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.

Correct answer: D. 568 lumen W^ -1 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 .

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Photometry

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Photometry

A source emits light of wavelengths 555 nm and 600 nm . The radiant...

A source emits light of wavelengths

555 \text{ nm}

and

600 \text{ nm}

. The radiant flux of the

555 \text{ nm}

part is

40 \text{ W}

while that of the

600 \text{ nm}

part is

30 \text{ W}

. The relative luminosity at

600 \text{ nm}

0.6

. Find the luminous efficiency.

Options

A $685 \text{ lumen W}^{-1}$
B $411 \text{ lumen W}^{-1}$
C $397 \text{ lumen W}^{-1}$
D $568 \text{ lumen W}^{-1}$

Correct answer

D. $568 \text{ lumen W}^{-1}$

Step-by-step solution

Total radiant flux is given by

\Phi_{e} = 40 + 30 = 70 \text{ W}

.

The relative luminosity of

555 \text{ nm}

light is

1

, and the maximum luminous efficacy is

685 \text{ lumen W}^{-1}

.

Luminous flux of the

555 \text{ nm}

part is

\Phi_{v1} = 40 \times 685 \times 1 = 27400 \text{ lumen}

.

Luminous flux of the

600 \text{ nm}

part is

\Phi_{v2} = 30 \times 685 \times 0.6 = 12330 \text{ lumen}

.

Total luminous flux is

\Phi_{v} = 27400 + 12330 = 39730 \text{ lumen}

.

Luminous efficiency is

\dfrac{\Phi_{v}}{\Phi_{e}} = \dfrac{39730}{70} \approx 568 \text{ lumen W}^{-1}

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