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A source emits light of wavelengths 555 nm and 600 nm. The radiant flux of the 555 nm part is 40 W and of the 600 nm part is 30 W. The relative luminosity at 600 nm is 0-6. Find (a) the total radiant flux, (b) the total luminous flux, (c) the luminous efficiency.

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The correct Answer is:
A, B, C

the radian flux of 555 m part is 40 W and of the 600 nm part is 30 W.
a total radiant flux
`=40W+30W=70W`
b. Luminous flux
`=(L.Flux)555nm+(L.Flux)600nm`
`=1xx40xx685+(0.6)xx30xx685`
`=39730lumen
c. Luminous efficiency
`=(Total luminous flux) /(total radiant flux) `
`=39730/70=567.6 lumen /W`
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