Explanation 1) Concept: We first find the rate of absorption and rate of emission in terms of photon flux. After taking the ratio of this two quantities, we can find the rate of emission in terms of rate of absorption. Substituting this value in the rate of emission formula, we can find the power of the emitter. 2) Formula: i) E p h = h c λ ii) R e m i t = P e m i t E p h 3) Given: i) Wavelength of sun light λ = 500 n m ii) Area A a b s = 2 × 10 - 6 m 2 iii) Distance of source from detector r = 3 m iv) Rate of absorption R a b s = 4 p h o t o n s / s 4) Calculations: The rate at which photons are absorbed by the detector is related to the rate of photon emission by the light source as follows:. Let I be the photon flux a distance r from the source. Since the photons are emitted uniformly in all directions, we have R e m i t = I A R e m i t = 4 π r 2 I … … … … … … … … … … … … … … … … … … … … … … … … … . 1 ) The detector absorbs the 80 % of flux. R a b s = 0.80 I A a b s … … … … … … … … … … … … … … … … … … … … … … … … … … 2 ) Taking the ratio of 1) and 2) and solving for R e m i t , we get, R a b s = 0.80 A a b s 4 π r 2 R e m i t R e m i t = 4 π r 2 0

Fundamentals of Physics, Volume 1, Chapter 1-20

10th Edition

ISBN: 9781118233764

Author: David Halliday

Publisher: WILEY

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Chapter 38, Problem 7P

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The power with which the emitter emits light.

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A light detector (your eye) has an area of 2.00*10-6 m2 and absorbs 80% of the incident light, which is at wavelength 500 nm. The detector faces an isotropic source, 3.00 m from the source. If the detector absorbs photons at the rate of exactly 4.000 s-1, at what power does the emitter emit light?

A light detector has an area of 3.1m2 and absorbs 53.9% of the incident light, which is at wavelength 682.3nm. The detector faces an isotropic source, 2.1m from the source. If the detector absorbs photons at the rate of exactly 6photons/s , at what rate( in photons/s) does the emitter emit light?

A flashlight emits 1.0W of light energy. Assuming a frequency of 5.2 x 10^14 Hz for the light, determine the number of photons given off by the flashlight per second.

Chapter 38 Solutions

Fundamentals of Physics, Volume 1, Chapter 1-20

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The power with which the emitter emits light.

Solution Summary: The author explains how the power of the emitter is 1.1times 10-10W.

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Chapter 38 Solutions

To find:

The power with which the emitter emits light.