens goes on undeviated. Where those two emergent rays cross is the image. If they actually cross on the far side, the image is real and $v>0$ ; if they diverge and only their backward extensions (drawn dashed) meet on the near side, the image is virtual and $v<0$ . That dashed-ray discipline for virtual images is the same craft the sign-convention technique article drills, and it is what turns "the object is inside the focus" into "so the image is a magnified virtual one" without any algebra at all. The formula and the diagram are the same statement told two ways; the diagram is your sanity check on the sign of $v$ .

Combining lenses: add the powers

When two thin lenses sit in contact, their combined behaviour is not found by adding focal lengths — it is found by adding powers. The power of a lens is $P = 1/f$ with $f$ in metres, measured in dioptres (D), positive for converging and negative for diverging. For lenses in contact,

P = P_1 + P_2, \qquad \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \qquad (3)

where: $P_1, P_2$ are the individual powers and $P$ the power of the combination, $F$ its equivalent focal length. This is why an optician stacks trial lenses and simply sums the numbers on them: two converging lenses of $+5$ D and $+2$ D make a $+7$ D system, stronger than either. If one lens diverges, its power enters negative and subtracts — a $+5$ D lens in contact with a $-2$ D lens gives only $+3$ D. When the two lenses are separated by a distance $d$ (also in metres), the combination weakens by a cross-term:

P = P_1 + P_2 - d\,P_1 P_2 \qquad (4)

where: $d$ is the separation. The subtraction says that pulling the lenses apart reduces the system's power; it is the working equation behind eyepieces and camera lens groups, and it reduces to equation (3) when $d \to 0$ .

The displacement method: a focal length with no object distance

One board favourite refuses to tell you $u$ or $v$ at all. An object and a screen are fixed a distance $D$ apart, and a converging lens between them is found to throw a sharp image at two different positions, a distance $d$ apart. The focal length follows from the symmetry of conjugate points — object and image distances simply swap between the two positions — as

f = \frac{D^2 - d^2}{4D} \qquad (5)

where: $D$ is the fixed object-to-screen distance and $d$ the separation between the two sharp-image lens positions. It is the exam's favourite way to "find $f$ without knowing $u$ or $v$ ", and it works only when $D \ge 4f$ , so that two real positions exist at all. The second figure shows the bench.

Figure

The displacement-method bench: an object and a screen are held a fixed distance D apart, and a converging lens forms a sharp image at two positions separated by d, from which the focal length is D squared minus d squared all over four D

Derivation AThe master recipe on a real image

The goal is to run the four-step recipe end to end on the board staple — a converging lens of focal length $20$ cm with an object $30$ cm in front — and land the locked anchor $v = +60$ cm, $m = -2$ , showing exactly how each New Cartesian sign is placed so that the arithmetic classifies the image for you. This is the template every other single-lens problem is a copy of, so doing it once with the signs shown in full is the whole investment.

Step 1. Sign the data. The lens is converging, so $f = +20\ \mathrm{cm}$ . The object is real, on the incoming-light side, so $u = -30\ \mathrm{cm}$ . Nothing else is needed; the recipe supplies the rest.

Step 2. Solve for $v$ with $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} + \dfrac{1}{-30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$ , so $v = +60\ \mathrm{cm}$ . The plus sign is the physics: the image sits on the far side of the lens, where a screen would catch it.

Step 3. Read the magnification: $m = \dfrac{v}{u} = \dfrac{+60}{-30} = -2$ . The minus says inverted, the magnitude $2$ says twice the object's height. If the object is $4\ \mathrm{cm}$ tall, then $h' = m\,h = -2 \times 4 = -8\ \mathrm{cm}$ — an inverted image $8\ \mathrm{cm}$ high.

Step 4. State the nature. $v > 0$ : real, on the far side. $m < 0$ : inverted. $|m| > 1$ : magnified. So the answer is a real, inverted, magnified image, $60\ \mathrm{cm}$ beyond the lens and $8\ \mathrm{cm}$ tall — exactly what a projector does with a slide held just outside the focus. Every mark in the question is earned by naming those three properties, not by the number $60$ alone. Verified: $v = +60$ cm, $m = -2$ and $h' = -8$ cm for $f = +20$ , $u = -30$ are recomputed in verify_hs2-lens-numericals.py.

