Question #108200
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.
1
Expert's answer
2020-04-08T10:28:34-0400

(a) The irradiance EeE_e at the surface


Ee=Ier2=Φ4πr2=10043.14222E_e=\frac{I_e}{r^2}=\frac{\Phi}{4\pi r^2}=\frac{100}{4\cdot 3.14\cdot 2^2}\approx2 W/m2W/m^2


(b) The illuminance EvE_v at the surface


V(640nm)=0.107V(640 nm)=0.107


Ev=K(λ)Ee=685V(λ)Ee=6850.1072147E_v=K(\lambda)\cdot E_e=685\cdot V( \lambda )\cdot E_e=685\cdot 0.107\cdot 2\approx147 lxlx.








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