Topic 13.4: Images Formed by Lenses

What Topic 13.4 requires

Topic 13.4 closes Unit 13, Geometric Optics, which the CED weights at 12-15% of the multiple-choice section and estimates at about 8 to 12 class periods. It is the refraction counterpart of Topic 13.2: same equation, same magnification relation, different geometry.

13.4.A, describe the image formed by a lens. Seven essential knowledge statements sit under it.

• 13.4.A.1 Incident rays parallel to the principal axis of a thin convex (converging) lens are refracted and converge toward a common location on the transmitted side of the lens, the focal point.
• 13.4.A.2 Incident rays parallel to the principal axis of a thin concave (diverging) lens are refracted and diverge as if they originated from a focal point on the incident side of the lens.
• 13.4.A.3 A real image is formed when rays originating from a common point are refracted such that they intersect at another common point.
• 13.4.A.4 A virtual image is formed when refracted rays diverge such that they appear to have originated from a common point.
• 13.4.A.5 For a thin lens, the image location depends on the focal length and on the distance between the object and the midline of the lens, as given by the thin-lens equation $\dfrac{1}{s_i} + \dfrac{1}{s_o} = \dfrac{1}{f}$. 13.4.A.5.i adds that the focal point, object and image "follow sign conventions that are used to determine those locations relative to the lens itself"; 13.4.A.5.ii that lenses have a focal point on both sides, depending on the shape of the respective side.
• 13.4.A.6 For a thin lens, magnification is the ratio of image size to object size, with relevant equation $|M| = \left| \dfrac{h_i}{h_o} \right| = \left| \dfrac{s_i}{s_o} \right|$.
• 13.4.A.7 Ray diagrams determine the location, type, size and orientation of images formed by lenses. 13.4.A.7.i names the three principal rays; 13.4.A.7.ii states that images formed by a lens can be upright or inverted, virtual or real, and reduced, enlarged, or the same size as the object.

Two structural details matter. First, 13.4 carries no boundary statement. Of the four topics in Unit 13, only 13.2 does. What limits this one is the phrase "thin lens", written into 13.4.A.1, 13.4.A.2, 13.4.A.5 and 13.4.A.6 rather than fenced off separately. A thin lens is one whose thickness you can ignore, so all the bending happens at one plane through its midline, which is why 13.4.A.5 measures the object distance to that midline and not to a surface.

Second, the CED lists five suggested skills here, one more than for 13.2: 1.C create qualitative sketches of graphs that represent features of a model or the behavior of a physical system; 2.B calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.D predict new values or factors of change of physical quantities using functional dependence between variables; 3.A create experimental procedures that are appropriate for a given scientific question; and 3.B apply an appropriate law, definition, theoretical relationship, or model to make a claim.

That list is more quantitative and more experimental than the mirror topic's: 2.B says you will be asked for numbers, 1.C and 2.D for a graph, 3.A for a measurement design. Three of the CED's five sample activities for the unit belong to 13.4, one of them finding a magnifying glass's focal length from a graph.

The sign convention, declared before anything else

Signs are the whole difficulty of this topic, so here is the convention this page uses, stated once and held to the end. A lens transmits light: it arrives on one side and leaves by the other. Call the side it comes from the incident side and the side it leaves by the transmitted side.

The rule underneath is the one that governs mirrors too: positive means the side real light actually reaches after the optic has acted on it. For a mirror that is back toward the object, because light bounces; for a lens it is straight on through. The geometry inverts between the two topics even though the principle does not.

The consequences then follow from the arithmetic rather than being asserted alongside it.

• If the algebra returns $s_i > 0$, refracted rays are genuinely crossing on the far side of the lens. That is the definition of a real image in 13.4.A.3, and a screen placed there catches it.
• If the algebra returns $s_i < 0$, the meeting point is back on the incident side. Refracted rays are diverging and only their backward extensions meet, which is 13.4.A.4 word for word. Nothing lands on a screen there.

The reason to trust this convention rather than one carried in from a textbook: the CED works an example with it. The Instructional Approaches section prints a thin-lens case study, a converging lens with an object at a point P on its left, and asks a chain of questions about that one setup.

