A thin converging lens with aperture ratio `D : f = 1 : 35 (D` is the lens diameter, `f` is its focal length) provided the image of a sufficiently distance object on a photographic plate. The object luminance is `L = 260 cd//m^(2)`. The losses of light in the lens amount to `alpha = 0.10` . Find the illuminance of the image.

Explanation

Refer to problem. If `a` is the area of the obect, then provided the angular diameter of the object at the lens is much smaller then other relevent angles like `(D)/(f)` we calacualte the light falling on the lens as `LA (piD^(2))/(4s^(2))`
where `u^(2)` is the object distance squared. If `beta`is the transverse magnification `(beta = (s')/(u))` then the area of the image is `beta^(2) A`. Hence the illuminance of the image (also taking account of the light lost in the lens)
`E =(1 - alpha) LA (piD^(2))/(4s^(2))(1)/(beta^(2)A) = ((1- alpha)piD^(2)L)/(4f^(2))`
since `s' = f` for a distant object. Substituation gives `E = 15 1x`.