Focal Length vs Radius of Curvature: f = R/2

The radius of curvature is the distance from a spherical mirror to the centre of the sphere it was cut from. The focal length is the distance to the point where parallel rays converge, and it is half the radius. Only the focal length goes in the mirror equation, so halve a given radius first.

AP Physics: Unit 13 (topics 13.2 Images Formed by Mirrors). This pair sits entirely in AP Physics 2 Topic 13.2, Images Formed by Mirrors, whose single learning objective is 13.2.A, describe the image formed by a mirror. The focal point is defined by 13.2.A.1 for a concave, converging, mirror, where rays parallel to the principal axis reflect toward a common location, and by 13.2.A.2 for a convex, diverging, mirror, where they reflect such that they appear to have originated from a common location behind the mirror. Essential knowledge 13.2.A.3 places the focal point of a plane mirror an infinite distance from the mirror. The relation between the two quantities is 13.2.A.4, and its exact wording matters: the focal point of a spherical mirror MAY BE APPROXIMATED as a point located on the principal axis of the mirror halfway between the surface of the mirror and the center of the mirror's radius of curvature. It is a locating statement with no equation attached, so the CED never writes f = R/2 as an equation. The AP Physics 2 equation sheet does not print it either: the waves, sound, and optics block prints 1/s_i + 1/s_o = 1/f as the relevant equation for 13.2.A.7 and the magnification magnitude for 13.2.A.8, and no relation between focal length and radius appears anywhere; the symbol list for that block defines f as frequency or focal length and contains no R at all. Both facts were checked by rendering the appendix page. Sign conventions are required by 13.2.A.7.i and not printed by the framework, so this page declares one before its first calculation: distances from the mirror surface, object distance positive, image distance positive in front and negative behind, focal length positive for concave and negative for convex, radius quoted as a magnitude, and magnification equal to minus the image distance over the object distance. Ray diagrams are required by 13.2.A.9, with three principal rays named at 13.2.A.9.i and the three image choices at 13.2.A.9.ii. The Topic 13.2 boundary statement limits the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors, and it is the only boundary statement in Unit 13, verified across all four topic pages. There is no lens counterpart: Topic 13.4 defines lens focal points at 13.4.A.1 and 13.4.A.2 and says at 13.4.A.5.ii that lenses have a focal point on both sides depending on the shape of the respective side, with no lensmaker's equation in the CED or on the sheet. Unit 13, Geometric Optics, is weighted at 12 to 15 percent of the multiple-choice section over a suggested 8 to 12 class periods, and Topic 13.2 lists suggested skills 1.A, 2.A, 2.C and 3.C.

The distinction, stated once

Both are distances measured from the same point, the middle of the mirror's surface, along the same line, the principal axis. They differ by a factor of two and by what they describe.

The radius of curvature is a fact about the glass. A spherical mirror is a piece cut from a sphere, and $R$ is that sphere's radius: the distance from the mirror's surface to the centre of the sphere. It is geometry. You could measure it without ever shining a light on the mirror.

The focal length is a fact about the light. AP Physics 2 essential knowledge 13.2.A.1 says incident light rays parallel to the principal axis of a concave, converging, mirror will be reflected toward a common location, called the focal point. Essential knowledge 13.2.A.2 says the same rays on a convex, diverging, mirror are reflected such that they appear to have originated from a common location behind the mirror, also called the focal point. The focal length is the distance from the mirror to that point. It is optics.

The CED connects the two in one sentence, 13.2.A.4: the focal point of a spherical mirror may be approximated as a point located on the principal axis of the mirror halfway between the surface of the mirror and the center of the mirror's radius of curvature. Halfway, so in magnitude

\lvert f \rvert = \frac{R}{2}

The reason this matters on an exam is one-directional. The mirror equation contains $f$ and never contains $R$ . So a question that hands you a radius of curvature has hidden a division by two in the first line, and a question that hands you a focal length has already done it for you. Everything below is about not missing that step and not applying it twice.