• The calculation question gives $f = 0.40 \, \mathrm{m}$ with the object $0.30 \, \mathrm{m}$ to the left, and marks "1.2 m to the left of the lens" correct. Light runs left to right, so left is the incident side, and $\dfrac{1}{s_i} = \dfrac{1}{0.40} - \dfrac{1}{0.30} = -\dfrac{5}{6}$ per metre gives $s_i = -1.2 \, \mathrm{m}$. A negative image distance put the image back on the incident side, exactly as the table says.
• The claim question asks what characterises that image, and marks "Virtual, upright, and larger than the object". So a negative image distance is the virtual one, and it is upright.

That is the convention pinned by the CED's own answer key rather than by assertion. Note what the CED does not supply: 13.4.A.5.i says only that these locations "follow sign conventions", printing no table, and none of the nine bullets in the exam's appendix conventions list fixes an optics sign either. The nearest hint on the reference sheet is its symbol list, which defines $s$ as position rather than distance. A position is a signed coordinate.

Convex converges, concave diverges, and a lens has two focal points

The naming is the trap, and it runs the opposite way from mirrors. The CED writes both words every time: "thin convex (converging) lens" and "thin concave (diverging) lens". Set that beside the mirror topic's "concave (converging) mirror" and "convex (diverging) mirror" and the crossover is plain.

The words concave and convex describe geometry, not behaviour. What fixes the behaviour is whether light bounces off the surface or passes through it. Memorise the behaviour, not the word.

Convex lens, $f > 0$. Essential knowledge 13.4.A.1 puts the focal point "on the transmitted side", which the section above made the positive side. Real light converges there: the bright dot a magnifying glass makes in sunlight is that focal point, which is why it can burn paper.

Concave lens, $f < 0$. Essential knowledge 13.4.A.2 says the refracted rays "diverge as if they originated from a focal point on the incident side", the negative side, and "as if" is the CED telling you nothing is really there. A concave lens cannot burn anything.

Both lenses have two focal points. Essential knowledge 13.4.A.5.ii states that lenses have a focal point on both sides, depending on the shape of the respective side, which is why the CED's case study figure labels two points $F_1$ and $F_2$. For the thin symmetric lenses this topic uses, both sit the same distance $|f|$ from the midline, so one focal length describes the lens. It matters in ray diagrams, where the parallel ray needs the far focal point and the focal ray the near one.

One more contrast with the mirror topic, where $|f| = R/2$ tied focal length to curvature in essential knowledge 13.2.A.4. Nothing equivalent appears for lenses: no relationship to surface curvature, no lensmaker's equation, and nothing on the reference sheet. A lens's focal length is something a question tells you or something you measure.

Real and virtual, inverted and upright: how the signs produce the words

The CED's two definitions differ by one word, as they did for mirrors. 13.4.A.3: rays from a common point are refracted so that they intersect at another common point. 13.4.A.4: refracted rays diverge such that they appear to have originated from a common point.

Orientation is where students guess. Essential knowledge 13.4.A.6 prints magnification as

and those absolute value bars are on the reference sheet too, not added here for safety. Both ratios are wrapped, so the printed relation returns a magnitude only: enlarged, reduced, or the same size. It cannot answer upright or inverted, because the sign that would carry that has already been discarded.

The CED's case study makes the same point from the other direction. Its derivation question asks for "an expression for the magnification of the image" and marks $\dfrac{f}{d_o - f}$, which is exactly the ratio $s_i/s_o$ with no minus sign attached. Take its magnitude and you have the size; do not read its sign as orientation.

Many textbooks define a signed $M = -s_i/s_o$ and read the minus as "inverted". The AP Physics 2 sheet does not print that version, so writing it means defending a convention your reference sheet did not give you.

Orientation comes from the routes the CED does provide. First the diagram: 13.4.A.7 says ray diagrams determine "the location, type, size, and orientation" of the image. Second, a pairing that falls out of the geometry for a single thin lens with a real object:

• A real image ($s_i > 0$) is inverted.
• A virtual image ($s_i < 0$) is upright.

