CHEM 4525: Physical Chemistry Laboratory I
Experiment 1
Experiment 1 – Wave Phenomena: Interference
1. Objectives
Observe the wave-characteristic of light using diffraction and interference. Apply the law of diffraction
to find the wavelength of the light and spacing between grooves on a “diffraction grating” (CD). Practice
data and error analysis and writing a laboratory report.
2. Background
2.1. Waves and the Quantum World
One of the most important concepts in quantum theory is that objects behave like waves: the waveparticle duality. First proposed by Louis de Broglie in 1924 (Nobel prize in 1929), it is the core of the wave
mechanics and the mathematical formalism that is integral to the description of the microscopic world. As
you (hopefully) learned in your Physical Chemistry class, the states of microscopic particles, electrons,
atoms, molecules, as well as sub-nuclear particles are described by a wavefunction, which obeys the
Schrodinger equation. The wavefunction, as its name along with the origins stemming from the waveparticle duality suggest, has properties that are characteristic of waves: namely, it can experience a wavelike interference. The interference is again crucial to many problems of quantum mechanics. In this
laboratory, you will use a visible light from a laser source to explore the phenomenon of interference. Note
that the wave-particle duality also applies to light: although the wave character was well known for centuries
before the dawn of the quantum theory, the explanation of the Photoelectric effect caused by particles of
light, called photons, by Einstein in 1905 [1] (which earned him his Nobel prize in 1921) was also one of
the foundational discoveries that inspired the development of quantum mechanics.
2.2. Interference
The electromagnetic waves (i.e. light), just like the quantum mechanical wavefunction, have two
essential parts: amplitude and phase. Mathematically, the light wave is often described as a complex
function:
E ( x, t )  E0 ei k x xt 
(1)
In (1) the E(x,t) is the electric field of the electromagnetic wave as a function of x (position - for simplicity
we assume the wave propagates in the x-direction) and time t, E0 is the amplitude and the term in the
exponent (kxx - ωt) is the phase, which we will collectively denote φ. Other symbols are: i = √-1, kx is the
x-component of the wave-vector,  is the angular frequency, x is position, and t is time. The wave-vector
is related to the wavelength of the light (), while the angular frequency is related to the number frequency
() and period of oscillation (T). These relationships should be familiar (see Prelab quiz question 1).
It is the phase of the wave that is responsible for interference. When waves combine, the resultant
wave depends on the phase difference of the component waves. For example, they may combine in such a
way that crests meet crests and valleys meet valleys or “in-phase”. The resultant wave has twice the
amplitude of the constituent wave (Figure 1 on the right). This is called constructive interference. It should
be clear (See Prelab question 2) that the same resultant wave is obtained if the component waves are shifted
with respect to each other by 2, 3, 4 or any integer multiple of . This is called the path difference Δx.
We can therefore write the condition for the constructive interference as:
Path difference for constructive interference = m (where m is an integer).
The phase difference Δφ as you will show in problem 4 is related to the path differnce by:
2x
 

1
(2)
(3)
CHEM 4525: Physical Chemistry Laboratory I
Experiment 1
where λ is the wavelength. In problem 4 you will also derive the condition for the phase difference to
achieve constructive interference.
+
=