Side by side

	Focal length $f$	Radius of curvature $R$
What it describes	Where parallel rays converge or appear to diverge from	The sphere the mirror surface belongs to
CED statement	13.2.A.1 and 13.2.A.2 define the focal point; 13.2.A.7 uses $f$	13.2.A.4, only as the thing $f$ is half of
Measured from	The mirror surface, along the principal axis	The mirror surface, along the principal axis
Relation	$\lvert f \rvert = R/2$ , from 13.2.A.4	$R = 2\lvert f \rvert$
In the mirror equation	Yes, it is the $f$ in $1/s_i + 1/s_o = 1/f$	No, never. It has to be converted first
On the AP Physics 2 equation sheet	Yes, inside the mirror and lens equation	No, and $R$ is not even in the symbol list for that block
Sign	Signed. Positive for concave, negative for convex, by convention	Usually quoted as a magnitude in AP problems
For a plane mirror	Infinite, 13.2.A.3, so $1/f = 0$	Infinite. A flat surface is a sphere of unbounded radius
Exact or approximate	The focal point is defined by where the rays actually go	Exact geometry of the surface
Applies to lenses	Yes, the same $f$ in the same equation, 13.4.A.5	Not in AP Physics 2. No lens relation uses a radius
Object placed there gives	$s_o = f$ means no image forms, since $1/s_i = 0$	$s_o = R$ means $s_i = R$ and $M = -1$ exactly

Two rows carry the exam risk.

On the equation sheet is the first. The mirror equation is printed and the halving relation is not, so the conversion is something you supply. It is easy to forget under time pressure precisely because the printed equation looks complete.

Object placed there is the second, and it is the fastest way to check that you have used the right one. Put the object at the centre of curvature and the image lands back at the centre of curvature, the same size and inverted. Put it at the focal point and no image forms at all. Those two results are completely different, so if a calculation puts an object at a special distance and the answer looks wrong, check which special distance the problem meant.

What the CED prints, and what it does not

This section is here because the answer to "is $f = R/2$ on the sheet" decides how you write a solution, and the answer is no in a slightly interesting way.

The CED states the relation, in words, hedged. Essential knowledge 13.2.A.4 says the focal point of a spherical mirror may be approximated as a point located on the principal axis of the mirror halfway between the surface of the mirror and the center of the mirror's radius of curvature. Read that phrasing carefully. It is a locating statement about a point, not an equation, and it says may be approximated rather than is.

The CED does not print it as an equation. Topic 13.2 attaches a relevant equation to 13.2.A.7, the mirror equation, and to 13.2.A.8, the magnification. Essential knowledge 13.2.A.4 has no equation attached at all. So the halving lives in a sentence rather than in a box.

The equation sheet does not print it either. The waves, sound, and optics block of the AP Physics 2 Table of Information prints fifteen lines, and the two optical ones relevant here are

\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f} \qquad \lvert M \rvert = \left\lvert \frac{h_i}{h_o} \right\rvert = \left\lvert \frac{s_i}{s_o} \right\rvert

No relation between $f$ and a radius appears anywhere on the sheet. That was checked by rendering the appendix page rather than by memory, and the check went further: the symbol list beside that block defines $a$ as width, $A$ as amplitude, $d$ as separation, $D$ as path length, $f$ as frequency or focal length, $F$ as force, $h$ as height, $\ell$ as length, $L$ as distance, $m$ as order or mass, $M$ as magnification, $n$ as index of refraction, $s$ as position, $t$ as time, $T$ as period, $v$ as speed, $x$ and $y$ as positions, $\lambda$ as wavelength, $\theta$ as angle and $\omega$ as angular frequency. There is no $R$ in it. The letter is not a defined symbol for optics on that sheet at all.

The practical consequence: if a problem gives a radius of curvature, halve it in a labelled line of your own working before touching the printed equation, and say where the halving came from. Citing 13.2.A.4 costs nothing and makes the step visible.

