Physics › Waves › Lenses and images
Lenses and images
A lens is refraction put to work, glass shaped so that every ray leaving a point meets again at another point. Get the geometry of that meeting under control and cameras, spectacles, magnifying glasses and the eye itself all reduce to one exam question.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on Refraction and total internal reflection.
IN THIS TOPIC
- Move between focal length and power in dioptres with P = 1/f, carrying the sign for a diverging lens.
- Draw ray diagrams for a thin converging lens, locating real and virtual images.
- Use the thin lens equation and m = v/u to find image positions and sizes.
- Read a negative v as a virtual image, and say what a viewer would see.
COMMON MISCONCEPTION
Cover half a lens and you lose half the image.
Two shapes, one job
A converging lens bulges outwards and bends parallel rays inwards to a real meeting point, the principal focus F, a focal length f from the lens. A diverging lens is thinnest in the middle and spreads parallel rays apart, as if they had come from a focus on the near side: its focal length is counted negative.
Opticians quote power instead of focal length, because the powers of thin lenses in contact simply add.
with f in metres and P in dioptres (D). A +4.0 D lens converges with f = 0.25 m. A −2.0 D lens diverges with f = −0.50 m. Hold the two together and the pair behaves as +2.0 D. That additivity is the reason the unit exists, because a prescription then becomes a sum. Note that lenses sit on the Edexcel specification and not in AQA's core content, so the equations here carry no data-sheet badge.
Ray diagrams: three rays, one image
Every point of an object sends rays through every part of the lens, and the lens returns them all to one image point. To find it you need only the three rays whose paths you already know.
A ray arriving parallel to the axis leaves through the far focus. A ray through the centre of a thin lens passes straight on. A ray through the near focus leaves parallel. Any two of them locate the image and the third checks your work. With the object beyond F the rays really do cross again, and a screen placed there catches a sharp, inverted picture, a real image. That is the geometry inside every camera and every eye.
Cover half the lens and every image point still receives rays through the remaining half, so the whole image survives, only dimmer.
The thin lens equation
Measure the object distance u and the image distance v from the lens, and the whole ray diagram compresses into one line.
Use it with the real is positive convention. Distances to real objects and real images count as positive, and a virtual image on the object's side of the lens gives a negative v. The image height follows from the similar triangles around the central ray, and that ratio is the magnification.
One convention, used consistently on this page: put the sizes of v and u into the magnification, and read the orientation separately, from the ray diagram or from the sign of v. A real image (positive v) is inverted; a virtual one (negative v) is upright. Mixing signed and unsigned habits in one calculation is how a +2 ends up labelling an upside-down image.
WORKED EXAMPLE
Where does the image form?
An object stands 0.30 m from a converging lens of focal length 0.20 m. Find the image position and magnification.
1/v = 1/f − 1/u = 1/0.20 − 1/0.30 = 5.0 − 3.33 = 1.67 m−1, so v = 0.60 m beyond the lens.
m = |v|/u = 0.60/0.30 = 2.0, twice the size; the positive v says the image is real, and a real image is inverted. Put a screen 60 cm behind the lens and it lands there sharp.
Slide the object inside the focal length and the equation answers with a negative v: the rays on the far side now diverge, and only their backward projections meet.
GUIDED PRACTICE
The magnifying glass
The same f = 0.20 m lens now holds an object at u = 0.12 m. Find the image position and magnification, and state what kind of image this is.
Show the working
1/v = 1/0.20 − 1/0.12 = 5.0 − 8.33 = −3.33 m−1, so v = −0.30 m.
Magnification takes the sizes, not the signs, so m = |v|/u = 0.30/0.12 = 2.5. The negative v is the part that matters here. It says the image is virtual, upright, magnified, sitting on the object's own side and visible only by looking through the lens.
ASSESSMENT FOCUS
- State the convention before you substitute. Real is positive, and a negative v is the equation telling you the image is virtual, so read it as information and not as a slip.
- Power questions are unit questions. Get f into metres before P = 1/f, keep the negative sign on a diverging lens for both f and P, and add powers for lenses in contact.
- Ray diagrams score for standard rays drawn with a ruler, arrowheads on the rays, and the image labelled real or virtual, upright or inverted, magnified or diminished. Magnification is a ratio, so it never carries a unit.
- When a question covers part of the lens, say that the brightness falls and the full image survives, because every image point still receives rays through the uncovered part.
CHECK YOURSELF
A student's spectacles use a −2.5 D diverging lens. Find its focal length, and state what kind of image this lens forms of a distant lamp.
Show a hint
P = 1/f still holds; carry the sign.
Show the answer
f = 1/P = 1/(−2.5) = −0.40 m: a diverging lens of focal length 40 cm.
Parallel rays from the distant lamp leave the lens spreading apart, so they never meet. Their projections meet at the near focus: a virtual, upright, diminished image 0.40 m from the lens, on the lamp's side.
Power is one over the focal length, in metres.
Powers of lenses in contact add.
Real is positive, so a negative v is a virtual image.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the lenses and images questions page.
WHERE TO GO NEXT
- Rearranging equations is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
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- Move between focal length and power in dioptres with P = 1/f, carrying the sign for a diverging lens.
- Draw ray diagrams for a thin converging lens, locating real and virtual images.
- Use the thin lens equation and m = v/u to find image positions and sizes.
- Read a negative v as a virtual image, and say what a viewer would see.
Open the full revision checklist to track your progress across the whole unit.