A source emits light of wavelengths 555 nm and 600 nm . The radiant flux of the 555 nm part is 40 W while that of the 600 nm part is 30 W . The relative luminosity at 600 nm is 0.6 . Find the luminous efficiency.

Step-by-step solution

Total radiant flux is given by _ e = 40 + 30 = 70 W . The relative luminosity of 555 nm light is 1 , and the maximum luminous efficacy is 685 lumen W ⁻¹ . Luminous flux of the 555 nm part is _ v1 = 40 685 1 = 27400 lumen . Luminous flux of the 600 nm part is _ v2 = 30 685 0.6 = 12330 lumen . Total luminous flux is _ v = 27400 + 12330 = 39730 lumen . Luminous efficiency is _ v _ e = 39730 70 568 lumen W ⁻¹ .