One more absence worth recording, because it is the natural next question. There is no lens equivalent. Topic 13.4 defines a lens focal point at 13.4.A.1 and 13.4.A.2, and 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. That is as close as AP Physics 2 comes to a curvature statement for lenses, and it stays qualitative. No lensmaker's equation appears in the CED or on the sheet, so a lens focal length is always given or measured, never computed from a radius.

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

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. The framework requires a convention and does not print one, and the sheet prints magnification with absolute value bars on every term, so no sign information comes from it. The signs have to be declared.

Here is the one this page uses, unchanged to the last line.

All distances are measured from the reflecting surface of the mirror, along the principal axis.
Object distance $s_o > 0$ for a real object. Every object here is real.
Image distance $s_i > 0$ in front of the mirror, on the side the light is on, meaning real. $s_i < 0$ behind the mirror, meaning virtual.
Focal length $f > 0$ for a concave mirror, whose focal point is in front. $f < 0$ for a convex mirror, whose focal point is behind, exactly as 13.2.A.2 describes.
Radius of curvature $R$ is quoted as a magnitude, and the sign is attached when it is halved: $f = +R/2$ for a concave mirror, $f = -R/2$ for a convex one.
Magnification $M = -s_i/s_o$ . Positive is upright, negative is inverted, and $\lvert M \rvert > 1$ is enlarged.

The last of these is where the two quantities interact with signs most often. A radius of curvature quoted in a problem is almost always a plain positive number, because it is describing a physical sphere, and the mirror type is stated separately in words. So the sign of $f$ comes from the words "concave" or "convex", not from the number you were given. Reading a bare " $R = 30$ cm" on a convex mirror and writing $f = +15$ cm is the most common way this goes wrong.

For the two mirror types side by side, with the same convention and a full set of image classifications, see concave vs convex mirror. For what real and virtual mean, see real vs virtual image.

The case that separates them: the object at the centre of curvature

There is one situation where the difference between $f$ and $R$ is not a detail but the entire answer, and it doubles as a proof that the halving is right.

Put an object exactly at the centre of curvature of a concave mirror, so $s_o = R$ . Use $f = R/2$ and the mirror equation:

\frac{1}{s_i} = \frac{1}{f} - \frac{1}{s_o} = \frac{2}{R} - \frac{1}{R} = \frac{1}{R}

so $s_i = R$ . The image forms at the centre of curvature too. And the magnification is

M = -\frac{s_i}{s_o} = -\frac{R}{R} = -1

Same size, inverted, real, in the same place as the object. No numbers were needed; the result is exact for any $R$ , and it comes out clean only because $f$ is exactly half of $R$ . Any other ratio would put the image somewhere else.

Now contrast the other special distance. Put the object at the focal point, $s_o = f$ :

\frac{1}{s_i} = \frac{1}{f} - \frac{1}{f} = 0

so $s_i$ is infinite and no image forms. The reflected rays leave parallel, which is 13.2.A.1 run backwards.

Two distances, both special, both on the principal axis, and the outcomes could hardly be further apart: an exact same-size inverted image at one, and no image at all at the other. That is the practical reason to keep them separate, and it is also a fast self-check. If your answer says a same-size inverted image, the object was at $R$ . If it says no image, the object was at $f$ .

Why the CED says may be approximated

The hedge in 13.2.A.4 is real physics and it is worth knowing why it is there, even though AP Physics 2 never asks you to work with the error.

A spherical mirror does not bring every parallel ray to one point. Rays close to the principal axis do converge to a common location, and rays that strike far out toward the rim cross the axis nearer the mirror. So "the focal point" is a single point only for rays close to the axis, and the halfway rule is the limit that approach gives. That is why the CED writes may be approximated rather than is.

