Lenses and ray diagramsEdexcel A-Level Physics: Revision notes
Section 1
Converging and diverging lenses
A converging (convex) lens is thicker at the centre and refracts parallel rays so that they come together. A diverging (concave) lens is thinner at the centre and refracts parallel rays so that they spread out.
The principal axis is the line through the optical centre of the lens at right angles to its surface. A ray passing through the optical centre goes straight on, undeviated, for a thin lens.
Section 2
Focal length and the principal focus
For a converging lens, rays parallel to the principal axis are brought to a real focus at the principal focus on the far side. For a diverging lens, parallel rays diverge and appear to come from a virtual principal focus on the same side as the incoming light.
The focal length f is the distance from the optical centre to the principal focus. A lens has a principal focus on each side, at equal distances.
In calculations, f is positive for a converging lens and negative for a diverging lens.
A distant object sends almost parallel rays, so a converging lens focuses its image at the principal focus. This gives a quick measurement of f.
Section 3
Ray diagrams: locating an image
Draw two rays from the top of the object. The image of the top is where they meet (or appear to meet).
- A ray parallel to the axis: refracted through the principal focus (converging) or so that it appears to come from the principal focus on the object side (diverging).
- A ray through the optical centre: undeviated.
- Optionally, a ray through the principal focus on the object side emerges parallel to the axis.
The image of a point on the axis lies on the axis, directly below or above the image of the top.
For a thin lens, draw the change of direction on a single vertical line through the optical centre, not at the curved surfaces.
Section 4
Image positions for a converging lens
The image depends on the object distance u compared with f:
- u greater than 2f: real, inverted, diminished, between f and 2f
- u between f and 2f: real, inverted, magnified, beyond 2f
- u less than f: virtual, upright, magnified, on the object side
A real image can be caught on a screen because the refracted rays actually meet. A virtual image cannot, because the refracted rays only appear to diverge from it.
A diverging lens always gives a virtual, upright, diminished image of a real object.
Section 5
Lens power
The power of a lens is P = 1/f, with f in metres. The unit is the dioptre (D), where 1 D = 1 m⁻¹. A converging lens has positive power and a diverging lens has negative power. A shorter focal length means a more powerful lens, which refracts light more strongly.
Example: f = 0.25 m gives P = +4.0 D; f = −0.50 m gives P = −2.0 D.
Convert focal lengths to metres before using P = 1/f. A focal length of 20 cm is 0.20 m, giving 5.0 D, not 0.05 D.
Section 6
Thin lenses in combination
For thin lenses placed in contact, the total power is the sum: P = P₁ + P₂ + ...
Worked example: a converging lens with f = 0.20 m (P₁ = +5.0 D) in contact with a diverging lens with f = −0.50 m (P₂ = −2.0 D):
P = 5.0 − 2.0 = +3.0 D, so f = 1/3.0 = 0.33 m.
The diverging lens weakens the convergence, so the combined focal length is longer than that of the converging lens alone.
Always include the sign of each power. Forgetting that a diverging lens is negative is the usual error in combination questions.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Lenses and ray diagrams
- A student wants to find the focal length of a thin converging lens. She holds the lens so that it forms a sharp image of a distant window on a white screen and measures the distance from the lens to the screen. The distance is 0.125 m.Explain why the distance from the lens to the screen is equal to the focal length of the lens.2 marks
- A diverging lens of focal length 0.40 m is used to view a small illuminated object placed on its principal axis. A student plans to locate the image by drawing a scale ray diagram of the arrangement.Describe how the student should use two rays from the top of the object to locate the top of the image.2 marks
- An optician combines two thin lenses in contact on a common principal axis. The first is a converging lens of focal length 0.20 m. The second is a diverging lens of focal length 0.50 m (magnitude). A parallel beam of light is sent through the combination along the axis.Calculate the power of each lens and the power of the combination.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).