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Lenses and images
A lens is refraction with a purpose: a piece of glass shaped so that every ray from 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 become one exam question.
Builds on Refraction and total internal reflection.
IN THIS TOPIC
- Use P = 1/f to move between focal length and power in dioptres, for converging and diverging lenses.
- 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.
WHAT YOU PROBABLY THINK
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 work in power rather than focal length, because 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. Stack them and the pair behaves as +2.0 D. That additivity is the whole reason the unit exists: a prescription is just a sum.
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 only need 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 locate the image; the third is the check. With the object beyond F the rays really do cross again, and a screen placed there catches a sharp, inverted picture: a real image, the geometry inside every camera and every eye.
This is also why the claim above fails. Cover half the lens and every image point still receives rays through the remaining half: the whole image survives, just dimmer.
The thin lens equation
Measure the object distance u and image distance v from the lens, and the geometry of the ray diagram compresses into one line, on the data sheet as
with the real is positive convention: distances to real objects and images are positive, and a virtual image on the object's side of the lens carries a negative v. The image height follows from the central ray's similar triangles, the magnification:
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: real, inverted and twice the size. A screen 60 cm behind the lens shows it 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.
YOUR TURN
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.
m = v/u = 0.30/0.12 = 2.5. The negative v says the image is virtual: upright, magnified, on the object's own side, visible only by looking through the lens.
THE EXAM BIT
- State the convention before you substitute: real is positive. A negative v in your answer is not an error, it is the equation telling you the image is virtual.
- Power questions are unit questions: f must be in metres before P = 1/f, and diverging lenses carry negative f and negative P. Lenses in contact: add the powers.
- Ray diagrams score for the standard rays drawn with a ruler, arrows on rays, and the image labelled real or virtual, upright or inverted, magnified or diminished.
- Magnification is a ratio of distances or heights, so it has no unit. Quoting it in metres throws away a mark.
- If a question covers part of the lens, say brightness falls but the full image remains: 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 focal length, and powers in contact add.
Real is positive: a negative v means the image is virtual.
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.
CHECK YOUR PROGRESS
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- Use P = 1/f to move between focal length and power in dioptres, for converging and diverging lenses.
- 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.
Open the full revision checklist to track your progress across the whole unit.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.