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INTERFERENCE AND DIFFRACTION OF LIGHT WAVES
The point of this experiment is:
•
To observe diffraction and interference of light waves.
•
To measure the wavelength of light from a helium-neon laser.
APPARATUS: Laser, slit-film with assorted slit sets, meter stick, vernier caliper,
cylindrical holder, and flat observation screen
Diffraction
L
ight is an electromagnetic wave consisting of oscillating electric (E) and magnetic
fields (B) at right angles to each other and to the direction of motion. As with any
wave that passes an object, it diffracts or bends around corners into the geometric
shadow region. The shadow does not have a sharp edge, but a fuzzy ripple pattern. This
becomes very pronounced if the wavelength is near the size of the object. For a narrow
slit of width b the diffraction intensity minima occur at angles (left or right of the normal)
given by
Eq. 1
Single Slit Pattern:
b sin(θ) = mλ
where m = ±1, ±2, ±3,...
λ/2
Screen
Incident
Waves
b/2
b
2
8
°
θ
Diffracted
Waves
FIGURE 1
Figure 1 shows waves moving to the right and passing through a slit of size b. As they
move through the slit they fan out in all directions. In the figure two waves, one from the
slit edge and the other from the center, interfere destructively because they are out of
phase by 180˚ (π radians). At this angular position one would see little light. Equation 1
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explains that for multiples (m) of the ratio, λ/b, one sees low intensity (minima) at those
angles. If the wave-peaks (intensity) exactly line up, the waves are said to be in phase. In
this case the amplitudes add "constructively" and the resulting intensity is enhanced. If
they are out of phase by π
radians or 180C, the amplitudes cancel and add
"destructively": the intensity will be reduced to nothing. Figure 3(a) shows the resulting
intensity pattern. The light from a slit is not a simple spot but a pattern with side lobes.
Note that different colors (different wavelengths) will have separate phase differences.
Therefore, different colors will appear enhanced at different angles.
Screen
n=2
Diffracted
Light Waves
n=1
n=0
Incident Light Waves
Bright
Spots
x +1
x-1
n=1
Slits
Separation: d
n=2
Screen Distance
D
Light Intensity
on the Screen.
FIGURE 2
In this experiment we will also examine the case involving two or more regularly spaced
parallel slits. Figure 2 shows how a wave coming from the left passes through two close
slits or holes. The waves diffract through each slit: they fan outwards and cross each
other's path. A screen on the right shows how the intensity maxima add up to give a
bright spot pattern. Mathematically, the positions of the maxima are defined by
Eq. 2
Double Slit Pattern:
nλ = d sin θ
where n= 0, ±1, ±2, ...
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Interference diffraction-3/10
where d is the separation between the adjacent slits. Fig. 3b (bottom) shows the
theoretical pattern for a double slit.
Although Equations 1 and 2 appear similar, they describe very different effects.
Review them carefully and understand the differences before continuing.
The angle, θ, in Fig. 2 is measured from the central line above or below to the bright spot.
The angles at which constructive interference occur are exaggerated and made large to
make the figure legible. In our experiment the angle, θ, is quite small; thus, we can make
the approximation that sin θ ≈ tan θ = x D, where D is the distance to the screen and x
the position above or below the central point. The positions xn of intensity maxima are
thus given by
xn ≈ n
Eq. 3
D
λ;
d
where n = 0, ±1, ±2,...
The interference intensity minima will be midway between the maxima.
A "REAL" DOUBLE SLIT
Figure 3 (a) Single Slit Profile
Figure 3 (b) Single Slit Plus Double Slit Profile
(NOT TO SCALE)
Now, we have to combine the single and double slit effects to understand what the
experiment will really show. The interference pattern for the double-slit aperture
described by Eq. 2 was derived for an "ideal" case in which the slits were imagined to be
infinitesimally small. In the real world, the slits have finite width; therefore, the effects
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Interference diffraction-4/10
of the single-slit diffraction pattern (Eq. 1) must also be taken into account.
The
combination of the effects leads to the pattern shown in Fig. 3b (bottom) which is what
we will see in this experiment. Note that Fig. 3b is just a composite or convolution of Fig.
3a (single slit) and Fig. 2 (ideal double slit). Remember that Fig. 2 showed only a few
spots to keep things simple; our experiment will show a dozen or more. Understand that:
•
The outside shape or "envelope" is given by the single-slit
diffraction pattern (governed by width b of a single slit (Eq.
1)).
•
The closely-spaced peaks are given by double-slit
interference (governed by separation d of adjacent slits
(Eq. 2)).
DIFFRACTION GRATING
A diffraction grating has a large number of parallel slits separated by a small distance d.
The intensity maxima occur at the same angular positions as for the double slit:
Eq. 4
Diffraction Grating Eq.:
nλ = d sin θ
where n = 0 , ±1, ±2, ...
For the grating the maxima are, however, much sharper (more intense) than for the
double slit because we have more slits contributing to the constructive interference. Also,
d will be much smaller than that for the double slit; thus, we will get larger angles, θ.
And, because the slit width is very small the single slit effect predicted by Eq. 1 is greatly
reduced.
PROCEDURE
In the experiment you will use the intense beam of light produced by a helium-neon laser
designed for classroom use.
