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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Solution

The radian flux of 555 nm part is 40 W and of the 600 nm part is 30 w.

(a) Total radiant flux

=40 W +30 W =70 W

(b) Luminous flux

=(L.Flux) 555 nm +(L.Flux) 600 nm

= $1 \times 40 \times 685 + (0.6) \times 30 \times 685$

=39730 lumen

(c) k=Luminous efficiency

$= \frac{T o t a l l u m i n o u s f l u x}{T o t a l r a d i a n t f l u x}$

$= \frac{39730}{70} = 567.6 l u m e n / W$

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