# 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. - Physics

Sum

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.

#### Solution

Given,

The radiant flux of the light of wavelength 555 nm is 40 W.

The radiant flux of the light of wavelength 600 nm is 30 W.

The relative luminosity at 600 nm is 0.6.

(a) Total radiant flux = radiant flux of 555 nm part of light + radiant flux of 600 nm part of light

= 40 W + 30 W = 70 W

(b) Total luminous flux = luminous flux of 555 nm part of light + luminous flux of 600 nm part of light

= 1 × 40 × 685 + 0.6 × 30 × 685

= 39730 lumen

(c)

"Luminous efficiency"="Total luminous flux"/"Total radiant flux"

=39730/70=567.6 "lumen/W"

So, the luminous efficiency is 568 lumen/W.

Concept: Light Process and Photometry
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 22 Photometry
Q 6 | Page 455