Converging vs Diverging Lens: The Difference

The distinction, stated once

AP Physics 2 gives both lenses in the same sentence shape, one after the other, so read them side by side and watch which words change.

Converging. Essential knowledge 13.4.A.1: incident light rays parallel to the principal axis of a thin convex (converging) lens will be refracted and converge toward a common location on the transmitted side of the lens, called the focal point.

Diverging. Essential knowledge 13.4.A.2: incident light rays parallel to the principal axis of a thin concave (diverging) lens will be refracted and diverge as if they originated from a focal point on the incident side of the lens.

Three differences are carried in those two sentences and everything else on this page is a consequence of them.

Converge against diverge. One lens gathers a parallel bundle; the other spreads it.

Transmitted side against incident side. The converging lens's focal point is on the far side, downstream of the light. The diverging lens's is on the near side, upstream, where the light came from.

"Converge toward" against "diverge as if they originated from". That is the CED choosing its words carefully. A converging lens's focal point is a place light actually arrives at. A diverging lens's is a place no light ever reaches; the rays merely point back at it.

The third difference is the one with consequences. Real light crossing at a point is what makes a real image possible, so a converging lens can produce one and a diverging lens cannot. The sign convention then encodes the same fact numerically: a focal point on the transmitted side gives $f > 0$, one on the incident side gives $f < 0$. From there, one equation does the rest.

Note that the CED supplies both names for both lenses, in parentheses, so you are never expected to infer converging or diverging from the word convex or concave. That is worth holding on to, because those two words mean the opposite thing for mirrors, and the confusion is covered below.

The sign convention used on this page, declared before any number

The CED requires a sign convention and does not print one. Essential knowledge 13.4.A.5.i says only that the locations of a lens's focal point, an object, and the image of the object formed by the lens follow sign conventions that are used to determine those locations relative to the lens itself. The AP Physics 2 equation sheet then prints magnification wrapped in absolute value bars,

so the printed relation gives sizes and not orientations. Everything with a sign has to come from a convention you state. Here is the one this page uses, and it does not change between sections.

• Distances are measured from the midline of the lens, which is what 13.4.A.5 specifies for the object distance.
• Object distance $s_o > 0$ for a real object. Every object on this page is real, so $s_o$ is positive throughout.
• Image distance $s_i > 0$ when the image forms on the transmitted side, where the light goes. $s_i < 0$ when it forms on the incident side. Positive means real, negative means virtual.
• Focal length $f > 0$ for a converging lens, whose focal point is on the transmitted side per 13.4.A.1. $f < 0$ for a diverging lens, whose focal point is on the incident side per 13.4.A.2.
• Magnification $M = -s_i/s_o$. Positive means upright, negative means inverted, and $\lvert M \rvert > 1$ means enlarged. The AP Physics 2 sheet does not print this signed form, so quote $\lvert M \rvert$ for size on the exam and take orientation from the sign of $s_i$.

The sign of $f$ is the only input that distinguishes the two lenses. Fix it and the printed relation from 13.4.A.5 does everything else:

Rearranged for calculation, $\dfrac{1}{s_i} = \dfrac{1}{f} - \dfrac{1}{s_o}$. Put $f$ in with its sign, read the sign of the answer, classify the image. That is the whole method and this page uses no other.

Two features of lenses that mirrors do not share are worth noting while the convention is fresh. 13.4.A.5.ii says lenses have a focal point on both sides of the lens that depends on the shape of the respective side of the lens, so a lens has two focal points where a mirror has one. And no equation relating focal length to any radius of curvature appears anywhere in AP Physics 2 for lenses: the CED gives the halfway approximation only for a spherical mirror, and no lensmaker's equation appears in the framework or on the sheet. A lens focal length is measured or given, never computed from the glass. Concave against convex mirror works the mirror side of the same convention.

Side by side

The row that decides how much work a question needs is does the image type change with $s_o$. For a diverging lens the answer is fixed before you compute anything: virtual, upright, reduced. For a converging lens you have to compare $s_o$ with $f$ first, and only then does the classification follow.

The row worth arguing with is can it burn paper in sunlight, because it is the same fact as "does light reach the focal point" in a form you can test outdoors. A converging lens concentrates sunlight at a point real light passes through. A diverging lens has no such point available to it at any distance.

One row deliberately does not appear: which lens is stronger, or which has the shorter focal length. Neither statement is meaningful, because the sign is a direction and the magnitude is independent of it. A diverging lens of $\lvert f \rvert = 5 \ \mathrm{cm}$ bends light more sharply than a converging lens of $f = 50 \ \mathrm{cm}$, in the other direction.

