Lenses
Lenses affect the light that passes through them making an image of the object
from which the light waves come. Sometimes this image is real (you can form it on
a surface such as a piece of paper) and sometimes virtual (it cannot be formed on
a surface)
Curvature and the sign convention
The radius of curvature (R) of a surface is taken as a positive number if it
increases the curvature of the waves – i.e. converges the waves and negative if it
decreases the curvature of the ways – i.e. diverges the waves.
A surface that converges a wavefront is taken a positive e.g. a convex lens
A surface that diverges a wavefront is taken as negative e.g. a concave lens
Figure 1
Figure 2
The curvature of the surface, or wave, is defined as 1/radius of that surface or wave. A
surface with a small radius of curvature therefore has a large curvature.
Waves and curved surfaces
+
+
+
∞
-
+
-
-
-
∞
Figure 3
bi-convex
plano-convex
meniscus
bi-concave
plano-concave
When light waves fall on a curved surface that surface changes the curvature of the wave.
Lenses are made with at least one curved surface (convex or concave) and so lenses simply
change the curvature of the waves that fall on them.
Therefore:
Curvature of waves leaving the lens = curvature of waves hitting the lens + curvature added by the lens
1
Effect of a lens
A lens adds curvature to a wavefront. As the waves enter the lens their curvature is changed.
In the example below they are bent from a plane wave to form a converging wave in the lens.
This wave is further converged when it leaves the lens.
Wavelength in air

Wavelength in glass
’
Figure 4
Focus
Wavelength in air

You can follow this change of curvature as the waves meet the lens and then pass through it,
emerging at the other side.
As the wave hits the lens it is the centre of the wave that meets the glass first and so this part
of the wave is slowed down first (light waves move slower in glass than they do in air). This
means that the outer portions of the wave ‘catch up’ so increasing the curvature to form a
converging beam.
As the wave leaves the lens the outer portions move into the air first and so speed up first.
This means that the outer portions move off more rapidly first and so the curvature of the
wave is further increased so converging the light more strongly.
The overall action of the convex lens is therefore to converge light waves to a focus.
Looking at the diagram you can see that waves with no curvature have been converged to a
point a distance f (the focal length) from the lens. This means that they have been given a
curvature 1/f. The shape of the incident waves have no effect on how much the lens changes
their curvature and so we can say that a lens increases the curvature of waves that pass
through it by a constant amount 1/f.
(Remember that 1/f will be positive for a convex lens and negative for a concave lens)
The example has been done with a plane wave but the same would occur with a spherical
wave spreading outwards from a point object.
Wavefronts
Wavefronts spreading outwards from a point source are taken as having a negative
curvature and so the distance from the lens to the object is negative
Wavefronts that converge to an image are taken as having a positive curvature and so the
distance from the lens to the image is positive.
2
Lens equation
The curvature of the waves spreading outwards towards a lens from an object distance u
from the lens have a curvature 1/u and those converging to a point image on the other side
of the lens have a curvature 1/v. The lens adds a constant curvature (1/f) to the wave and so
the equation relating the object distance (u) image distance (v) and the focal length (f) is:
Lens equation:
1/v = 1/u + 1/f
using the above sign convention called the Cartesian sign convention
This can be shown in the following diagram.
I
O
Figure 5
v
u
Negative distances
Positive distances
(for objects on the left of the lens)
(for images on the right of the lens)
Example
1/v = 1/u + 1/f
P
For a lens of focal length +20 cm (convex), object distance – 50 cm (on left of the lens)
1/v = 1/20 – 1/50 = 0.03
Therefore: v = 33.3 cm (positive so on the right hand side of the lens)
ositive
distances
3