Physics & Color
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Physics 102
Physics & Color
Suzanne Amador Kane, Physics Department, Haverford College © 2011
Introduction
Today you will explore some of the physics that underlies the science of color.
The absorption of light & Beer’s law:
We quantify the absorption of light by matter using spectrometers, devices that measure the absorption
(or the emission) of light by a well-defined sample of the light-absorbing material of interest. The
operation of one type of spectrometer is shown in Fig. 1 below. (Other spectrometers operate using the
same Fourier transform ideas you encountered in our Sound Lab. If you measure instead the timevarying intensity of light, you can use a Fourier transform to determine the frequency spectrum of the
light; this is currently feasible only for lower frequencies.) ). The basic idea is simple. You start with a
source of light with many different wavelengths. A monochromator is then used to select only a limited
range of wavelengths (say,  to  + ). This light’s intensity, Io, is measured immediately after the
monochromator. The light is passed through the sample, leaving only intensity I to be measured by a
light detector. The ratio I/Io is plotted as a function of wavelength,  to make up what is called an
absorption (or emission) spectrum (plural spectra).
Figure 1. Schematic drawing of the operation of a spectrometer
This ratio depends on many things: the concentration of absorbing molecules in the sample, how thick
and sample is and how absorbing each molecule is. The mathematical relationship that defines the way
in which all these quantities determine how light is absorbed is called Beer’s Law (also called the Beer
Lambert Law; it’s a scientist’s name, not the beverage!)
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Let’s assume we have a dilute solution of our absorbing molecules upon which we shine an initial
intensity I. Light is partly absorbed and partly transmitted by the solution. For a very dilute solution
that is very weakly absorbing, the amount of light absorbed depends upon the number of molecules in
solution. This is proportional to concentration, c, and the length of the light’s path in the solution, x.
(Fig. 2) The intensity of light is diminished by:
I = - c x  I
Eq. 1
Where  (called the extinction coefficient) is some constant of proportionality that describes how
absorbing our individual molecules are at this wavelength of light.
Figure 2: Geometry for Beer’s Law.
This relationship holds for so few molecules in solution that the absorption of light is very tiny. What
about the case where appreciable absorption takes place? In that case we can rewrite our equation as:
I / x = - c  I
Eq. 2
Which becomes in the limit as x  0,
dI / dx = - c  I
Eq. 3
The solution to this equation is an exponential:
I(x) = Io exp(-c  x) ,
Eq. 4
where Io = the incident intensity (the value when x = 0). This relationship describes how intensity, I(x),
varies with distance x inside an absorbing solution with concentration c and extinction coefficient . The
quantum physics (and chemistry!) of the absorbing molecules is described by , while c and x define how
many molecules the light encounters.
This same exponential fall-off of radiation with distance holds true in x-ray imaging, nuclear medicine
and radiation therapy, laser surgery and ultrasound imaging. As a result, it is of widespread importance
in medicine as well as physics.
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Problem 1: (2 pts) (a) Show that you get Eq. 3 when you take the derivative of Eq. 4. (b) Then, solve for
the absorbance, A, a quantity used often in spectroscopy, defined as: A = - ln (I/Io) ; find an equation for
A in terms of x, c and . (Different scientific publications report spectroscopic results in terms of A,  and
cso it is useful to get these definitions straight.)
Some background on color vision
After what you’ve learned in class, it is nature to assume that color is a physical phenomenon. After all,
all textbooks reproduce a spectrum of electromagnetic radiation that assigns colors to each wavelength
in the visible region. Really, though, color is a psychophysical phenomenon: a complex physiological
and mental reaction to physical inputs. It depends on the organism perceiving the light (down to which
organism or even which individual person we’re talking about) and the conditions in which the light is
perceived (low vs. high intensity, etc.)
Below (Fig. 3 (a)) we see a plot of the extinction coefficient vs. wavelength (another way to plot the
absorption spectrum) for the visual receptor cells of the human eye: the three types of cone cells
(plotted in red, green and blue) and rod cells (in black). To be quite clear, the absorption spectrum is
greatest (smallest) where the molecules are most (least) absorbing. Our perception of color depends in a
complex way upon the absorption of light of all of the three types of cone cell exposed to a given set of
illuminating wavelengths of light.
