Lenses and Mirrors: A Problem-Solving Method
Geometric optics is unusual among introductory topics because there is barely anything to memorise. One equation relates focal length, object distance and image distance; a second gives the magnification; a third connects a mirror's focal length to its radius. That is the whole toolkit for a large fraction of exam questions.
What trips people up is not the algebra but the signs. A negative image distance is not an error message - it is the equation telling you the image is virtual. Once you adopt a single sign convention and apply it without exception, lens and mirror problems become almost mechanical. This guide sets out that convention, gives a repeatable procedure, and works four examples through completely.
One equation for both
The thin lens equation and the mirror equation are identical in form: 1/f = 1/d_o + 1/d_i, where f is the focal length, d_o the object distance and d_i the image distance. This is not a coincidence of notation. Both devices take rays diverging from a point and redirect them so that they converge on, or appear to diverge from, another point, and the same geometry of similar triangles produces the same relationship.
The practical consequence is that learning one case teaches you the other. What differs is only how focal length is determined. For a spherical mirror it comes straight from the geometry, f = R/2, where R is the radius of curvature. For a thin lens it depends on the glass and the curvature of both surfaces, which is what the lensmaker's equation handles, and it is often quoted instead as lens power P = 1/f in diopters when f is in metres.
The sign convention, stated once
Every worked example below uses the convention in which real is positive. Write these rules at the top of your page and treat them as non-negotiable; nearly every wrong answer in this topic comes from applying a rule inconsistently rather than from arithmetic.
- d_o is positive for a real object, which is the only case in introductory problems.
- d_i is positive when the image is real: on the far side of a lens from the object, or in front of a mirror. It is negative when the image is virtual.
- f is positive for a converging element - a convex lens or a concave mirror - and negative for a diverging element, meaning a concave lens or a convex mirror.
- For mirrors, f = R/2, with R positive for concave and negative for convex.
- Magnification is m = −d_i/d_o. A positive m means an upright image, a negative m means inverted, and the size of |m| gives the scaling.
- Heights follow the same signs, since m = h_i/h_o.
What magnification is telling you
The minus sign in m = −d_i/d_o is doing real work, not decoration. A real image, with d_i positive, comes out with negative magnification and is therefore inverted - which is why the image a projector throws on a screen is upside down until you flip the slide. A virtual image, with d_i negative, gives positive magnification and stays upright, which is what you see through a magnifying glass or in a bathroom mirror.
So you can read the answer without drawing anything. The sign of d_i tells you real or virtual, the sign of m tells you upright or inverted, and |m| tells you bigger or smaller. Those three facts are usually exactly what the question asks for, and they are already contained in the two numbers you calculated.
A repeatable procedure
Work in the same order every time. The discipline matters more than the speed, because each step catches a different kind of mistake.
- Identify the element and get the sign of f right: converging positive, diverging negative. For a mirror, compute f = R/2 first.
- Write down d_o as a positive number and keep all lengths in the same unit.
- Rearrange to 1/d_i = 1/f − 1/d_o and solve for d_i, keeping the sign the algebra gives you.
- Compute m = −d_i/d_o.
- Translate into words: real or virtual from the sign of d_i, upright or inverted from the sign of m, enlarged or reduced from |m|.
- Sanity-check with a rough ray diagram, or against the known behaviour of that element.
Four worked examples
Converging lens, distant object. A convex lens has f = +10 cm and an object sits at d_o = 30 cm. Then 1/d_i = 1/10 − 1/30 = 2/30, so d_i = +15 cm, and m = −15/30 = −0.50. The image is real, inverted and half size, formed 15 cm beyond the lens - a camera forming a picture on its sensor.
Same lens, object inside the focal length. Now d_o = 5 cm with f still +10 cm. Then 1/d_i = 1/10 − 1/5 = −1/10, so d_i = −10 cm, and m = −(−10)/5 = +2.0. The negative d_i says the image is virtual, sitting 10 cm on the same side as the object, and the positive m of 2.0 says it is upright and twice as tall. That is a magnifying glass, and it explains why the effect disappears if you hold the lens too far from the page.
Convex mirror. A shop security mirror bulges outward with a radius of curvature of 40 cm, so f = R/2 = −20 cm. A customer stands at d_o = 30 cm from it. Then 1/d_i = −1/20 − 1/30 = −5/60, so d_i = −12 cm, and m = −(−12)/30 = +0.40. Virtual, upright and reduced to 40 percent - smaller images, but a much wider field of view, which is the entire point of the mirror.
Concave mirror beyond the centre of curvature. A concave mirror has R = 30 cm, so f = +15 cm, and an object stands at d_o = 45 cm. Then 1/d_i = 1/15 − 1/45 = 2/45, so d_i = +22.5 cm, and m = −22.5/45 = −0.50. Real, inverted and half size, and the image lands between the focal point and the centre of curvature, exactly where the standard ray diagram puts it.
Ray diagrams and the traps to avoid
A quick ray diagram is the cheapest check available, and three rays are enough for a converging lens. A ray arriving parallel to the axis leaves through the far focal point; a ray through the centre of the lens carries straight on; a ray passing through the near focal point emerges parallel to the axis. Where they meet is the image. If they diverge instead, trace them backwards and their apparent meeting point is the virtual image - the geometric picture of a negative d_i.
Two edge cases are worth knowing before they appear on a paper. An object placed exactly at the focal point gives 1/d_i = 0, so the rays leave parallel and no image forms at any finite distance. And a diverging lens or convex mirror always produces a virtual, upright, reduced image, whatever the object distance, because 1/f is negative and so 1/d_i can never come out positive.
Beyond that, the common traps are small and repeatable: mixing centimetres with metres, forgetting to invert after solving for 1/d_i, dropping the minus sign in the magnification, and using R instead of f for a mirror. Working in the fixed order above catches all four. When you want the underlying relations side by side, the optics category on PhysRef lists the thin lens and mirror equations, both magnification forms, and the focal-length-to-radius relation with their variables spelled out.
Frequently asked questions
What does a negative image distance mean?
That the image is virtual. The outgoing rays are diverging, so they only appear to come from a point on the same side of a lens as the object, or behind a mirror. You can see a virtual image with your eye, but you cannot catch it on a screen placed there.
Why is the same equation used for lenses and mirrors?
Because both take rays from an object point and redirect them toward an image point, and the similar-triangle geometry is identical in the two cases. Only the way the focal length arises differs: f = R/2 for a spherical mirror, and surface curvature plus refractive index for a lens.
How do I know whether the image is upright or inverted?
From the sign of the magnification, m = −d_i/d_o. Positive means upright, negative means inverted. Real images from a single converging element always come out inverted, and virtual images always come out upright.
Can a diverging lens ever produce a real image?
Not from a real object. With f negative, 1/d_i = 1/f − 1/d_o is always negative, so d_i is always negative. The image is invariably virtual, upright and smaller than the object.
What happens if the object is exactly at the focal point?
The equation gives 1/d_i = 0, meaning the image distance is infinite. The rays leave the lens or mirror parallel to each other, so no image forms at any finite distance. This is how a lamp at the focus of a reflector produces a collimated beam.