Question

A surface oriented perpendicularly is positioned 2 m away from a lightbulb emitting 100 W of radiant flux. (a) Calculate the irradiance Ee at the surface. b) If all 100 W is emitted from a red bulb at λ = 650 nm, calculate the illuminance Ev at the surface. Hint: use the following equation: photometric unit = K(λ)×radiometric unit. Where K(λ) is the luminous efficacy given by : K(λ) = 685V (λ) and V (λ) is the luminous efficiency obtained from a CIE luminous efficiency curve.

Answer 1

(a) The irradiance $E_e$ at the surface

$E_e=\frac{I_e}{r^2}=\frac{\Phi}{4\pi r^2}=\frac{100}{4\cdot 3.14\cdot 2^2}\approx2$ $W/m^2$

(b) The illuminance $E_v$ at the surface

$V(640 nm)=0.107$

$E_v=K(\lambda)\cdot E_e=685\cdot V( \lambda )\cdot E_e=685\cdot 0.107\cdot 2\approx147$ $lx$.