Figure 1. Constructive interference of two waves
The opposite scenario, called destructive interference, occurs when the constituent waves are “out-ofphase”, which means that crests meet valleys and valleys meet crests (Figure 2) The resultant wave (on the
right) has zero amplitude. Again, it should be obvious that the same resultant wave is obtained if the phase
difference of the component waves is 32 , 52 , 72  or any half odd integer multiple of :
Phase difference for total destructive interference =  m  12   (where m is an integer)
+
=
Figure 2: Destructive interference of two waves
Again, in problem 4 you will derive the corresponding condition for the phase difference.
2
(4)
CHEM 4525: Physical Chemistry Laboratory I
Experiment 1
The above examples illustrate what happens to the amplitude of the resulting wave upon superposition
of two waves, either in-phase (constructive) or out-of-phase (destructive). In reality, however, we are not
interested in the amplidute as much as in intensity. Intensity is the sqare absolute value of the electric field
(1) and is what is actually measureable in practice:
I  E ( x, t )
2
(5)
Note the correspondence between the amplitude E(x, t) of the electric field and the quantum mechanical
wavefunction ψ, which are not observable, and between their square absolute values | ψ|2 and I = |E(x, t)|2
which correspond to observable quantities. As you will show (problem 5) for a single wave the intensity is
just the square absolute value of its amplitude:
I 0  E0
2
(6)
While for superposition of two waves, with the same amplitudes E0 and frequencies ω the resulting intensity
is:
I  2I 0 1  cos    2I 0 1  cosk x x 
(7)
where Δφ is the phase difference between the two waves, and Δx is the path difference. In problem 5 you
will also show that intensity maxima occur for Δφ= m.
2.3. The double slit experiment
The first wave theory of light was proposed by Christian Huygens (1629 – 1695). It was not until 1801,
however, that the English scientist Thomas Young (1773 – 1829) performed an experiment that
demonstrated the wave nature of light. In Young’s double slit experiment monochromatic light illuminates
two slits in an opaque barrier. The slits are separated by a distance d. Light from the slits falls on a screen
a distance D away. It is assumed that D >> d. The image that appears on the screen is a set of light and dark
bands called interference fringes. The pattern the fringes form is called an interference pattern. The
following figure (Figure 3) shows a typical double-slit pattern. The arrows mark the central bright fringe,
which is taken to be bright fringe 0.
Figure 3. The fringe pattern of a typical double slit diffraction experiment.
The interference here arises due to another important characteristic of waves: diffraction. Diffraction
essentially means that the light can change direction. The diffracted light from a slit then combines
(interferes) with the light diffracted from the other slit. We can use geometry to locate the bright and dark
fringes of the double-slit diffraction pattern (Figure 4):
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CHEM 4525: Physical Chemistry Laboratory I
Experiment 1
m
3
2
l1
1

d
l2
0
-1
Screen
-2
D
Monochromatic
light of
wavelength .
-3
Figure 4: The double slit experiment.
We see that sin  
x
(if you don’t see it, it might be a good idea to review basic trigonometry). From
d
the discussion of the conditions for the constructive and destructive interference above, it can be shown
(see Prelab quiz question 3) that the bright fringes occur when:
sin   m

m  0,  1,  2,  3, ...
d
(8)
Likewise, for the dark fringes (destructive interference) one finds:
1