The framework does not name this effect, and the phrase does not appear in Topic 13.2, so do not introduce vocabulary the course has not given you. What the course does instead is arrange the problems so the approximation is never stressed:

The Topic 13.2 boundary statement limits the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors. Nothing else, and no other shape of curve. It is the only boundary statement in Unit 13, checked across all four topics.
Essential knowledge 13.2.A.9.i names three principal rays for ray diagrams, all drawn near the axis: 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.
Every calculation is done with the single relation 13.2.A.7 gives, which has one focal length in it and no room for a ray-by-ray correction.

So the honest summary for an exam is: treat $\lvert f \rvert = R/2$ as exact when you use it, because the course does, and know that the CED's own wording marks it as an approximation. If a question asks you to comment, the phrase to use is the CED's own, that the focal point may be approximated as the halfway point.

The plane mirror is the limiting case and a good consistency check. 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 $R$ is unbounded as well. Flatten a curved mirror and both quantities run off to infinity together, keeping their ratio of two the whole way.

When it costs a mark

Substituting a radius of curvature straight into the mirror equation. The equation takes $f$ . A given $R$ has to be halved first, using 13.2.A.4. This is the error the whole page exists for, and it produces an answer that is wrong by a clean factor with no warning sign.

Halving twice. The mirror image of the same error. If the problem already gave you a focal length, it has done the halving, and dividing again is just as wrong.

Giving a convex mirror a positive focal length. A radius is quoted as a magnitude and the mirror type is stated in words, so the sign comes from the words. Concave gives $f = +R/2$ and convex gives $f = -R/2$ under the convention declared above, matching 13.2.A.1 and 13.2.A.2.

Confusing the object at $R$ with the object at $f$ . One gives a real inverted image the same size, at the centre of curvature. The other gives no image at all. They are not near neighbours.

Writing $f = R/2$ as though the equation sheet printed it. It does not, and $R$ is not in the symbol list for the sheet's waves, sound, and optics block. Show the halving as its own line and cite 13.2.A.4.

Using a radius of curvature on a lens. AP Physics 2 gives no relation between a lens focal length and any radius, and no lensmaker's equation appears in the CED or on the sheet. Essential knowledge 13.4.A.5.ii goes as far as saying a lens has a focal point on each side depending on the shape of the respective side, and stops there.

Calling the relation exact without qualification when a question asks about it. Essential knowledge 13.2.A.4 says may be approximated. Use the number as if exact, describe it as the CED does.

Measuring $R$ from the centre of curvature to the focal point. That distance is also $R/2$ , which makes the error invisible in a numerical answer and visible in a diagram. $R$ runs from the mirror surface to the centre of curvature, and $f$ runs from the mirror surface to the focal point, so both start at the mirror.

When they behave alike, and why that lulls you

Three things make these two easy to blur.

They are the same kind of quantity, measured along the same line. Both are distances in metres from the mirror surface along the principal axis, both grow when the mirror flattens, and both are infinite for a plane mirror by 13.2.A.3. Nothing about the units or the geometry warns you that they are different numbers.

They are strictly proportional. Doubling $R$ doubles $f$ , always. So any question about how something changes when the mirror is made shallower gets the same qualitative answer either way, and a student can reason correctly about trends while holding the two quantities as one idea. The trend questions are the ones skill 2.D asks, so this can go undetected for a long time.

The CED mentions $R$ once and then never uses it. After 13.2.A.4, every statement in Topic 13.2 is written in terms of $f$ : the mirror equation at 13.2.A.7, the magnification at 13.2.A.8, the principal rays at 13.2.A.9.i. A student who works from the equation list alone will never see a radius, which is fine until a question quotes one.

The one place they genuinely coincide is the plane mirror, where both are infinite and the distinction has nothing to bite on. The shared equation collapses to $1/s_i = -1/s_o$ , giving $s_i = -s_o$ : virtual, the same distance behind the mirror as the object is in front. Essential knowledge 13.2.A.7.ii states that independently, which is a useful confirmation that the convention is being used consistently.

The question that separates them in one step: is this number about the shape of the mirror, or about where the light goes? The shape gives $R$ . The light gives $f$ , and $f$ is the one the equation wants.

