Imagine a point source of light at a distance d1 from a convex lens. 
The light is visible on a screen at distance d2 on the other side of
the lens.  The focal length f of the lens is less than d1.

All is well.

Now, if we move the light towards the lens so that d1 < f, the image
on the screen (which has not moved) will become blurred.
What I'd like to do is describe mathematically the distribution of
light in this blurred image.
To put it another way: the brightness on the screen will vary with
distance from the centre of the image (call it x), but is there a
formula to decribe the brightness at a point on the screen as a
function of that distance x?

Many thanks in advance, and I hope my maths is enough to understand the answer!

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Request for Question Clarification by pafalafa-ga on 18 Aug 2004 04:12 PDT

This chapter on image blurring likens blurring to heat diffusion, and
uses similar math to describe the blurring process:


http://www4.ncsu.edu/eos/users/w/white/www/white/ma325/MVlec3.pdf


It may offer you some insight to your question.  Let us know if this
is close to what you need, or if it misses the mark.  Your feedback
will help the researchers here to "focus" their efforts.

pafalafa-ga

If you are ok with the following 2 approximations, the answer is very simple.

1) Thin lens approximation.  Basically we are assuming that the light
rays involved travel nearly parallel to the axis of the lens.  This
approximation is not valid if the lens has an extremely short focal
length.

2) No diffraction.  Diffraction is the 'scattering' effect of light as
it passes through an aperature.  It means that there is no such thing
in real life as a perfectly focussed image--there's always some
blurring.  This blurring is bigger for small aperatures.  If you have
a relatively large lens, you don't have to worry about it.

With these assumptions, the distribution of light is uniform.  The
point source lights up a disk on the screen.  The disk has sharp
edges, and the intensity is the same everywhere on the disk.  You can
convince yourself of this by drawing some ray diagrams:  a light ray
that that passes through the lens at a given radial distance from the
axis of the lens (i.e. from the center) will strike the screen a
proportional distance from the center of the disk.  All the light that
passes through an annular ring of the lens ends up at a corresponding
annular ring on the screen.  Since the lens is uniformly illuminated,
the screen is as well.