Go deeper · Derivation B · 1 minThe displacement method for the focal length

Derivation BThe displacement method for the focal length

The goal is to prove the displacement-method formula $f = (D^2 - d^2)/4D$ and use it on the locked anchor $D = 100\ \mathrm{cm}$ , $d = 40\ \mathrm{cm}$ to get $f = 21\ \mathrm{cm}$ , using nothing but the thin-lens formula and the symmetry of conjugate points — no calculus, only the fact that swapping object and image distances leaves the lens equation unchanged. This is the exam's standard route to a focal length when the object and image distances are deliberately hidden.

Step 1. Fix the geometry. The object and screen are a fixed distance $D$ apart, and the lens throws a sharp image for two positions. Because the thin-lens formula is symmetric under exchanging the roles of object and image, if one position has object distance $a$ and image distance $b$ (with $a + b = D$ ), the other sharp position simply has them swapped, object distance $b$ and image distance $a$ .

Step 2. Relate the shift $d$ to $a$ and $b$ . Moving the lens from the first position to the second shifts it by $d = |b - a|$ . Together with $a + b = D$ , adding and subtracting gives $a = (D - d)/2$ and $b = (D + d)/2$ .

Step 3. Feed these into the lens relation. In magnitude the thin-lens formula for a real object and real image reads $\dfrac{1}{f} = \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{a + b}{ab} = \dfrac{D}{ab}$ . Now $ab = \dfrac{(D - d)}{2}\cdot\dfrac{(D + d)}{2} = \dfrac{D^2 - d^2}{4}$ , so $\dfrac{1}{f} = \dfrac{D}{(D^2 - d^2)/4} = \dfrac{4D}{D^2 - d^2}$ , which inverts to $f = \dfrac{D^2 - d^2}{4D}$ .

Step 4. Put in the numbers: $f = \dfrac{100^2 - 40^2}{4 \times 100} = \dfrac{10000 - 1600}{400} = \dfrac{8400}{400} = 21\ \mathrm{cm}$ . Notice that the object distance and image distance never appear in the answer — you measured $D$ and $d$ with a ruler and got $f$ directly, which is exactly why examiners love this method and why it is the standard laboratory route to a lens's focal length. Verified: $f = 21$ cm for $D = 100$ , $d = 40$ , and the conjugate-symmetry algebra, are recomputed in verify_hs2-lens-numericals.py.

Variants and where they trap you

variant	how to handle it	the trap
converging lens, object beyond $F$	$1/v = 1/f + 1/u$ ; expect $v>0$ , $m<0$	forgetting to state real and inverted
converging lens, object inside $F$ (magnifier)	same formula; expect $v<0$ , $m>0$ , $\|m\|>1$	calling the virtual image real
diverging lens, any real object	same formula, $f<0$ ; always $v<0$ , $0<m<1$	expecting a real image from a diverging lens
two lenses in contact	add powers, $P = P_1 + P_2$	adding focal lengths instead of powers
two lenses separated by $d$	$P = P_1 + P_2 - dP_1P_2$	dropping the cross-term or mixing cm and m
object and screen, two sharp positions	$f = (D^2 - d^2)/4D$	using it when $D < 4f$ , where no two positions exist

Common trapA lens uses m = v/u, not the mirror's m = −v/u

The single most expensive habit in lens numericals is importing the mirror's magnification formula. A mirror uses $m = -v/u$ ; a lens uses $m = +v/u$ , with no leading minus, and the two conventions are built so that each still reports a real image as inverted and a virtual image as erect. Bring the mirror's minus sign into a lens problem and every orientation you state flips — you will call an inverted projector image erect and lose the nature marks even though your image distance was right. Two more slips ride alongside it. First, sign the focal length from the lens type: converging $f>0$ , diverging $f<0$ ; a diverging lens fed in with $f>0$ manufactures an impossible real image. Second, combine lenses by adding powers, never focal lengths: $f_1 = 20\ \mathrm{cm}$ and $f_2 = 50\ \mathrm{cm}$ in contact do not make a $70\ \mathrm{cm}$ lens; their powers $+5$ D and $+2$ D add to $+7$ D, an $F = 14.29\ \mathrm{cm}$ lens. Sign first, read the nature back, and add powers not lengths, and the whole family behaves.