Where this sits on the AP exam

Both belong to Topic 13.2, Images Formed by Mirrors, the only topic in Unit 13, Geometric Optics that mentions a radius of curvature. The unit is weighted at 12 to 15 percent of the multiple-choice section over a suggested 8 to 12 class periods, and Topic 13.2 has a single learning objective, 13.2.A, describe the image formed by a mirror.

Topic 13.2 is also the only topic in Unit 13 that carries a boundary statement, and it is the one quoted above: AP Physics 2 limits the study of mirrors to plane mirrors, convex spherical mirrors, and concave spherical mirrors. That was verified against all four topic pages in the unit.

The suggested skills for the topic are 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.A, derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway; 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 from experimental data, physical representations, or physical principles or laws.

Skill 2.A is the one this pair is built for. Deriving that an object at the centre of curvature produces an image at the centre of curvature with $M = -1$ , in symbols and with no numbers, is exactly the pathway 2.A describes, and it uses the halving relation as its only extra ingredient. Skill 1.A is why the radius matters in the first place: on a ray diagram, the centre of curvature is a labelled point and drawing it at the wrong distance makes every ray wrong.

Ray diagrams are required rather than optional. Essential knowledge 13.2.A.9 says they can be used to determine the location, type, size, and orientation of images formed by mirrors, and 13.2.A.9.ii lists the three choices a complete description makes: upright or inverted, virtual or real, and reduced, enlarged, or the same size as the object.

For the definitions on their own, see focal length, concave mirror, convex mirror and plane mirror. For the shared equation applied to lenses as well, see thin lens equation. For the mechanism behind mirrors, which is reflection rather than refraction, see reflection vs refraction and reflection vs total internal reflection.

A concave mirror given by its radius, at two object distances

A concave spherical mirror has a radius of curvature of $40 \ \mathrm{cm}$ . (a) Find its focal length. (b) An object is placed $60 \ \mathrm{cm}$ in front of it. Find the image distance and magnification, and classify the image. (c) The object is moved to $10 \ \mathrm{cm}$ in front of the mirror. Repeat. (d) Say what would have gone wrong had the $40 \ \mathrm{cm}$ been used as the focal length.

(a) Essential knowledge 13.2.A.4 places the focal point halfway between the mirror surface and the center of the mirror's radius of curvature, so $\lvert f \rvert = R/2 = 40/2 = 20 \ \mathrm{cm}$ . The mirror is concave, so under the convention declared above $f = +20 \ \mathrm{cm}$ .
(b) The mirror equation, the relevant equation for 13.2.A.7: $1/s_i = 1/f - 1/s_o = 1/20 - 1/60$ . Over a common denominator of $60$ : $3/60 - 1/60 = 2/60 = 1/30$ , so $s_i = +30 \ \mathrm{cm}$ .
Positive, so the image is real and in front of the mirror. $M = -s_i/s_o = -30/60 = -0.50$ , so it is inverted and half the size. Real, inverted, reduced.
(c) At $s_o = 10 \ \mathrm{cm}$ , which is inside the focal point: $1/s_i = 1/20 - 1/10 = 1/20 - 2/20 = -1/20$ , so $s_i = -20 \ \mathrm{cm}$ . Negative, so the image is virtual and behind the mirror. $M = -(-20)/10 = +2.0$ , so it is upright and twice the size. Virtual, upright, enlarged.
(d) Using $f = 40 \ \mathrm{cm}$ in part (b) would give $1/s_i = 1/40 - 1/60 = 3/120 - 2/120 = 1/120$ , so $s_i = +120 \ \mathrm{cm}$ and $M = -2.0$ : real, inverted and enlarged instead of real, inverted and reduced. The classification itself flips, not just the number, which is why the halving is not a cosmetic step.

