Question

Oxygenated hemoglobin absorbs weakly in the red (hence its red color) and strongly in the near infrared, whereas deoxygenated hemoglobin has the opposite absorption. This fact is used in a "pulse oximeter" to measure oxygen saturation in arterial blood. The device clips onto the end of a person's finger and has two light-emitting diodes—a red (632 nm) and an infrared (930 nm)—and a photocell that detects the amount of light transmitted through the finger at each wavelength.

(a) Determine the frequency of each of these light sources.

red Hz

infrared Hz

(b) If 66% of the energy of the red source is absorbed in the blood, by what factor does the amplitude of the electromagnetic wave change? Hint: The intensity of the wave is equal to the average power per unit area as given by I = Emax2 2μ0c = c 2μ0 Bmax2.

Answer 1

Lets calcualte the frequency of both red and infrared, we know that light is emitting from emitting diodes so

formula here propose by c = ( c = 3.8 * 10^8 m/s ) c is stand for speed light in vaccum

therefore

fred = c/ red ( f = frequency, = wavelength )

fred =  3.8 * 10^8 /632 * 10^-9 m = 6* 10^14 Hz

similarly

finfard =  3.8 * 10^8 /930* 10^-9 m = 4 * 10^14 Hz

b) 66% energy absord of red source therefore first calculate the energy which absorb in blood

E = hf ( E = energy , h = plank constant , f = frequency )

E = 6.63*10^-34 * 6* 10^14

= 39.78 * 10-20 joule

only 66% energy has been absorb in blood

= 40 *66/100 = 26.4 * 10^-20joule ( 39.78 ~ 40)

Now Intensity porportional to square of amplitude

I

Now calculate the Intensity is I in question already given formula of intensity

therefore

I = Emax2 2μ0c

= (26.4* 10^-20 ) ^2 2* 4*3.1410^-7*3.8 *10^8

= 6.65*10^-35 W/m2

therefore Amplitude is

A = = 8.15 * 10^-17 meter^2