The CED's answer key gives one instance of the second half: the case-study image with $s_i = -1.2 \, \mathrm{m}$ is marked "Virtual, upright". One ray shows why it holds generally. Take the ray through the centre of the lens, which passes straight through undeviated. If the image forms on the transmitted side, that ray crossed the axis at the lens and arrives on the opposite side from the object: inverted. If the rays instead diverge and you run them backwards, the centre ray's extension stays on the object's side: upright.

So every lens question has a fixed reading order: get $f$ with its sign, solve for $s_i$, read the sign for real or virtual, convert to inverted or upright, then take $|M|$ for the size.

The three principal rays, exactly as the CED names them

Essential knowledge 13.4.A.7.i names them: "The three principal rays are typically used to find the images formed by lenses. The principal rays are 1) the ray parallel to the principal axis, 2) the ray that passes through the center of the lens where the principal axis intersects the lens, and 3) the ray that passes through the focal point of the lens."

Set that against the mirror version in 13.2.A.9.i and one word changes everything. For a mirror, ray 2 "reflects at" the centre of the mirror; for a lens it "passes through" the centre. The mirror's middle ray turns around, the lens's carries straight on, and that is the ray to draw first every time because it needs no construction.

Start every ray at the tip of the object.

Convex (converging) lens.

1. Parallel ray: travels parallel to the axis, refracts through the far focal point. This is 13.4.A.1 used directly.
2. Centre ray: aims at the point where the axis meets the lens and continues undeviated, because a thin lens has parallel faces at its midline.
3. Focal ray: passes through the near focal point on the way in and emerges parallel to the axis. This is ray 1 reversed.

Concave (diverging) lens. The same three, with the focal points working in reverse because there is nothing real to pass through: the parallel ray refracts away from the axis so its backward extension passes through the focal point on the incident side (13.4.A.2 word for word), the centre ray is undeviated as before, and the focal ray heads toward the far-side focal point and emerges parallel to the axis.

Two rays locate an image, so the third is a check, and a mistake shows up as three lines failing to meet at one point. When the refracted rays diverge, extend them backwards as dashed lines and mark the image where those meet. Dashed for virtual, solid for real.

For a concave lens the diagram comes out the same way whatever you do with the object: rays diverge, the dashed extensions meet on the incident side between the lens and its focal point, and the image is small and upright. That is a peephole.

Every case, in one table

Throughout, $f$ is the signed focal length from above and the object is real and on the incident side ($s_o > 0$).

Four readings of that table are worth having.

Only the converging lens has a story. It runs through five behaviours as the object walks in, and the switch happens at the focal point: outside $f$ real and inverted, inside $f$ virtual and upright. At the switch, $s_o = f$, we get $1/s_i = 1/f - 1/f = 0$ and the image distance runs to infinity, because the emergent rays leave parallel and never meet. That is a collimator, the same trick as a bulb at a headlamp's focal point.

The fifth row is the magnifying glass, and it is the CED case study's setup. A converging lens with the object closer than the focal length gives a virtual, upright, enlarged image on the object's side. Hold a magnifier too far from the page and the image flips and goes real; that flip is the object crossing $f$.

The diverging lens has no story at all. One line proves it: with $f$ negative, $1/s_i = 1/f - 1/s_o$ is a negative number minus a positive one, so it is negative for every positive $s_o$. Virtual and upright, always. A negative $f$ also makes $\lvert 1/s_i \rvert$ larger than $1/s_o$, so $\lvert s_i \rvert < s_o$ and $|M| < 1$.

Compare this table with the mirror one and the rows line up. A convex lens behaves like a concave mirror and a concave lens like a convex mirror, because converging is converging whichever way the light goes. Only the side that "positive" points to has changed, so learning one table properly gives you both.

Reading the graph: how a lens experiment finds f

Two of the five suggested skills here are graph skills (1.C sketch a qualitative graph, 2.D predict values using functional dependence) and a third (3.A) is designing the procedure. The CED's sample activity for 13.4 has students find a magnifying glass's focal length from several pairs of object and image distances, so this is worth doing once properly.

