If `s =` object distance, `s'` average distance, `l =` luminance of the sounce, `Delta S =` area of the source assumed to be a plane surface held normal to the principal axis, then we find for the flux `Delta Phi` incident on the lens
`Delta Phi = int L Delta S cos theta d Omega`
`= L Delta S int_(0)^(oo) cos theta 2pi sin theta d theta = L Delta S pi sin^(2) alpha = L Delta S (pi D^(2))/(4s^(2))`
Here we are assuming `D lt lt s`, and ignoring the variation of `L` since `alpha` is small
Then if `L`' is the luminance of the image, and `Delta S' = ((S')/(S))^(2) Delta S` is the area of the image then similarly
`L'Delta S' (D^(2))/(4s'^(2)) pi = L' DeltaS(D^(2))/(4s^(2))pi = L Delta S(D^(2))/(4s^(2)) pi`
(b) In this case the image on the white screen from a Lambert source. Then if its luminance is `L_(0)` its luminosity will be the `pi L_(0)` and
`piL_(0)(s'^(2))/(s^(2))Delta S = L Delta S (D^(2))/(4s^(2)pi`
or `L_(0) propD^(2)`
since `s'` depends on `d, s` but not on `D`.
