05.05 Lens and Mirror Equation

The big idea The lens/mirror equation gives you exact numbers where a ray diagram only gives you a picture. The one piece the source lesson never teaches directly is what a negative answer means, so this page adds the sign convention and the magnification equation that go with it.

Working with reciprocals without losing track

The math here is just fractions, but the two-step reciprocal (solve for 1/x, then flip it) is where most mistakes happen, not the physics.

Not sure about the basics underneath this lesson?

Variables

Focal length, f: for a lens, the distance from its center to either focal point; for a mirror, the distance from its reflective surface to the focal point. Object distance, dₒ: distance from the object to the lens's center or the mirror's surface. Image distance, dᵢ: distance from the image to the lens's center or the mirror's surface. All three are measured in the same distance unit, usually centimeters or meters; never mix units within one problem.

The lens and mirror equation

Lens/mirror equation

\dfrac{1}{f}=\dfrac{1}{d_o}+\dfrac{1}{d_i}

f, dₒ, dᵢ all in the same distance unit. Usually two variables are given and you solve for the third.

Magnification

m=-\dfrac{d_i}{d_o}

m has no unit. |m| > 1 means larger than the object, |m| < 1 means smaller. A positive m means upright; a negative m means inverted.

Sign convention: what a negative answer means

This is the part the equation alone doesn't tell you. dₒ is positive for an ordinary object in front of the lens or mirror, in every problem in this course. dᵢ positive means a real image: opposite side of the lens from the object, or same side of the mirror as the object. dᵢ negative means a virtual image: same side of the lens as the object, or behind the mirror's surface. Nothing went wrong if your algebra produces a negative dᵢ. It's telling you the image type, exactly like the ray diagrams in 05.04 already showed for an object inside F.

For an ordinary object, object distance d sub o is positive. A converging lens has positive image distance on the side opposite the object for a real image and negative image distance on the object side for a virtual image. A concave mirror has positive image distance in front of the mirror for a real image and negative image distance behind the mirror for a virtual image. — For these course cases, dᵢ is positive for a real image and negative for a virtual image; the diagram maps the correct side for each optical element.

Worked example: lens, real image

Candle flame through a convex lens

Object distance. $d_o=25.0\,\text{cm}$
Focal length. $f=18.4\,\text{cm}$

Find the image distance and describe the image.

Solve the equation for 1/dᵢ.
$\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o}=\dfrac{1}{18.4}-\dfrac{1}{25.0}$
Evaluate, then take the reciprocal.
$\dfrac{1}{d_i}=0.0143\,\text{cm}^{-1}\ \Rightarrow\ d_i=69.7\,\text{cm}$
Check magnification.
$m=-\dfrac{69.7}{25.0}=-2.79$

dᵢ = 69.7 cm, positive, so the image is real, on the opposite side of the lens. m = −2.79: negative means inverted, and |m| > 1 means about 2.8 times larger than the object.

Worked example: mirror, and reading a negative result

Concave mirror, short focal length

Object distance. $d_o=2.50\,\text{m}$
Focal length. $f=0.050\,\text{m}$

Find the image distance and describe the image.

Solve for 1/dᵢ.
$\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o}=\dfrac{1}{0.050}-\dfrac{1}{2.50}$
Evaluate and take the reciprocal.
$d_i\approx0.051\,\text{m}$

dᵢ = 0.051 m, positive, so the image is real, on the same side of the mirror as the object (the general behavior for an object beyond C).

What if dᵢ comes out negative?

Sample result. $d_i=-12\,\text{cm}$

Interpret this result for a lens.

Read the sign, not just the magnitude.
For a lens, negative dᵢ places the image on the same side as the object.

The image is virtual, 12 cm from the lens, on the object's own side. This is exactly the case 05.06's lab asks you to describe.

Common mistakes

Forgetting the reciprocal step

Solving 1/f = 1/dₒ + 1/dᵢ for dᵢ gives you 1/dᵢ first. You still have to take the reciprocal of that number to get dᵢ itself. Use the calculator's 1/x or x⁻¹ key.

Mixing units

If f is given in centimeters and dₒ in meters, convert one before substituting. The source's own worked examples stay in one unit throughout for exactly this reason.

Treating a negative dᵢ as an error

A negative image distance is a real answer, not a mistake to erase. It means virtual image; reread the sign convention instead of redoing the arithmetic.

Skipping magnification's sign

m = −dᵢ/dₒ. Forgetting the negative sign in front will flip upright and inverted in your answer even when the size is right.

Task-fit practice

Work every problem to a decimal answer, including the reciprocal step, before checking.

Final checklist

Tick these off as you go. Saved in this browser.

I keep all three distances in the same unit before substituting.
I can solve 1/f = 1/dₒ + 1/dᵢ for any one of the three variables.
I can say what a negative dᵢ means without treating it as an error.
I can find magnification and use its sign to state upright or inverted.
I completed the 05.05 quiz.