Radiance and Occular Hazards

[scherer@uiuc.edu (David Scherer)]

It's Friday afternoon, so let's try another approach to address this concept.

An extended source emits a certain radiant power, say 1 W, in a certian
area, say .01 m^2.  The radiant exitance is 100 W/m^2.  The power density
received at the retina will be less than this.  But how much less?

Well, the source spreads out into a solid angle, say 2 sr.  In this example
the randiance is R = 50 W/m^2-sr.

At a certian distance, xo, the pupil of an observer's eye intercepts a solid
angle, omega, equal to the pupil area divided by the distance squared: omega
= (pi) d^2 / (4 xo^2).  The irradiance at the pupil is R * omega.

The retinal image is minified, so the irradiance is increased by 1/M^2,
where M is the magnification.  But M is related to the focal length,f, and
the object distance, xo:  |M| = f / xo.

Finally the eye will only transmit a fraction T of the incident light.  (And
T depends on wavelength.)

So the retinal irradiance is I = R * T * omega / |M2| =   (pi) d^2 R T /4 f^2.

This whole process is analogous to the f-stop on a camera lens.  In this
case the pupil is the aperture stop and (f/d) is the f-number.  Note the
implicit assumption that the source exposes the entire pupil.  For small,
collimated sources, the total radiant power will reach the retina and the
source radiance may not be a useful concept.

As Paul says, Fun Stuff.

Dave Scherer
scherer@uiuc.edu