Question

Problem 3: Intensity from a distant light source (25 points) Consider an isotropic light source that emits EM radiation with wavelength A and power Po. (Isotropic- uniformly in all directions). In addition, consider a detector with area a, facing the source which is a long distance R away. (a) Derive a formula for the amplitude of the electric field at the detector (b) Derive a formula for the number of photons per unit time that strike the detector. What is the relationship between the rate (number per second) at which photons hit the detector, and the amplitude of the E-field at the detector? (c) Suppose the source is a 1.5-watt green LED bulb (that's about the brightest single LED bulb commercially available). Also suppose the detector is a human eye that is one mile from the source on a dark night. About how many photons per second enter the eye? What is the amplitude of the E-field at the eye.

Answer 1

3]

a]

The intensity received is:

This intensity is the time averaged Poynting flux magnitude

$\frac{1}{2\mu_oc}E_o^2=\frac{P_o}{4\pi r^2}$

=> $E_o=\sqrt{\frac{P_o\mu_oc}{2\pi r^2}}$

this is the amplitude of the electric field at the detector.

b]

Power at the detector is:

$P=\frac{P_o}{4\pi r^2}(a)$

and each photon has an energy, $E=\frac{hc}{\lambda}$

so, the number of photons striking the detector per unit time is:

$n = \frac{P}{E}=\frac{\frac{P_o}{4\pi r^2}a}{\frac{hc}{\lambda}} = \frac{P_oa\lambda}{4\pi r^2hc}$

but $E_o^2=\frac{P_o\mu_oc}{2\pi r^2}$

so, $n=\frac{E_o^2a\lambda}{2\mu_oc^2h}$

c] P_o = 1.5 W

r = 1 mile = 1609 m  

so, calculating the amplitude gives, E_o = 0.005896 V/m

therefore, the number of photons reaching the eye is:

$n=\frac{(0.005896)^2(\pi [1\times 10^{-3}]^2)(550\times 10^{-9})}{2\mu_oc^2h}\approx 400$

where the area is calculated by taking the diameter of a human eye to be 2 mm (hence radius = 1 mm) and the wavelength is selected to be 550 nm.