Worked examples

1 · Board level — the projector image (CBSE and NEET pattern)

A converging lens of focal length $20\ \mathrm{cm}$ forms an image of a $4\ \mathrm{cm}$ tall object placed $30\ \mathrm{cm}$ in front of it. Find the image distance, the magnification, the image height and the full nature of the image.

Step 1. Sign the data: converging, so $f = +20\ \mathrm{cm}$ ; real object, so $u = -30\ \mathrm{cm}$ .

Step 2. $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} - \dfrac{1}{30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$ , so $v = \boxed{+60\ \mathrm{cm}}$ .

Step 3. $m = \dfrac{v}{u} = \dfrac{60}{-30} = \boxed{-2}$ , and $h' = m\,h = -2 \times 4 = \boxed{-8\ \mathrm{cm}}$ .

Step 4. Nature: $v>0$ real and on the far side, $m<0$ inverted, $|m|>1$ magnified — a real, inverted, magnified image, $8\ \mathrm{cm}$ tall, $60\ \mathrm{cm}$ past the lens. This is the projector geometry: a slide held just outside the focal length throws an enlarged inverted picture onto the far wall.

2 · Board level — the magnifying glass (CBSE and ICSE pattern)

The same lens ( $f = 20\ \mathrm{cm}$ ) now has the object placed only $15\ \mathrm{cm}$ in front — inside the focus. Find the image and its nature.

Step 1. Signs: $f = +20\ \mathrm{cm}$ , $u = -15\ \mathrm{cm}$ (still a real object, now inside $F$ ).

Step 2. $\dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{15} = \dfrac{3 - 4}{60} = -\dfrac{1}{60}$ , so $v = \boxed{-60\ \mathrm{cm}}$ .

Step 3. $m = \dfrac{v}{u} = \dfrac{-60}{-15} = \boxed{+4}$ .

Step 4. Nature: $v<0$ virtual and on the same side as the object, $m>0$ erect, $|m|>1$ magnified — a virtual, erect, magnified image four times the object's size. This is the magnifying glass: bring the object inside the focal length and a converging lens makes a large upright image you view through the lens. Notice how the single sign of $v$ flipped the entire story as the object crossed the focus — the same lens, moved $15\ \mathrm{cm}$ closer, switched from projector to magnifier.

3 · Competitive level — two lenses in contact (JEE and NEET pattern)

A converging lens of power $+5\ \mathrm{D}$ is placed in contact with a converging lens of power $+2\ \mathrm{D}$ . Find the power and focal length of the combination. What if the second lens were diverging, $-2\ \mathrm{D}$ ?

Step 1. In contact, powers add: $P = P_1 + P_2 = (+5) + (+2) = \boxed{+7\ \mathrm{D}}$ .

Step 2. The equivalent focal length is $F = \dfrac{1}{P} = \dfrac{1}{7}\ \mathrm{m} = \dfrac{100}{7}\ \mathrm{cm} = \boxed{14.29\ \mathrm{cm}}$ — shorter than either component, because two converging lenses reinforce.

Step 3. Check against focal lengths: $f_1 = 1/5\ \mathrm{m} = 20\ \mathrm{cm}$ and $f_2 = 1/2\ \mathrm{m} = 50\ \mathrm{cm}$ ; a naive "add the focal lengths" would give a nonsensical $70\ \mathrm{cm}$ . The correct route, $1/F = 1/20 + 1/50 = (5 + 2)/100 = 7/100$ , gives $F = 14.29\ \mathrm{cm}$ , matching the power sum.