Figure 3. (a) Absorption spectra of the visual receptor molecules in the cone cells (colored lines) and rod
cells (dashed line) of the human eye. (b) Emission spectra of an argon laser (green) and white lamp
source (similar to the Sun’s emission spectrum).
Along the x-axis (wavelength) is a set of labels that indicate what color a human would sense if only that
wavelength were present. The first thing you ought to note is that the absorption spectrum for each
type of cell is very broad. Each of the rod and cone cell types responds to a wide range of wavelength of
light, and there is considerable overlap. For example, 500 nm light is detected by all of them, at varying
levels of sensitivity. However, each type of cell also has a peak absorption range of wavelengths. The
light labeled “red” (650 to 700 nm) is strongly absorbed by the red cone receptor pigments. We call
light ranging from 400 nm to about 475 nm “violet” (not purple, but intensely blue) “indigo” and “blue”,
and indeed that is where the blue cone cells are most strongly absorbing. To the surprise of most
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people, the middle cone cell pigment is centered on the “green” part of the visible spectrum. Based on
their experiences with mixing paints, they are expecting it to be centered on “yellow” in the spectrum,
which is actually about 570 nm—close to the peak absorption of the red cone pigments! What gives?
Stage lighting indeed uses three separate lightbulbs, with red, green and blue light, to reproduce all
possible colors, as do television and computer monitors. We will explore this surprising fact in our lab.
(If you are confused about why the third lightbulb isn’t yellow, you will see why in lab!) What we call
“white” light is in fact any mixture of wavelengths that stimulates our eye’s cones the same way that
sunlight does. We can see this by examining a plot of the intensity of light emitted by a source such as a
lamp, laser or the Sun. (Fig. 3(b)). Now the peak of the emission curve indicates the wavelength at
which the most light is emitted. For a laser, this emission spectrum is a very sharp peak at the laser’s
well-defined wavelength. For a source of white light such as the Sun or a white incandescent lightbulb,
the emission spectrum has intensity at many wavelengths, with the characteristic shape shown in Fig.
3(b). Our eyes responds to this by absorbing light with all our cone cells (to varying degrees). It’s that
exact ratio of how much each color of cone cells responds that makes us think of Sunlight as “white”.
“Black” is the absence of any significant response of any of our visual receptor cells to light. There may
be light present, but it would have to be at a wavelength at which our eyes do not respond! For
example, we cannot see ultraviolet or infrared light, since these forms of electromagnetic radiation have
wavelengths too short or too long (respectively) for our eyes to detect. Note that the laser emission
spectrum in Fig. 3(b) is sharply peaked around a single wavelength; humans with normal color vision
consequently perceive laser light as having a very pure color determined by the correspondence of
wavelength to color on the x-axis of Fig. 3(a).
Problem 2: (1 pt) (a) While the cone cells have different color responses, we have only one type of rod
cell. The rod cells are, however, more sensitive at low light levels, so our vision gradually shifts over
from relying on the response of cone cells to light in daylight to the response of rod cells at dusk in dim
lighting and at night. What of the following color objects will appear “brightest” in dim lighting? Blue,
green or red objects of equal intensity in the daylight.
(b) Another effect that is easy to observe yourself involves exposing your eyes to an intense color of
light, so that it exhausts the color receptor molecules stimulated.
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If you exhaust the blue cone cells, they do not respond the next time you see white light, so you see only
the white light minus its blue component. We can use this phenomenon to explain why operating room
staff often wear drab green or blue “scrubs” (operating room outfits), rather than nice clean-looking
white ones. To see why, try staring hard for about a minute (really—do it for a long time!) at a bloodlike red image like this one (cover up the green square first), then staring first at a piece of white paper,
then second at the green region below. Explain what would happen to a surgeon who looked at a
bloody operating scene, then at either white or drab green clothing. What advantage does the drab
green have? (Just describe what you see—you don’t have to explain it in terms of the cone cells
absorption spectra!)
Experiment 1: Color Mixing experiments do only this experiment today!
