Find the mean illuminance of the irradiated part of an opaque sphere receiving
(a) a parallel luminous flux resulting in illuminance `E_(0)` at the point of normal incidence,
(b) light from a point isotropic source located at a distance `l = 100 cm` from the cnetre of the sphere, the radius of the sphere is `R = 60 cm` and the luminous intensity is `I = 36 cd`.

Explanation

Mean illuminance
`= ("Total luminous flux incident")/("Total area illuminated")`.
Now, to calacuate the total luminous flux incident on the sphere, we note that the illuminance at the point of normal incidence is `E_(0)`. Thus the incident flux is `E_(0).piR^(2)`. Thus
Mean illuminance `= (pi R^(2).E_(0))/(2piR)`
or `lt E gt = (1)/(2) E_(0)`.
(b) The sphere subtends a solid angle
`2pi (1-cos alpha) = 1pi (1-sqrt(l^(2)-r^(2))/(l))`
at the point source and therefore recives a total flux of
`2pi I (1-sqrt(l^(2)-r^(2))/(l))`
The area irradiated is : `2pi R^(2) int_(0)^(90-alpha) sin theta d theta = 2pi R^(2) (1-sin alpha) = 2piR^(2) (1-(R )/(l))`
thus `lt E gt = (I)/(R^(2)) = (1-sqrt(1-(R//l)^(2)))/(1-(R )/(l))`
Subsitituting we get `lt E gt = 50` lux.