SAFETY PRECAUTION:
DO NOT LOOK INTO THE LASER. NEVER LIFT UP THE
LASER OR POINT IT AT SOMEONE. TURN OFF THE
LASER WHEN IT IS NOT BEING USED.
The information for the various target slits (widths for single, separations for double and
grating lines per millimeter) is at the end in an Appendix. The slits are mounted on 35mm
projection slides which your instructor will hand out and collect as you finish. Treat them
with care.
PART A -SINGLE SLIT
Arrange the laser and paper hanger with the distance (D) between the single-slit target
and screen as large as possible. The information identifying the dimensions of each slit
are in the Appendix and should be on the 35mm slides of the slits. Record observations
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Interference diffraction-5/10
for x for one set.
Notice the periodic variation of intensity of the maxima. This is a
difficult measurement, so be careful that D is large!! What do you see? From your
observation, determine the wavelength of light from the helium-neon laser.
PART B - DOUBLE SLIT
Now, use the double-slit target and observe the pattern of light from each pair of slits.
Again the information about the slit widths and slit spacing are given on the target and in
the Appendix. Remember that n is positive and negative. Also, D >>x for sin θ ≈ tan θ.
Determine λ.
PART C - DIFFRACTION GRATINGS
You will be given 4 different diffraction gratings of 80, 100, 300, and 600 lines per
millimeter. For a certain measured distance, "D", to the screen, measure, x, the positions
of each bright order. Calculate the wavelength of laser light for each.
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APPENDIX
Use these tables to identify the slits on the slide sets.
Single Slit Target:
PATTERNS -----------------NO. SLITS ------------------SLIT WIDTH ----------------
A
1
.02 mm
B
1
.04 mm
C
1
.08 mm
D
1
.16 mm
A
B
C
D
2
.04 mm
.240 mm
2
.04 mm
.500 mm
2
.08 mm
.250 mm
2
.08 mm
.500 mm
Double Slit Target:
PATTERN ------------------NO. SLITS ------------------SLIT WIDTH ---------------SLIT SPACE ---------------Diffraction Gratings:
100, 300, and 600 lines/mm. Calculate, d, the line spacing for each in
nanometer units (1nm = 10-9m).
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INTERFERENCE AND DIFFRACTION OF LIGHT WAVES
Name:
Partner:
Section:
Date:
Express all wavelengths in nanometers, nm = 10-9 meter. You may or may not need
to use all the table entries.
1. Single slit minima positions
Measure the left and right ( ±xm) positions of the minima with respect to the 0˚ pattern
center that is the brightest spot. See Figure 3(a). Remember m = ±1, ±2, ... . Record the
slit pattern and consult the Appendix for widths(b).
Pattern
, D:
Order m
, b:
Position xm
tanθ = x/D
λ
Average = λA =
Pattern:
, D:
Order m
, b:
Position xm
tanθ = x/D
λ
Average = λB =
Pattern:
, D:
Order m
, b:
Position xm
tanθ = x/D
Average = λC =
λ
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Summary of the three average wavelengths from the single slits:
λA
λB
λC
Average λ
2. Double Slit
Maxima positions
Measure the left and right ( ±xm) positions of the maxima with respect to the 0˚ pattern
center that is the brightest spot. See Figure 3(b). Remember m = ±1, ±2, ... . Record the slit
pattern and consult the Appendix for widths(b) and separations(d).
Pattern:
, D:
Order m
, b:
Position xm
, d:
tanθ = x/D
λ
Average = λA =
Pattern:
, D:
Order m
, b:
Position xm
, d:
tanθ = x/D
λ
Average = λB =
, D:
, b:
, d:
Pattern:
Order m
Position xm
tanθ = x/D
Average = λC =
λ
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Summary of the three average wavelengths from the double slits:
λA
λB
λC
Average λ
For the double slit pattern A, how many peaks do you expect due to the double slit effect
within the central “envelope” arising from the single slit effect? Show your work.
How many peaks did you see?
3. Diffraction grating
Grating Lines/millimeter:
Order n
, Spacing, d:
, D:
Position x
tanθ = x/D
λ
Average = λA =
Grating Lines/millimeter:
Order n
, Spacing, d:
, D:
Position x
tanθ = x/D
λ
Average = λB =
Grating Lines/millimeter:
Order n
, Spacing, d:
Position x
, D:
tanθ = x/D
Average = λC
λ
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Summary of the three average wavelengths from the diffraction gratings:
λA
λB
λC
Average λ
λ(Single Slit)
λ(Double Slit)
λ(Grating)
Average λ
Calculate the standard deviation of this mean value.
Of the three wavelength determinations, which is your best measurement? Why?
Remember that we used two speakers to produce interference in your first lab. Do you see
the similarity in the formulas for light and sound interference? The distance between the
sound sources is analogous to the distance between the slits in the double slit pattern.
Compare the distances between the two speakers and the two slits to produce the first
maximum at the same location for a given “screen” distance. Use the wavelength of sound
to be 0.5mm and that of light to be 500nm.
Even massive particles like electrons have wavelength associated with them. It is called
the de Broglie wavelength. It is given by λ = h/p. Where h is the Planck constant and p is
the momentum of the particle. Does this mean that you can use electrons to produce
interference?