Thin lenses

Thin lenses redirect rays by refraction. Ray diagrams give the geometry first; the thin-lens equation then turns that geometry into image distance and magnification, provided one sign convention is chosen and used consistently.

Converging and diverging lenses

A converging lens has positive focal length in the convention used here and brings parallel rays to a real focal point. A diverging lens has negative focal length and makes parallel rays appear to come from a focal point on the incident side.

Ray tracing comes before signs. A converging lens sends an incident parallel ray through the far focal point, while a central ray continues approximately straight in the thin-lens model.

Three principal rays

For a converging lens: a ray parallel to the axis refracts through the far focus; a ray through the near focus emerges parallel; a ray through the optical center is approximately undeviated in the ideal thin-lens model.

For a diverging lens, a parallel ray emerges as if it came from the near focus. Two well-chosen rays are enough to locate an image.

A diverging lens produces rays that leave diverging. Their backward extensions meet on the object side, giving a virtual upright image for a real object.

Thin-lens equation and magnification

1/f = 1/dₒ + 1/dᵢ and m = hᵢ/hₒ = −dᵢ/dₒ

With the sign convention used on this page, positive dᵢ means a real image and negative dᵢ a virtual one. A negative magnification means inverted; positive means upright.

Thin-lens calculator

Sign convention here: f > 0 converging, f < 0 diverging; d_o > 0 for a real object; d_i > 0 real image.

f (cm) dₒ (cm)

Object position changes the image type

For a converging lens, an object outside the focal length can produce a real inverted image; an object inside the focal length produces a virtual upright magnified image. A diverging lens with a real object produces a virtual upright reduced image in the ideal model.

Real images can be projected onto a screen because rays physically converge there. Virtual images cannot be caught on a screen at their apparent location.

Use the diagram to predict the image before calculating

Before inserting numbers, trace at least two principal rays and predict whether the image should be real or virtual, upright or inverted, enlarged or reduced. The signs returned by the equation should agree with that geometry. If they do not, revisit the sign convention rather than forcing a preferred answer.

The thin-lens model neglects lens thickness and aberrations. Real lenses can have spherical and chromatic aberration, and compound optical systems often use several elements to control them. The simple equation remains an excellent first model when lens thickness is small compared with the relevant object and image distances.

Worked examples

1. Converging real image

A converging lens has f = +10 cm and dₒ = 30 cm. Find dᵢ and magnification.

Solution

1/dᵢ = 1/10 − 1/30 = 1/15, so dᵢ = +15 cm. m = −15/30 = −0.50: real, inverted and half-size.

2. Object inside focal length

For the same f = +10 cm lens, place the object at dₒ = 5.0 cm.

Solution

1/dᵢ = 1/10 − 1/5 = −1/10, so dᵢ = −10 cm. The image is virtual. m = −(−10)/5 = +2, upright and magnified.

3. Diverging lens

A diverging lens has f = −12 cm and dₒ = +24 cm. Find dᵢ.

Solution

1/dᵢ = 1/(−12) − 1/24 = −1/8, so dᵢ = −8.0 cm, a virtual image on the object side.