Step 4. If the second lens diverges, its power enters negative: $P = (+5) + (-2) = +3\ \mathrm{D}$ , so $F = 100/3 = 33.3\ \mathrm{cm}$ — a weaker converging system. The diverging lens partly undoes the converging one. Timing note: about thirty seconds when the powers are given in dioptres — just add them. The whole speed advantage of working in powers is that contact combinations become a single sum; convert any focal lengths to powers first ( $P = 100/f_{\mathrm{cm}}$ in dioptres) and never add focal lengths directly.

4 · Competitive level — the displacement method (JEE pattern)

An object and a screen are fixed $100\ \mathrm{cm}$ apart. A converging lens between them forms a sharp image on the screen at two positions, $40\ \mathrm{cm}$ apart. Find the focal length of the lens.

Step 1. Identify the method: object and screen fixed, two sharp-image positions — this is the displacement (Bessel) method, $f = (D^2 - d^2)/4D$ , with $D = 100\ \mathrm{cm}$ and $d = 40\ \mathrm{cm}$ .

Step 2. Compute the numerator: $D^2 - d^2 = 100^2 - 40^2 = 10000 - 1600 = 8400\ \mathrm{cm}^2$ .

Step 3. Divide by $4D$ : $f = \dfrac{8400}{4 \times 100} = \dfrac{8400}{400} = \boxed{21\ \mathrm{cm}}$ .

Step 4. Sanity check the feasibility: two positions exist only when $D \ge 4f$ , and here $4f = 84\ \mathrm{cm} \le 100\ \mathrm{cm}$ , so the two sharp images are genuine. Had the screen been closer than $4f = 84\ \mathrm{cm}$ , no sharp image would form at all, and the question would have no answer. Timing note: under a minute. The trap is misreading $d$ as the object-to-lens distance instead of the separation between the two lens positions; $D$ is object-to-screen, $d$ is position-to-position, and mixing them wrecks the formula.

Exam traps

Using the mirror's magnification for a lens. A lens is $m = +v/u$ ; the mirror is $m = -v/u$ . Import the minus sign and every orientation flips — the number-one lost mark on lens numericals across CBSE, ICSE, JEE and NEET.

Wrong sign on the focal length. Converging $f>0$ , diverging $f<0$ , read from the lens type before any arithmetic. A diverging lens entered with $f>0$ returns an impossible real image of a real object.

Adding focal lengths instead of powers. Lenses in contact add powers: $P = P_1 + P_2$ . Two lenses of $20$ and $50\ \mathrm{cm}$ make a $14.29\ \mathrm{cm}$ combination, not $70\ \mathrm{cm}$ . Convert to dioptres ( $P = 100/f_{\mathrm{cm}}$ ) and sum.

Not stating the nature. Boards award explicit marks for "real or virtual, erect or inverted, magnified or diminished". Read it off the signs — $v$ for the side, $m$ for orientation and size — and never leave a bare number.

Centimetres in a power. Power is $P = 1/f$ with $f$ in metres: $f = 20\ \mathrm{cm} = 0.20\ \mathrm{m}$ gives $P = +5\ \mathrm{D}$ , not $+0.05$ . Convert to metres before taking the reciprocal.

Board scope. These lens numericals are set by CBSE, ICSE, JEE-Main, JEE-Advanced, NEET and AP Physics 2 (Unit 13, which does the thin-lens equation as a calculation). IGCSE 0625 treats a converging lens only qualitatively, with $m = \text{image height}/\text{object height}$ , and does not set the $1/v - 1/u = 1/f$ lens-formula or combination numericals — so if you are an IGCSE candidate, read this for the ideas, not the drill. IB SL/HL and A-Level 9702 carry no geometric-optics lens content at all; the whole page is enrichment for them. One caution for AP students: many American texts use a "real-is-positive" sign convention rather than New Cartesian — the discipline is identical, but apply whichever single convention your board uses without a mid-problem switch.