sin    m  
2 d

m  0,  1,  2,  3, ...
(9)
2.4. Many slits: optical grating
If one has more than two slits with equal spacing (d), the same equations for the maxima (eqn. 8) and
minima (eqn. 9) apply. The difference is that multiple slits will make the maxima much sharper and they
will appear much farther apart than for just two slits. A variation on a multiple slit setup, is the so-called
diffraction grating, where the light is not transmitted through slits, but reflected off of periodically spaced
grooves on a surface (Figure 5).
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CHEM 4525: Physical Chemistry Laboratory I
Experiment 1
Figure 5: Laminar diffraction grating.
If the grooves are sufficiently small compared to the wavelength of the light, the light will diffract and
interference maxima and minima will be observed. An easy to come-by diffraction grating is a compact
disc (CD).
3. Experimental
You will use a laser and three double slits with different inter-slit spacing to find the wavelength of
the laser light. Then you will use this information to determine the spacing between the grooves on the CD
and calculate how many grooves the CD has.
Warning: Laser light can hurt your eyes, even when reflected. Follow the common sense and other safety
precautions. Do not lean into the beam and in any way put your eyes into the beam line. Remove all objects
that can reflect the light from your hands, such as watches and rings. The CD experiment requires that we
reflect the light (off of the CD). Be very careful and make sure that either you or anybody else is not in the
reflected beam path.
1. Set up the double slit experiment with the He-Ne laser, one of the three double slits on the slide and a
screen far down-field (on the wooden desk near the door) to allow precise measurement of the interference
fringe spacing. When you observed the interference pattern, use a paper and pencil to mark the fringe
positions. You will have to measure the slit spacing and the distance between the double slit and the screen.
Use a ruler for the former and yard stick or tape measure for the latter. Note the uncertainties of your
measurements as you will need them for the error analysis.
2. Repeat for the other two double slits on the slide.
3. Set up the CD experiment (Figure 6). Fix the CD in the holder so that the laser beam hits it to the left
(when facing the CD) of the center and that the beam reflects off the CD and onto the paper taped to the
two wooden stands. (see CD setup). Be extremely careful not the catch the reflected light in your face.
The two boards should be separated by a small slit to let the laser light through. Adjust the CD so that the
reflected laser light hits the front of the laser directly below the aperture where the laser light comes out.
Make sure the light DOES NOT go into the laser (into the hole).
4. Mark the centers of the two diffuse maxima on either side of the laser, turn the laser off and measure
their distance. Again note the uncertainties of your measurements. You will need a few more distances to
be able to calculate the groove density and a number of grooves on the CD, but these should be obvious
and easily done with a ruler.
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CHEM 4525: Physical Chemistry Laboratory I
Experiment 1
Figure 6. Experimental setup for the diffraction off of the CD.
4. Data Analysis
1. Use equation (8) with the known slit spacing and distance from the slits to the board to determine the
wavelength of the laser for all three values of d (three double slits) along with the estimated errors.
2. Compare your results for the three different double slits (if you see any systematic differences in the
precision, depending on the slit spacing, pin it out and rationalize in the Discussion of your report). Compare
your result with the correct, known wavelength of the He-Ne laser. Discuss the accuracy of your results or
lack thereof and potential reasons for any discrepancies.
3. Using the known laser wavelength and the results of the CD measurement, calculate the groove spacing
of the CD. Make sure you use the formula (eqn. 8) correctly.
4. Calculate the number of grooves on the CD.
References:
1. Albert Einstein: “Concerning an Heuristic Point of View Toward the Emission and Transformation of
Light. Annalen der Physik 17 (1905): 132-148.
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CHEM 4525: Physical Chemistry Laboratory I
Experiment 1
Pre-lab quiz questions
1. What is the de Broglie wavelength for an object? Calculate a de Broglie wavelength for an electron with
the kinetic energy of 10 eV (eV = electronvolt) and for a benzene molecule undergoing thermal motions at
300 K (vrms ≅ 300 m.s-1).
2. Write down the relationship between:
a) The angular frequency and number frequency, including units of each.
b) Wave vector (k) and wavelength
c) Wavelength and number frequency (for light, the speed of light = c).
3. a) What frequency (in Hz) corresponds to the wavelength of 450 nm? What color of light is that?
b) What frequency (in Hz) corresponds to the wavelength of 650 nm? What color of light is that?
4. Using (1) and the results of problem 2., derive the relationship between the phase difference and path
difference eqn. (3). Then using the conditions for the constructive and destructive interference (eqns. (2)
and (4)) derive the corresponding conditions for the phase difference.
5. Derive equations (6) and (7). Using the results of problem 2, show that intensity maxima occur when two
interfering waves satisfy the condition in eqn. 3 and are equal to 4I0. Also show that the minima of zero
intensity occur when the condition in eqn. 4 holds.
6. Using the geometry in Figure 4, derive the condition for the diffraction maxima (eqn. 8) and minima
(eqn. 9).
7. What is the difference between light diffraction and interference? Why is Young diffraction experiment
called “diffraction experiment” when we actually observe bright and dark spots due to interference?
8. Pick the correct statement(s). Using the same light source in the diffraction experiment, if the distance
between slits is doubled while everything else stays the same, the spacing between interference fringes will:
a) Double
b) Stay the same
c) Decrease by a factor of two
d) Can’t answer because the wavelength of light must be known
e) Can’t answer because the distance between the double slit and the screen has to be known
f) Can’t answer because the slit width has to be known.
g) Can’t answer because all the factors in d) – f) have to be known.
9. Pick the correct statement(s). Using the same double slit for the diffraction experiment, if you double the
frequency of the light and everything else stays the same, the spacing between interference fringes will:
a) Double
b) Stay the same
c) Decrease by a factor of two
d) Can’t answer because the wavelength of light must be known
e) Can’t answer because the distance between the double slit and the screen has to be known
f) Can’t answer because the slit width has to be known.
g) Can’t answer because all the factors in d) – f) have to be known.
10. Why is it not prudent to look into the laser beam? Why should you remove any shiny objects, such as
watches, rings etc. from your hands when you experiment with lasers?
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CHEM 4525: Physical Chemistry Laboratory I
Experiment 1
11. Why in the setup on Fig. 6 does the light have to hit the laser head below the output aperture? (Hint:
how can you tell if the CD is squarely facing the laser beam?)
12. Why in the setup on Fig. 6 does the laser beam have to hit the CD to the side of center and not above,
below or anywhere else? (Hint: it has something to do with the direction of the grooves).
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