Real vs Virtual Image: What Is the Difference?

The distinction, stated once

AP Physics 2 defines both kinds twice, once for mirrors and once for lenses, and the four statements say the same thing in two settings.

Real. Essential knowledge 13.2.A.5: a real image is formed by a mirror when light rays emanating from a common point are reflected and then intersect at a common point. Essential knowledge 13.4.A.3: a real image is formed by a lens when light rays originating from a common point are refracted such that they intersect at another common point.

Virtual. Essential knowledge 13.2.A.6: a virtual image is formed by a mirror when reflected light rays diverge such that they appear to have originated from a common point. Essential knowledge 13.4.A.4: a virtual image is formed by a lens when refracted light rays diverge such that they appear to have originated from a common point.

The operative words are intersect and appear to have originated. For a real image the rays are genuinely there, crossing at a place in space, and light energy actually arrives at that place. For a virtual image nothing arrives; the outgoing rays are spreading apart, and it is only their backward extensions, drawn on paper and not travelled by any light, that meet.

That gives you the physical test in one sentence. Put a screen where the image is supposed to be. A real image appears on it, because light is landing there. A virtual image does not, because no light is landing there. Your eye still sees a virtual image perfectly well, since a diverging bundle of rays is exactly what your eye receives from any ordinary object, and it cannot tell whether the bundle really started at that point or only seems to have.

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

This pair cannot be discussed numerically without a sign convention, and the CED is deliberate about that. Essential knowledge 13.2.A.7.i says the locations of a mirror's focal point, an object near the mirror, and the image of the object formed by the mirror follow sign conventions that are used to determine those locations relative to the mirror itself, and 13.4.A.5.i says the same for a lens. So the framework tells you a convention is required, and tells you it is anchored to the optical element, and then does not print one. The AP Physics 2 equation sheet reinforces the gap: it prints magnification only as a magnitude,

with absolute value bars on every term, so the printed relation carries no sign information at all.

So here is the convention this page uses, and it does not change anywhere below.

• Distances are measured from the optical element, from the reflecting surface for a mirror (13.2.A.7) and from the midline of a thin lens (13.4.A.5).
• Object distance $s_o > 0$ for a real object, one that light genuinely comes from. Every object in AP Physics 2 is real, so $s_o$ is positive on every line of this page.
• Image distance $s_i > 0$ when the image forms on the side the light leaves from: in front of a mirror, on the transmitted side of a lens. $s_i < 0$ when it forms on the other side: behind a mirror, on the incident side of a lens.
• Focal length $f > 0$ for a converging element, meaning a concave mirror or a convex lens. $f < 0$ for a diverging element, meaning a convex mirror or a concave lens.
• Magnification $M = -s_i/s_o$. Positive $M$ means upright, negative means inverted. $\lvert M \rvert$ greater than one means enlarged, less than one means reduced. The minus sign is part of the convention, not part of the printed equation, and $\lvert M \rvert$ computed this way always agrees with the sheet's magnitude relation.

With that in place the arithmetic does all the classification for you, through the one equation both topics share:

Solve for $s_i$ and read its sign. Positive means real, negative means virtual. Nothing else on this page is needed to tell them apart.

One practical note for the exam. Because the CED does not print a convention, the safest habit is to state yours in one line before you substitute, and then keep it. A convention that changes halfway through a multi-part question produces answers that are internally inconsistent, and that is worse than a convention a grader would have written differently.

Side by side

The orientation row carries a condition, and the condition is load-bearing. For a single mirror or single thin lens acting on a real object, $s_o$ is positive, so $M = -s_i/s_o$ takes its sign from $s_i$ alone: a real image with positive $s_i$ is inverted and a virtual image with negative $s_i$ is upright. That is a theorem about this convention and this setup, not a universal law about images, and it is stated here with its conditions attached rather than as a slogan. AP Physics 2 does not ask about multi-element systems, so within the course the row holds throughout.

The row about visibility is the one that produces the most confusion, so it is worth saying plainly. Both kinds of image can be seen. What only a real image can do is land on something.

