A small lamp having the form of a unfromly luminous sphere of radius `R = 6/0 cm` is suspended at a height `h = 3.0 m` above the floor. The luminance o fthe lamp is equal to `L = 2.0. 10^(4) cd//m^(2)` and is independent of direction. Find the illuminance of the floor directly below the lamp.

Explanation

See the figure below. The light emitted by an element of the illuminant towards the point `O` under consideration is
`d Phi = LdS d Omega cos (alpha + beta)`
The element `ds` has the area
`dS = 2pi R^(2) sin alpha d alpha`
The disatnce
`OA = [h^(2) + R^(2) - 2hR cos alpha]^(1//2)`
We also have
`(OA)/(sin alpha) = (h)/(sin(alpha + beta)) = (R )/(sin beta)`
From the diagram
`cos (alpha + beta) = (h cos alpha - R)/(OA)`
`cos beta = (h - R cos alpha)/(OA)`
if we imagine a small area `d sum` at `O` then
`(d sum cos beta)/(OA^(2)) = d Omega`
Hence, the illuminance at `O` is
`int (d Phi)/(d sum) = int L2pi R^(2) sin alpha d alpha ((h cos alpha - R)(h - R cos alpha))/((OA)^(4))`

The limit of `alpha = 0` to that value for which `alpha + beta = 90^(@)`, for the light is emitted tangentially. Thus
`alpha_(max) = cos^(-1)(R)/(h)`.
Thus `E int_(0)^(cos^(-1)((R)/(h))) L.2 pi R^(2) sin alpha d alpha ((h - R cos alpha)(h cos alpha - R))/((h^(2) + R^(2) - 2h R cos alpha)^(2))`
we put `y = h^(2) +R^(2) - 2hR cos alpha`
So, `dy = 2hR sin alpha d alpha`
`E = underset((h - R)^(2))overset(h^(2) - R^(2))int L.2 pi R^(2) (dy)/(2hR) (((h - (h^(2).+R^(2) - y))/(2h))((h^(2) + R^(2) - y)/(2R)-R))/(y^(2))`
`= (L.2 pi R^(2))/(8h^(2) R^(2)) underset((h - R)^(2))overset(h^(2) - R^(2))int ((h^(2) - R^(2) + y)(h^(2) - R^(2) - y))/(y^(2)) dy`
`= (piL)/(4h^(2)) underset((h - R)^(2))overset(h^(2) - R^(2))int [((h^(2) - R^(2))^(2))/(y^(2))-1]dy = (piL)/(4h^(2)) [-((h^(2) - R^(2))^(2))/y -y]_((h - R)^(2))^(h^(2) - R^(2))`
`= (piL)/(4h^(2)) [(h + R^(2)) - (h^(2) - R^(2)) - (h^(2) - R^(2)) + (h - R^(2))]`
`= (piL)/(4h^(2)) [2h^(2) + 2R^(2) - 2h^(2) + 2R^(2)] = (piLR^(2))/(h^(2))`
Substitution gives: `E = 25.1` lux