The case that separates them: the same focal length magnitude, five object distances

Take a converging lens with $f = +20 \ \mathrm{cm}$ and a diverging lens with $f = -20 \ \mathrm{cm}$, so the only difference in every calculation is one minus sign. Put a $3.0 \ \mathrm{cm}$ object at $60$, $40$, $30$, $20$ and $10 \ \mathrm{cm}$ in front of each, using $\dfrac{1}{s_i} = \dfrac{1}{f} - \dfrac{1}{s_o}$ and $M = -s_i/s_o$ throughout.

Converging lens, $f = +20 \ \mathrm{cm}$:

Diverging lens, $f = -20 \ \mathrm{cm}$:

Look at the two right-hand columns. The converging lens's image column changes three times over five rows, including one row where no image forms at all. The diverging lens's image column says the same three words five times, and the only thing the object distance changes is how small the image is and how close behind the lens it sits.

The CED's own vocabulary anticipates this split. 13.4.A.7.ii says images formed by a lens can be upright or inverted, virtual or real, and reduced, enlarged, or the same size as the object, which is eight combinations on paper. The converging lens supplies most of them. The diverging lens supplies exactly one.

A note on the row where nothing happens. With the object at the focal point, $\dfrac{1}{s_i} = \dfrac{1}{20} - \dfrac{1}{20} = 0$, so $s_i$ is infinite and no image forms. The refracted rays leave the lens parallel to each other and never meet, on either side. That is not a gap in the table, it is the boundary between the converging lens's two regimes, and it is why the classification of a converging lens question always starts by comparing $s_o$ with $f$.

Why a diverging lens has no cases at all

"Always" is a word to distrust in physics, and here it survives a proof, so it is safe to use. Three claims, three lines of algebra, no cases.

Start from the printed relation rearranged, with $f$ negative. Write $f = -\lvert f \rvert$:

Claim one: the image is always virtual. For any real object, $s_o > 0$, so both terms in the bracket are positive and the bracket is positive. The whole expression is therefore negative, so $1/s_i < 0$ and $s_i < 0$ for every positive $s_o$. A negative image distance is a virtual image under the convention declared above, so a diverging lens can never form a real image from a real object.

Claim two: the image is always upright. With $s_i < 0$ and $s_o > 0$, $M = -s_i/s_o$ is a negative divided by a positive, negated, so $M > 0$ always. Positive magnification is an upright image.

Claim three: the image is always reduced. Take magnitudes in the same expression. Since $\left\lvert \dfrac{1}{s_i} \right\rvert = \dfrac{1}{\lvert f \rvert} + \dfrac{1}{s_o}$, which is strictly larger than $\dfrac{1}{s_o}$, inverting reverses the inequality and gives $\lvert s_i \rvert < s_o$. Therefore $\lvert M \rvert = \lvert s_i \rvert / s_o < 1$ for every object distance.

Two bounds fall out of the same line, and they are worth having because they let you check a numerical answer without redoing it.

$\left\lvert \dfrac{1}{s_i} \right\rvert$ is also strictly larger than $\dfrac{1}{\lvert f \rvert}$, so $\lvert s_i \rvert < \lvert f \rvert$. The virtual image of a diverging lens always sits between the lens and the near-side focal point. In the table above, every image distance is smaller in magnitude than $20 \ \mathrm{cm}$, and the largest, $15 \ \mathrm{cm}$, belongs to the most distant object.

And $\lvert M \rvert$ runs between $0$ and $1$ but reaches neither: it approaches $1$ as the object comes right up to the lens and approaches $0$ as the object goes to infinity. So a diverging lens can make an image very slightly smaller or almost vanishingly small, and nothing else.

This is why a diverging lens is the easy half of a lens question. Type, orientation and relative size are all answerable before you touch a calculator, and the calculator only supplies the two numbers.

Ray diagrams, and the three rays the CED names

Essential knowledge 13.4.A.7 says ray diagrams can be used to determine the location, type, size, and orientation of images formed by lenses, and 13.4.A.7.i names the three principal rays that are typically used: the ray parallel to the principal axis, the ray that passes through the centre of the lens where the principal axis intersects the lens, and the ray that passes through the focal point of the lens.

Those three descriptions are the same for both lenses. What differs is what each ray does when it gets there.

The middle row is the one to draw first, every time. It is identical for both lenses and for every object distance, so it costs nothing and it fixes one line of the diagram immediately.

The first row is where the sign of $f$ shows up geometrically. On a converging lens the refracted ray really goes through the focal point. On a diverging lens the refracted ray goes away from the axis and you extend it backwards, as a dashed line, to the focal point on the incident side. That dashed extension is the whole meaning of 13.4.A.2's phrase "as if they originated from".