Let’s do the simplest color perception experiment possible now. You have before you a black optical rail
on which you can mount a red and green diode laser, plus several plastic squares with different colors:
red, green, blue, yellow, orange, magenta (purple). Record what you see when you shine the red and
green laser through each square. Explain whether objects that appear (for example) red 1) absorb or 2)
transmit (or reflect) red light. Be sure you understand this basic fact well before proceeding! Confusion
about this simplest of color ideas can make the rest of the lab hard!
To understand how we relate our perception of color to the results of this last experiment, and to
absorption spectra, let’s now take absorption spectra of various solutions of food dye: red, green, blue
and yellow. Calibrate the spectrometer by choosing Experiment/Calibrate and inserting a cuvette with
water only first. Then, using the cuvettes provided, insert each in turn into your spectrometer (first
removing of course the fiber optic insert from the emission experiments.) Take an absorption spectrum
for each food dye and save each. Explain how your absorption spectrum relates to your observations for
the filters above.
Now, place on your optical rail the color mixing apparatus. This is a device with red, blue and green light
bulbs. A lens focuses the light from all three bulbs onto a white screen. (Fig. 4) The intensity of each
color can be separately controlled with knobs on the back of the apparatus. (Fig. 4) The colors mix by
overlapping on the white screen, which reflects part of the light back to your eye. You perceive colors
corresponding to mixtures of these three sources. Using this idea (and perhaps using objects to cast
shadows on the screen to block one or more of the red, green and blue light bulbs), answer the
questions on the report form.
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Figure 4: Left: Color mixing apparatus. Red, green and blue lights are emitted from a source and shine
onto a white screen. Colored shadows form when one or both of the light sources are blocked. Right:
knobs on the back of the sources allow you to adjust the intensities of each color separately. (Vernier
Instruments)
Experiment 1: Physics & Color (15 pts total)
Prelab Problem 1 (2 pts)
a)
b)
Prelab Problem 2 (2 pts):
a)
b)
3) (3 pts) How much laser light is transmitted for each laser/filter combination? H = high (almost all);
M = moderate (the intensity is reduced, but it’s still pretty bright); L = very little is transmitted
Use the colored plastic squares except where indicated (the square are better quality filters)
Color
Filter
Red
Blue
Green
Yellow
Orange
Magenta
(paddle)
Cyan
(paddle)
Red Laser
Green Laser
Blue Laser (you may have to
share!)
4) (1 pts) Now let’s reason based on these results. When an object has a particular color, such as red,
what does that mean in terms of the color absorbed by the object? Reflected or transmitted by the
object (i.e., does it or does it not absorb red appreciably? Does it or does it not absorb (visible) colors
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other than red?) Be specific. Make a very approximate sketch of the absorption spectrum of a red filter.
(Just show where the absorption is very high and very low)
5) (3 pts) Using your color mixing apparatus, explain what color you get when you mix the following
colors. The lightsource has three bulbs (blue, red, green) with intensities controllable by knobs in the
back. Start with the knobs fully on (fully clockwise). Place your color source at one far end of the silver
optical rail and the white screen at the other far end. Place the lens 25 cm from the source and the
screen about 95 cm from the source (approximately). Now you will see regions illuminated by one or
more lightbulbs. By turning off and on each lightbulb and by examining the regions of overlap,
determine which regions correspond to each bulb and mixtures of each bulb
(a) What light bulbs contribute to these mixtures of lightbulbs? Explain briefly using a drawing how you
determined your answers.
Red + Green =
Blue + Red =
Green + Blue =
Red + Green + Blue =
(b) Using your results from how you combined colors in this part to make yellow, explain why the
transmission of the yellow filter above makes sense. Also, explain how this agrees with light of the
wavelength around 575 nm also appearing yellow (see Figure 3a above—what is that wavelength doing
to your eye’s visual receptors?)
6) (4 pts) Concluding questions: You should be able to answer these based entirely on our exercises
in class today and this writeup.
(a) Why can’t you make a laser that emits a very well-defined wavelength that appears magenta
(purple) to the human eye? (Note that magenta light is not the same as short-wavelength “violet” light.