Origins · 1 minCarl Friedrich Gauss and how this was discovered

Carl Friedrich Gauss

1777–1855 · Göttingen

Gauss is remembered as the "Prince of Mathematicians" — number theory, the normal distribution, the geometry of curved surfaces — and as the astronomer who recovered the lost asteroid Ceres from a handful of observations. His tie to this page is his Dioptrische Untersuchungen (Investigations in Optics, 1841), where he recast the entire behaviour of a lens or mirror into the paraxial framework these numericals run on. Gauss showed that any well-behaved optical system, however many surfaces it has, can be reduced to a few cardinal points and a single pair of focal lengths, so that image location becomes the plug-and-play arithmetic of $1/v - 1/u = 1/f$ . That is why the thin-lens picture you drill here is called Gaussian optics. Honest scope: Gauss was a mathematician and astronomer, not a lens-grinder — he handed physics the imaging framework, not the optical-bench drill; the sign machinery and the dioptre bookkeeping are later refinements built on his foundation.

History

1611

— Johannes Kepler, in Dioptrice, gives the first systematic account of image formation by converging and diverging lenses.

1637

— René Descartes, in La Dioptrique, publishes the sine law of refraction that underlies every lens calculation.

1840

— Friedrich Bessel devises the conjugate-position (displacement) method for measuring a lens's focal length from two lens positions at a fixed object–screen separation.

1841

— Carl Friedrich Gauss casts paraxial lens and mirror imaging in the cardinal-point framework that becomes Gaussian optics.

1875

— The dioptre is adopted as the practical unit of lens power, so opticians add powers rather than focal lengths.

Going further

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), the thin-lens formula, magnification and combinations · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" — lens problems and the displacement method · D. C. Pandey / Cengage optics problem sets for the JEE and NEET variant families · AP Physics 2 (College Board CED, Unit 13) thin-lens-equation calculations.

Provenance. Standard geometric optics. Recompute: Board 1 ( $f = +20$ , $u = -30$ ) $\to v = +60$ , $m = -2$ , $h' = -8$ cm; Board 2 ( $f = +20$ , $u = -15$ ) $\to v = -60$ , $m = +4$ ; contact combination $+5\ \mathrm{D} + 2\ \mathrm{D} = +7\ \mathrm{D}$ , $F = 100/7 = 14.29$ cm, and the diverging case $+5 - 2 = +3\ \mathrm{D}$ ; separated $P = 5 + 5 - 0.10\times25 = 7.5\ \mathrm{D}$ ; displacement $f = (100^2 - 40^2)/400 = 21$ cm with the feasibility $4f = 84 \le 100$ ; the diverging single lens $f = -20$ , $u = -30 \to v = -12$ , $m = +0.4$ .

Prerequisites. The Mirror Formula and Sign Convention — the sign machinery this page shares, applied to mirrors first · Mirror and Lens Sign-Convention Discipline — the two-ray craft and the one origin rule that keeps every sign honest. Leads to. Lens–Mirror Systems and the Silvered Lens, where a lens and a mirror act together; and the university seams below.

Go deeper (UG). The thin-lens formula and its combinations are the paraxial corner of the full Thin Lens and Lensmaker's Equation and Image Formation and Magnification treatments in the university optics track, where finite thickness, aberrations and the cardinal points enter. Those are enrichment only; every result here follows from $1/v - 1/u = 1/f$ , the sign convention and adding powers, and no exam on this page asks for the matrix or aberration methods.

Why does a lens use m = v/u while a mirror uses m = −v/u?

Because the two conventions are set up on opposite sides of the element, and each is written so that the sign of $m$ still reports orientation correctly. For a mirror, a real image forms on the same side as the object (both distances negative), so the formula needs a minus, $m = -v/u$ , to come out negative and flag "inverted". For a lens, a real image forms on the far side (positive $v$ ) while the real object is negative, so $v/u$ is already negative on its own — no extra minus is needed, and $m = +v/u$ . In both cases a negative $m$ means inverted and a positive $m$ means erect; the two formulas just achieve that with the geometry each element actually has. Learn them as a matched pair and never carry one into the other's problem.

How do I know if the image is real or virtual without drawing the diagram?