The case that separates them: one mirror, two object distances

Take a single concave mirror, so $f = +20 \ \mathrm{cm}$ by the convention above, and put a $4.0 \ \mathrm{cm}$ tall object in front of it twice: once at $30 \ \mathrm{cm}$, outside the focal point, and once at $10 \ \mathrm{cm}$, inside it. Nothing about the mirror changes between the two. Only the object distance moves, and every property of the image flips.

Read the magnification row and the height row together. Both images are twice as tall, so $\lvert M \rvert = 2.0$ in both cases and the sheet's magnitude relation cannot distinguish them. The sign of $M$ can, and the sign came from the sign of $s_i$, which came from whether the object sat outside or inside the focal point. That is the whole mechanism.

The crossover is worth naming. At $s_o = f$ exactly, $1/s_i = 1/f - 1/f = 0$, so $s_i$ is unbounded and no image forms: the reflected rays leave parallel and never meet, in either direction. Objects further out than $f$ give real images; objects closer in than $f$ give virtual ones; the boundary between the two behaviours is the focal point itself, and at that one distance there is no image at all.

Which optical elements can produce which image

Only two of the five elements AP Physics 2 covers can make a real image, and the reason is structural rather than a fact to memorise. A real image needs the outgoing rays to converge. A diverging element never converges them, so it can never make a real image from a real object.

The CED's own descriptions make the converging and diverging split explicit. For mirrors, 13.2.A.1 says incident rays parallel to the principal axis of a concave, converging, mirror are reflected toward a common location called the focal point, while 13.2.A.2 says the same rays on a convex, diverging, mirror are reflected so that they appear to have originated from a common location behind the mirror. For lenses, 13.4.A.1 says parallel rays on a thin convex, converging, lens converge toward a common location on the transmitted side, while 13.4.A.2 says a thin concave, diverging, lens refracts them so they diverge as if they originated from a focal point on the incident side.

Notice how neatly the CED's own wording matches the convention declared above. The converging elements have their focal point on the side where a real image would form, so $f$ is positive. The diverging elements have theirs on the other side, so $f$ is negative. The signs are not arbitrary decorations; they encode which side the focal point is on.

The plane mirror is a good consistency check on all of it. Essential knowledge 13.2.A.3 says the focal point of a plane mirror is an infinite distance from the mirror, so $1/f = 0$ and the shared equation becomes $1/s_i = -1/s_o$, giving $s_i = -s_o$. Negative, so virtual; equal in magnitude, so the same distance behind the mirror as the object is in front, which is exactly what 13.2.A.7.ii states independently. And $M = -s_i/s_o = +1$, so upright and the same size. Three separate CED statements agreeing with one substitution is the sign that the convention is working.

For how the two curved mirrors differ in detail, see concave vs convex mirror. For the mechanism behind each family, mirrors reflect and lenses refract, see reflection vs refraction.

When it costs a mark

Not stating a convention, then flipping it. The CED requires a convention (13.2.A.7.i, 13.4.A.5.i) and prints none. If your $f$ is positive for a concave mirror in part (a) and negative in part (c), the two answers cannot both be right and a grader has no way to award either. Write the convention down once and keep it.

Reading the sheet's magnification as signed. The sheet prints $\lvert M \rvert = \lvert h_i/h_o \rvert = \lvert s_i/s_o \rvert$, with bars. It gives you the size ratio and says nothing about orientation. The orientation comes from the sign of $s_i$ through $M = -s_i/s_o$, or from your ray diagram.

Losing the minus sign in $M = -s_i/s_o$. This turns every inverted image upright and every upright one inverted, while leaving the magnitude correct, so the answer looks right and describes the opposite picture.

Reporting a virtual image as catchable on a screen. No light arrives at a virtual image's location, so no screen shows it. This is asked directly, and it is the single cleanest test of whether the definitions in 13.2.A.5 and 13.2.A.6 have landed.

Concluding that a virtual image cannot be seen. It can. Every mirror you have used showed you one.

Sign-slipping in $1/s_i + 1/s_o = 1/f$. The equation is a sum of reciprocals, so the algebra step is $1/s_i = 1/f - 1/s_o$, and then $s_i$ is the reciprocal of that result. Forgetting the final reciprocal produces a number in units of inverse centimetres that then gets reported as a distance.