The part of the spectrum called “violet” actually appears very deep blue (not purplish) to humans with
normal color vision; you can make violet lasers!)
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(b) In laser surgery, the goal is to use laser light to heat tissue in order to make an incision (cut) or to
destroy pathological tissue (such as a tumor) or to seal off (cauterize) blood-vessels. However, as you’ve
seen, laser light will only be absorbed by parts of the body with the correct absorption spectrum. What
laser would you choose to operate on a tissue that contains mostly blood as the pigment (lightabsorbing chemical) of interest? You can do this exercise using only your reasoning above; you do not
need any information about physiology, biology or medicine!
(c) Stare fixedly at the pattern below for a minute, then look at a white wall or piece of paper. Using
the ideas from Prelab problem 2(b), explain what you see. (If you can’t see a colored afterimage, ask a
lab partner or fellow student to do this for you, then explain their experience.)
This startling effect is apparently due to the eye responding by “seeing” the same pattern, but now in
the colors of the visual receptors that were not stimulated. (The white light is still stimulating all of the
visual receptors, but some are no longer responding.) Explain why the afterimage has the perceived
color, using your ideas from the color exercises. (Hint: One of the odd flag’s colors is cyan = blue +
green.)
(d) Printers use inks that are colored Cyan (bright whitish blue-green like in the flag above), Magenta
and Yellow to form all possible color combinations for the human eye. These inks take white light from
the room (that would ordinarily reflect as white light from the white paper) and filter them in the same
way that your color filters act on the lasers in this lab. Thus, they remove red, green and blue
components from the white light in varying combinations. Explain how you could combine these three
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colors (Cyan, Magenta, Yellow) to make up Red, Green and Blue colors. Give the relevant combinations
and explain why they work, using your observations above. Try out this out using combinations of the
color filters on the Color Paddle Sets (look at white light through magenta + yellow, etc.) Do your
theories agree with what you actually see?
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Experiment 2: Beer’s Law wait until I tell you whether we’ll do this experiment or not—no need to
put on the problem set
Here, rather than using a light-absorbing solution, we will use light-absorbing pieces of plastic:
transparent plastic filters that absorb a constant fraction of the light incident upon them. The number,
N of plastic filters you place in the beam of intensity I will play the role of c x in our equation for Beer’s
law.
I(x) = Io exp(-N 
If we take the logarithm of both sides of this equation, we get:
log (I(x)) = log (Io exp(-N log (Io - N 
If you have green filters, use a green laser; if you were given red filters, use a red laser. Measure the
intensity of transmitted light with 0, 1, 2, 3 and 4 filters inserted into the laser beam, using the light
detector provided and LoggerPro. Using Origin, enter your results in a spreadsheet. Now plot your
intensity vs. N (# filters), using a semi-log scale (that is, plot log(intensity) vs. N; this is a feature you can
select in Origin using the axis controls.) How linear is your data on the semilog scale? What other
effects (if any) might be determining the transmitted intensity of light?
Now take the cuvette (a special solution chamber for spectroscopy) filled with a concentrated solution
of red food dye. Hold the cuvette in the laser beam so you can see how the intensity of light falls off
with distance. (Make sure you shine the laser through the clear, transparent sides of the cuvette.) You
should be able to visualize what it looks like when the light’s intensity falls off exponentially with
distance as it is absorbed by the red food dye.
Experiment 2: Beer’s Law
Attach your plot and a fit to an exponential. Does your data agree with the predictions of Beer’s Law
over the measured range? What is the slope and what quantity in Eq. does it correspond to best?
What other effects (if any) might be determining the transmitted intensity of light? For example, would
you expect it to follow an exponential behavior down to extremely low light intensities? What
limitations might you encounter, and how would the curve deviate from exponential? (Hint: What if
the solution had dust in it?)
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Equipment list:
For each setup:





Optical rail (Vernier or PASCO)
Light sensor (Vernier or PASCO)
Red and green diode lasers (Vernier or PASCO)
Color mixer kit (Vernier)
Plastic color filter sets (Educational Innovations or Scientificsonline.com)
Physics 102