$f = +20 \ \mathrm{cm}$ . At $s_o = 60 \ \mathrm{cm}$ : $s_i = +30 \ \mathrm{cm}$ , $M = -0.50$ , real, inverted, reduced. At $s_o = 10 \ \mathrm{cm}$ : $s_i = -20 \ \mathrm{cm}$ , $M = +2.0$ , virtual, upright, enlarged. Skipping the halving would have made the first image enlarged instead of reduced.

A convex mirror, where the sign comes from the word not the number

A convex spherical mirror has a radius of curvature of $30 \ \mathrm{cm}$ . An object stands $20 \ \mathrm{cm}$ in front of it. (a) Find the focal length with its sign. (b) Find the image distance and magnification, and classify the image. (c) State which piece of information fixed the sign of $f$ .

(a) The magnitude is $\lvert f \rvert = R/2 = 30/2 = 15 \ \mathrm{cm}$ . Essential knowledge 13.2.A.2 says a convex, diverging, mirror reflects parallel rays such that they appear to have originated from a common location behind the mirror, so the focal point is behind and $f = -15 \ \mathrm{cm}$ under the declared convention.
(b) $1/s_i = 1/f - 1/s_o = 1/(-15) - 1/20$ . Over a common denominator of $60$ : $-4/60 - 3/60 = -7/60$ , so $s_i = -60/7 = -8.6 \ \mathrm{cm}$ .
Negative, so the image is virtual and behind the mirror. $M = -s_i/s_o = -(-8.6)/20 = +0.43$ , so it is upright and reduced. Virtual, upright, reduced, which is what a convex mirror always gives for a real object.
(c) The word convex, not the number $30$ . The radius was quoted as a plain magnitude describing a sphere, and nothing in it says which way the mirror faces. Writing $f = +15 \ \mathrm{cm}$ here would have given $1/s_i = 1/15 - 1/20 = 4/60 - 3/60 = 1/60$ , so $s_i = +60 \ \mathrm{cm}$ and $M = -3.0$ : a real, inverted, enlarged image, which a convex mirror cannot produce from a real object at all.
That last check is worth keeping. If a convex mirror calculation ever returns a positive image distance, the sign of $f$ went in wrong.

$f = -15 \ \mathrm{cm}$ , $s_i = -8.6 \ \mathrm{cm}$ , $M = +0.43$ : virtual, upright and reduced. The sign of the focal length came from the word convex, since the radius was given as a magnitude.

Measuring the focal length and predicting the radius

A student points a concave mirror at a distant streetlamp and finds the sharpest image on a card held $12.0 \ \mathrm{cm}$ in front of the mirror. (a) Find the focal length. (b) Find the radius of curvature. (c) Predict the image distance and magnification for an object placed at the centre of curvature, and check the prediction against the general result. (d) State what happens if the object is instead placed at the focal point.

(a) A distant object means $s_o$ is very large, so $1/s_o$ is effectively zero and the mirror equation reduces to $1/s_i = 1/f$ . The image distance is therefore the focal length: $f = +12.0 \ \mathrm{cm}$ , positive because the mirror is concave and the image formed in front of it on a card, which makes it real.
(b) Essential knowledge 13.2.A.4 gives $\lvert f \rvert = R/2$ , so $R = 2\lvert f \rvert = 2(12.0) = 24.0 \ \mathrm{cm}$ .
(c) At $s_o = R = 24.0 \ \mathrm{cm}$ : $1/s_i = 1/12.0 - 1/24.0 = 2/24 - 1/24 = 1/24$ , so $s_i = +24.0 \ \mathrm{cm}$ , and $M = -24.0/24.0 = -1.00$ . Real, inverted, the same size, at the centre of curvature.
General check, with no numbers: putting $s_o = R$ and $f = R/2$ into the mirror equation gives $1/s_i = 2/R - 1/R = 1/R$ , so $s_i = R$ and $M = -R/R = -1$ for every spherical mirror. The numerical result matches, which confirms both the halving and the arithmetic.
(d) At $s_o = f = 12.0 \ \mathrm{cm}$ : $1/s_i = 1/12.0 - 1/12.0 = 0$ , so $s_i$ is infinite and no image forms. The reflected rays leave parallel to the principal axis, which is 13.2.A.1 read in reverse. Compare with part (c): the same mirror, two special object distances, and one gives a same-size inverted image while the other gives none at all.

