To predict where a mirror forms an image, AP Physics 2 uses a compact algebraic relationship. Accurate answers depend on both the equation itself and a consistent choice of positive and negative distances.

Why the mirror equation matters

For spherical mirrors, the image location is determined by the object distance and the focal length. The mirror equation connects these quantities to the image distance, allowing you to decide where the image appears relative to the mirror.

When using this equation, the numbers cannot be treated as ordinary distances alone. Their signs carry physical meaning, telling you whether a point lies in front of the mirror or behind it.

If the sign convention is used incorrectly, the final numerical answer may look reasonable but describe the wrong physical situation.

The mirror equation

In AP Physics 2, image location for a mirror is found with the standard mirror equation.

All three distances must be measured along the principal axis and expressed in the same unit. You may use centimeters or meters, as long as you do not mix them in one calculation.

The equation can also be rearranged to solve directly for image distance: $d_i=\dfrac{f d_o}{d_o-f}$. This form can reduce algebra mistakes, but it only works correctly when the signs of $f$ and $d_o$ are assigned properly first.

Sign conventions for mirror problems

Object distance

For the standard situations in AP Physics 2, the object is a real object placed in front of the mirror, on the same side as the incoming light. In that common case, the object distance is positive, so $d_o>0$.

Because object placement is usually straightforward, most sign errors come from image distance or focal length rather than from $d_o$. Still, it is good practice to state the sign of the object distance before substituting into the equation.

Focal length

The sign of the focal length depends on the location of the focal point relative to the mirror.

Parallel rays reflected from a concave mirror converge at a real focal point in front of the mirror, giving a positive focal length ($f>0$). For a convex mirror, reflected rays diverge and only their backward extensions intersect behind the mirror, corresponding to a negative focal length ($f<0$). Source

• If the focal point is in front of the mirror, then $f>0$.

• If the focal point is behind the mirror, then $f<0$.

A positive focal length corresponds to a mirror that brings parallel incident rays together after reflection.

A spherical-mirror ray diagram shows how principal rays are constructed and where they intersect (real image) or appear to intersect when extended backward (virtual image). This supports interpreting negative results from the mirror equation as a virtual image behind the mirror rather than an algebra mistake. Source

A negative focal length corresponds to a mirror that causes reflected rays to spread out as though they came from a point behind the mirror. In standard mirror language, concave mirrors have $f>0$, and convex mirrors have $f<0$.

This sign is built into the geometry of the mirror, so it does not change when the object position changes.

Image distance

The sign of the image distance tells you where the image forms.

Ray diagrams compare image formation for (a) a concave mirror with the object outside the focal point (real, inverted image) and (b) inside the focal point (virtual, upright image). Panel (c) shows a convex mirror, which produces a virtual image behind the mirror for real objects in front. Source

• If the image is in front of the mirror, then $d_i>0$.

• If the image is behind the mirror, then $d_i<0$.

A positive image distance means the reflected rays actually meet in front of the mirror. A negative image distance means the reflected rays do not physically meet there; instead, they diverge and appear to come from a point behind the mirror.

Because of this, the sign of $d_i$ does more than locate the image. It also tells you whether the image is formed by actual ray intersection or by apparent ray origin.

Using the equation effectively

A reliable method is to separate the physics step from the algebra step.

• Identify the mirror type and assign the correct sign to $f$.

• Assign the object distance sign based on the object’s location.

• Substitute the signed values into the mirror equation.

• Solve for $d_i$.

• Interpret the sign of $d_i$ before describing the image location.

This last step is essential. A negative answer is not “wrong”; it is often the correct result and carries meaning about where the image appears.

It is also helpful to check whether the answer matches the physical situation. For example, if the algebra gives $d_i>0$, the image should be in front of the mirror. If the answer contradicts the known behavior of the mirror, recheck the signs before redoing the entire problem.

Interpreting special cases

Some results from the mirror equation are especially important.

• If $|d_i|$ is very large, the image forms very far from the mirror.

• If $d_o=f$, then the denominator in $d_i=\dfrac{f d_o}{d_o-f}$ becomes zero, which means the image is effectively at infinity.

• If the sign of $d_i$ changes as the object position changes, the image has moved from one side of the mirror to the other in the mathematical model.

These outcomes are not algebra accidents. They show that the mirror equation is encoding the geometry of reflection in a compact form.

Common mistakes to avoid

Several errors appear often in mirror problems:

• using an unsigned value for focal length

• treating every image distance as positive

• forgetting that the same unit must be used throughout

• solving correctly for $1/d_i$ but not inverting at the final step

• reporting only the number for $d_i$ without stating what its sign means

In AP Physics 2, the strongest answers connect the algebraic result to a physical statement. After finding $d_i$, always interpret it in words: in front of or behind the mirror, and therefore consistent with the sign convention used.