Read the sign of $v$ after you solve for it. For a lens in the New Cartesian convention, a positive image distance ( $v>0$ ) means the image is on the far side of the lens, where the refracted rays actually cross — it is real and can be caught on a screen. A negative image distance ( $v<0$ ) means the image is on the same side as the object, where the rays only appear to meet — it is virtual. A converging lens gives a real image for objects beyond the focus and a virtual one for objects inside it; a diverging lens gives a virtual image for every real object. So you never need the ray diagram to classify — solve for $v$ , look at its sign, and let $m$ tell you erect or inverted.

When do I add powers and when do I add focal lengths?

You add powers, never focal lengths, for lenses in contact: $P = P_1 + P_2$ . Adding focal lengths is simply wrong — two lenses of $20\ \mathrm{cm}$ and $50\ \mathrm{cm}$ do not make a $70\ \mathrm{cm}$ lens; their powers $+5\ \mathrm{D}$ and $+2\ \mathrm{D}$ add to $+7\ \mathrm{D}$ , an $F = 14.29\ \mathrm{cm}$ combination. The safe routine is to convert every focal length to a power first with $P = 100/f_{\mathrm{cm}}$ (or $1/f_{\mathrm{m}}$ ), sum the powers with their signs (diverging lenses negative), and convert back to a focal length only at the end. If the lenses are separated by a distance $d$ , use $P = P_1 + P_2 - dP_1P_2$ with $d$ in metres.

What is a dioptre, and why do opticians use it?

A dioptre (D) is the unit of lens power, defined as $P = 1/f$ with the focal length in metres. A $+2\ \mathrm{D}$ lens has $f = 0.5\ \mathrm{m}$ ; a $+5\ \mathrm{D}$ lens has $f = 0.2\ \mathrm{m}$ ; a stronger lens has a shorter focal length and a bigger number. Opticians work in dioptres because lenses placed together add their powers directly — the whole prescription is just a sum — so combining a distance correction with a reading add is a single addition rather than an awkward reciprocal calculation. Converging lenses take a positive power, diverging lenses negative. The one slip to avoid is putting the focal length in centimetres: $20\ \mathrm{cm}$ is $0.20\ \mathrm{m}$ , so its power is $+5\ \mathrm{D}$ , not $+0.05\ \mathrm{D}$ .

Why does the displacement method not need the object distance?

Because it exploits a symmetry that hides $u$ and $v$ entirely. The thin-lens formula is unchanged if you swap the object and image distances, so whenever a sharp image can be formed with the lens at one place, a second sharp image can be formed with the lens moved to another place where the two distances are swapped. All that separates the two positions is a measurable shift $d$ , and the object-to-screen distance $D$ is fixed. Algebra then gives $f = (D^2 - d^2)/4D$ purely in terms of $D$ and $d$ — both read off a ruler. You never measure how far the object is from the lens, which is exactly why the method is a laboratory standard: it turns a focal-length measurement into two length readings and one formula.

Can a diverging lens ever form a real image of a real object?

No — never. A diverging lens spreads every ray that passes through it, so the refracted rays can only be traced backwards to an apparent meeting point on the same side as the object. That makes the image virtual, erect and diminished for any real object, wherever you place it: with $f = -20\ \mathrm{cm}$ and $u = -30\ \mathrm{cm}$ the formula gives $v = -12\ \mathrm{cm}$ and $m = +0.4$ , a small upright image on the near side. The only way to get a real image out of a diverging lens is to feed it converging light (a "virtual object"), which happens inside multi-lens systems, not with an ordinary lit object. So if your arithmetic hands you a real image from a diverging lens and a real object, you have made a sign error, not a discovery.

The object is exactly at the focus of a converging lens. What happens?

The formula gives $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{f} - \dfrac{1}{f} = 0$ , so $v \to \infty$ : no image forms at any finite distance. Physically, a point source placed at the focus of a converging lens emerges as a beam of parallel rays that never reconverge, so there is nothing to catch on a screen and nothing to view. This is the boundary between the real-image regime (object beyond the focus) and the virtual, magnifier regime (object inside the focus), and it is exactly how a spotlight or a collimator works — put the source at the focus so the light leaves parallel. In an exam, an object "at the focus" is the signal that the image is at infinity, not an error to fix.