$f = +12.0 \ \mathrm{cm}$ and $R = 24.0 \ \mathrm{cm}$ . An object at the centre of curvature gives $s_i = +24.0 \ \mathrm{cm}$ and $M = -1.00$ , real, inverted and the same size, which the symbolic result confirms for any spherical mirror. An object at the focal point produces no image, since $1/s_i = 0$ .

Frequently asked questions

What is the difference between focal length and radius of curvature?

The radius of curvature is the radius of the sphere the mirror surface was cut from, measured from the mirror to the centre of that sphere. The focal length is the distance from the mirror to the point where rays parallel to the principal axis converge, or appear to diverge from. AP Physics 2 essential knowledge 13.2.A.4 places the focal point halfway between the mirror surface and the centre of the mirror's radius of curvature, so the focal length is half the radius in magnitude. Only the focal length appears in the mirror equation.

Is f = R/2 on the AP Physics 2 equation sheet?

No. The CED states the relation in words at essential knowledge 13.2.A.4, saying the focal point of a spherical mirror may be approximated as a point halfway between the mirror surface and the centre of its radius of curvature, and it attaches no equation to that statement. The waves, sound, and optics block of the Table of Information prints the mirror and lens equation and the magnification, and nothing relating focal length to a radius. The symbol R is not even defined in the symbol list for that block, so the halving is a step you supply and should show.

Why does the CED say the focal point may be approximated?

Because a spherical mirror does not bring every parallel ray to exactly one point. Rays close to the principal axis converge at the halfway point, and rays striking far out toward the rim cross the axis slightly nearer the mirror, so a single focal point is a good description only for rays near the axis. AP Physics 2 arranges its problems so this never matters: the Topic 13.2 boundary statement limits mirrors to plane, convex spherical and concave spherical, and the three principal rays named at 13.2.A.9.i are all drawn near the axis. Treat the relation as exact when calculating.

Do I use the radius of curvature or the focal length in the mirror equation?

The focal length. The relevant equation for essential knowledge 13.2.A.7 is one over the image distance plus one over the object distance equals one over the focal length, and no version of it takes a radius. If a problem gives a radius of curvature, halve it first using 13.2.A.4 and attach the sign from the mirror type, then substitute. If a problem gives a focal length, the halving has already been done and dividing again is an error.

Is the radius of curvature positive or negative?

In AP problems a radius of curvature is normally quoted as a plain magnitude, because it describes a physical sphere, and the mirror type is stated separately in words. The sign belongs to the focal length: positive for a concave mirror, whose focal point is in front of it by 13.2.A.1, and negative for a convex mirror, whose focal point is behind it by 13.2.A.2. So the sign comes from the word concave or convex, not from the number you were given, and getting it from the number is the usual way convex mirror problems go wrong.

What happens when an object is at the centre of curvature?

The image forms at the centre of curvature too, the same size and inverted. Substituting an object distance equal to R and a focal length of R over 2 into the mirror equation gives one over the image distance equal to two over R minus one over R, which is one over R, so the image distance is R and the magnification is minus one. The result is exact for any spherical mirror and holds only because the focal length is exactly half the radius, which makes it a good check that you halved correctly.

Do lenses have a radius of curvature in AP Physics 2?

Not in any usable form. AP Physics 2 Topic 13.4 defines a lens focal point at 13.4.A.1 and 13.4.A.2 and says at 13.4.A.5.ii that lenses have a focal point on both sides of the lens that depends on the shape of the respective side of the lens, and it stops there. No lensmaker's equation appears in the CED or on the equation sheet, so a lens focal length is always given in the problem or measured, never calculated from a radius. The halving relation belongs to spherical mirrors only.