The thin-lens equation is not linear in $s_i$ and $s_o$, but it is linear in their reciprocals. Rearranged with $1/s_i$ as the dependent variable:

Compare that with $y = mx + b$. Plot $1/s_i$ on the vertical axis against $1/s_o$ on the horizontal axis and you get a straight line with

• slope $= -1$, whatever the lens, and
• vertical intercept $= 1/f$.

The CED's case study asks exactly this and marks the option identifying the focal length with the vertical intercept. Be careful with the last step: the intercept you read off is $1/f$, so the focal length is its reciprocal. An intercept of $2.5 \, \mathrm{m^{-1}}$ means $f = 0.40 \, \mathrm{m}$.

Two further features fall out of the same line.

• The horizontal intercept is also $1/f$. Setting $1/s_i = 0$ gives $1/s_o = 1/f$, so the line cuts both axes at the same value, and that symmetry is the reversibility of object and image showing up in a graph.
• The slope being fixed at $-1$ is a check on your data, not a result. A best-fit slope well away from $-1$ means something is wrong with the measurements or with a sign, not with the lens.

For a converging lens, real-image data ($s_i > 0$) sits where both reciprocals are positive and virtual-image points fall below the horizontal axis. Keeping the object beyond the focal point keeps every point in the positive quadrant, which is worth saying if you are asked to design the experiment under skill 3.A: a real image can be caught on a screen and measured directly, while a virtual image gives you nothing to put a screen against.

Traps, and how 13.4 gets tested

Pair each suggested skill with the mistake it catches.

2.B, calculate. The trap is arithmetic on reciprocals. From $\dfrac{1}{s_i} = \dfrac{1}{f} - \dfrac{1}{s_o}$ you must combine the fractions before inverting, giving $s_i = \dfrac{f s_o}{s_o - f}$. Inverting term by term, turning $1/s_i = 1/f - 1/s_o$ into $s_i = f - s_o$, is wrong and easy to do at speed, and finishing at $1/s_i$ without taking the reciprocal costs the whole answer.

2.D, functional dependence. What happens to the image as something changes, without numbers. The case table beats recomputing, and so does the reciprocal graph: with slope fixed at $-1$, raising $1/s_o$ by an amount lowers $1/s_i$ by the same amount.

1.C, sketch a graph. Sketch the reciprocal plot, not the raw one. $s_i$ against $s_o$ is a hyperbola with an asymptote at $s_o = f$; $1/s_i$ against $1/s_o$ is a straight line of slope $-1$.

3.A, design a procedure. Say what you measure, how often, and what you plot: several pairs of object and image distances, plotted as reciprocals, focal length from the intercept.

3.B, make a claim. These ask you to characterise an image as real or virtual, upright or inverted, larger or smaller, all at once. Answer in that order and each part follows from the last.

Four errors are worth naming directly.

• Importing the mirror's geometry. For a mirror a positive image distance puts the image on the object's side; for a lens it puts it on the far side. "Positive means real" survives the crossing from 13.2; the side does not.
• Reading "convex" as diverging. True for a mirror, false for a lens. The CED writes "thin convex (converging) lens", so take the pairing from the CED, not the shape word.
• Losing the sign on $f$ for a diverging lens. It turns a virtual image into a real one and takes every later answer with it. Attach the sign when you read the word concave or diverging.
• Treating $|M| > 1$ as proof of a real image. A converging lens gives an enlarged real image with the object between $f$ and $2f$, and an enlarged virtual one with it inside $f$. Only the sign of $s_i$ separates them.

For the mechanism underneath all of this, Snell's law and the index of refraction, see Topic 13.3. For the reflection counterpart, same equation and reversed geometry, see Topic 13.2. The Unit 13 overview puts the four topics in order, and the AP Physics 2 equation sheet shows what you work from: the thin-lens equation and the magnification relation are two of the fifteen entries in its Waves, Sound, and Optics group, and the only two this topic uses.