Because a diverging lens's rays never cross, the image is always found by extending refracted rays backwards, which is 13.4.A.4's definition of a virtual image: a virtual image is formed by a lens when refracted light rays diverge such that they appear to have originated from a common point. A converging lens with the object beyond the focal point has the rays genuinely crossing, which is 13.4.A.3's real image: light rays originating from a common point are refracted such that they intersect at another common point. Real against virtual image works that distinction across both mirrors and lenses.

One practical rule for drawing: dashed lines are for light that is not there. If your diverging-lens diagram has no dashed lines in it, the image has been put in the wrong place.

When it costs a mark

• Carrying the words convex and concave across from mirrors. A convex lens converges; a convex mirror diverges. One transmits light and the other sends it back, so the same surface shape has opposite effects. The CED sidesteps this by printing both names together every time, at 13.4.A.1 and 13.4.A.2 for lenses and 13.2.A.1 and 13.2.A.2 for mirrors. Use the words converging and diverging in your answer and the trap disappears.
• Entering $f$ as a positive number for a diverging lens. This is the single most common way a lens question goes wrong, because every subsequent step then works perfectly and produces a real image that does not exist. If your diverging lens has produced $s_i > 0$, the sign of $f$ went in wrong.
• Assuming a converging lens always gives a real image. Only when $s_o > f$. Inside the focal length it gives a virtual, upright, enlarged image, which is the magnifying glass, and a CED instructional activity has students find the focal length of a magnifying glass from exactly this behaviour.
• Expecting an image when the object sits at the focal point. $1/s_i = 0$ and no image forms, because the refracted rays leave parallel.
• Reading a negative magnification as "smaller". The sign is orientation and the magnitude is size. $M = -2.0$ is an inverted image twice as tall.
• Quoting a signed magnification as if the sheet printed it. It prints $\lvert M \rvert$ with bars on every term. State $\lvert M \rvert$ for the size and derive orientation from the sign of $s_i$, or declare the signed convention you are using first.
• Looking for a lensmaker's equation. There is none in AP Physics 2, on the sheet or in the framework, and no radius-of-curvature relation for lenses either. The halfway approximation between surface and centre of curvature is stated for spherical mirrors only. A lens focal length is given or measured.
• Forgetting that a lens has two focal points. 13.4.A.5.ii says lenses have a focal point on both sides that depends on the shape of the respective side. For the thin symmetric lenses this course uses, both sit $\lvert f \rvert$ from the midline, so one number describes the lens, but a diagram may label $F_1$ and $F_2$ and expect you to know why there are two.
• Measuring the object distance from a lens surface. 13.4.A.5 says the midline of the lens, which is what makes the lens thin in the first place. For mirrors, 13.2.A.7 says the reflecting surface. Different reference points for the same symbol.

When they behave alike, and why that lulls you

There is exactly one region of overlap and it is a trap by design, because it is the region where both lenses produce a virtual, upright image on the same side as the object.

Put the object $10 \ \mathrm{cm}$ from each of the two lenses in the tables above. The converging lens gives $s_i = -20 \ \mathrm{cm}$ and $M = +2.0$. The diverging lens gives $s_i = -6.7 \ \mathrm{cm}$ and $M = +0.67$. Both images are virtual, both are upright, both are on the incident side, and a student who has only checked type and orientation cannot tell which lens is in the holder.

The size settles it, and it settles it in one direction only. A diverging lens has $\lvert M \rvert < 1$ always, proved above. So an image that is virtual, upright and enlarged can only have come from a converging lens with the object inside its focal length. There is no ambiguity at all in that case, which makes it a favourite of examiners.

Virtual, upright and reduced is the genuinely ambiguous case, and it points to a diverging lens by default because a converging lens cannot produce it: inside the focal length a converging lens always gives $\lvert M \rvert > 1$. So even the overlap resolves once you look at the third property, which is why the case table has three columns rather than two.

The other lull is language, and it is the one that survives into the exam room. "Concave lens" and "concave mirror" are the two halves of the same word doing opposite jobs, and a question that says concave without saying which optic is asking you to notice. The CED never leaves the reader in that position, always writing convex (converging) or concave (diverging), but a textbook, a lab handout or a question stem may.

What the exam does with all of this. A sample multiple-choice question in the AP Physics 2 exam information section is built directly on the distinction: an object sits $40 \ \mathrm{cm}$ from a lens and an enlarged image appears on a screen on the far side, and the four options offer a diverging lens of focal length $30 \ \mathrm{cm}$, a diverging lens of $50 \ \mathrm{cm}$, a converging lens of $30 \ \mathrm{cm}$ and a converging lens of $50 \ \mathrm{cm}$. Worked example three settles it, and both diverging options fall on one line of reasoning.