Expecting a real image from a diverging element. A convex mirror or a concave lens acting on a real object always gives $s_i < 0$. If your working produces a positive $s_i$ for one of those, the sign of $f$ went in wrong.

Treating a real image as automatically enlarged. Real and enlarged are independent. A concave mirror with the object well beyond $2f$ gives a real image that is inverted and reduced, and a convex lens does the same, which is how a camera works. Orientation is set by the sign of $M$; size by its magnitude.

When they look alike, and why that lulls you

Three situations make the two kinds of image hard to tell apart, and each one hides a different half of the definition.

Both look like an object when you look at them. Your eye receives a diverging bundle of rays in both cases, so the visual experience is the same. Nothing you can see tells you whether light is actually crossing at that location. The screen test exists precisely because sight does not settle the question.

The magnitudes can be identical. The concave-mirror pair in the section above gave $\lvert M \rvert = 2.0$ for both the real and the virtual image, and the sheet's magnification relation returns the same number for both. Any question that gives you only a size ratio has not given you enough to classify the image.

The same equation produces both. $1/s_i + 1/s_o = 1/f$ does not branch. You substitute once and the sign of the answer does the classifying. This is convenient and it is also why the classification step gets skipped: the arithmetic finishes before you have said what kind of image you found.

The distinction turns on exactly three questions, and they are the three that keep appearing on exams.

1. What is the sign of $s_i$? Positive is real, negative is virtual, under the convention declared above and no other.
2. Would a screen there show anything? Only if light arrives, which is only for a real image.
3. Is the element converging, and is the object outside its focal point? Both conditions are needed for a real image, and failing either gives a virtual one.

If you can answer those three, the pair is finished.

Where this sits on the AP exam

Both kinds of image are defined twice in Unit 13, Geometric Optics, which the CED weights at 12 to 15 percent of the multiple-choice section over a suggested 8 to 12 class periods: once for mirrors in Topic 13.2 and once for lenses in Topic 13.4. Both topics also state, in identical words at 13.2.A.9.ii and 13.4.A.7.ii, that images can be upright or inverted, virtual or real, and reduced, enlarged, or the same size as the object. Those are three independent binary or ternary choices, and a complete answer names all three.

Ray diagrams are required, not optional. Essential knowledge 13.2.A.9 and 13.4.A.7 both say ray diagrams can be used to determine the location, type, size, and orientation of images, and both list three principal rays. For a mirror (13.2.A.9.i) they are the ray parallel to the principal axis, the ray that reflects at the center of the mirror where the principal axis intersects the mirror, and the ray that passes through the focal point of the mirror. For a lens (13.4.A.7.i) they are the ray parallel to the principal axis, the ray that passes through the center of the lens where the principal axis intersects the lens, and the ray that passes through the focal point of the lens. A diagram is also the fastest way to check a sign: if your algebra says virtual and your diagram shows rays crossing in front of the mirror, one of the two is wrong.

The suggested skills differ slightly between the two topics and the difference is informative. Topic 13.2 lists 1.A, create diagrams, tables, charts, or schematics; 2.A, derive a symbolic expression from known quantities; 2.C, compare physical quantities between two or more scenarios or at different times and locations in a single scenario; and 3.C, justify or support a claim using evidence. Topic 13.4 lists 1.C, create qualitative sketches of graphs; 2.B, calculate or estimate an unknown quantity with units; 2.D, predict new values or factors of change using functional dependence; 3.A, create experimental procedures; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Skill 2.C on the mirror topic is why the two-object-distance comparison above is the shape of the question to practise.

Only one topic in Unit 13 carries a boundary statement, Topic 13.2, and it limits the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors. Topic 13.4 has none, and neither does 13.1 or 13.3, checked page by page. One of the CED's own sample instructional activities for Topic 13.4 is worth knowing about, because it is this page turned into an exercise: one student describes a situation where the image is real or virtual and larger or smaller than the object, giving eight possibilities, and the other decides whether the instrument is converging or diverging and where the object must be.