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A light detector (your eye) has an area of 2.0010-6m2and 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 exactly4.000s-1at what power does the emitter emit light?

Short Answer

Expert verified

The emitter emits power 1.1脳10-10W.

Step by step solution

01

Describe the expression of energy of the photon

The energy Eof a photon of wavelength is given by,

E=hc 鈥︹. (1)

Here, h is Planck鈥檚 constant, and c is the speed of light.

02

Determine the power that emitter emit

Let Remitbe the photon emission rate, then the absorption rate Rabs=4.00s-1in terms of the area of the eye Aabs=2.0010-6m2

Rabs=PabsEp

Since the detector absorbs only of the incident light, then the absorbed power is,

Pabs=80100IAabsRabsEp=0.8PemitAemitAabsPemit=RabsEp0.84r2Aabs....(2)

Substitute the below values in eq. 1

=500nm.h=6.62610-34J.sc=3108m/s

Ep=6.62610-34J.s3108m/s500nm=6.62610-34J.s3108m/s500nm1m109nm=3.9310-19J.

Substitute the below values in eq. 2.

Rabs=4.00s-1Ep=3.9310-19J.Aabs=2.0010-6m2r=3m

Pemit=4.00s-13.9310-19J0.843m22.0010-6m2=1.110-10W

Therefore, the emitter emits power 1.1脳10-10W.

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