Physics 101 Learning Objective 9.

1. Light: Nature’s Hidden Wave
A Look Into Double –Slit Interference and Its Applications
By Jesse Galay (LO9)

2. Goals of Learning Objective
The purposes of this Learning Objective are:
• Highlight the historical and scientific importance of Double-slit Interference
• Demonstrate how the experiment’s mathematical relationship can be applied.
• Provide examples of applications of Double-slit Interference

3. First, a Bit of History
Up into the late 1700’s many prominent scientists firmly
supported the idea that light consisted only of tiny particles.
One of these scientists was Isaac Newton, who attempted to
develop a particle based theory to explain the behavior of
light. However, this theory was found to be flawed as it
could not account for why light particles did not ricochet
and crash off of each other, along with other phenomenon.
Enter Thomas Young, an English scientist who believed that
the behavior of light could be described by considering that
in addition to behaving as particles, light could also behave
as a wave. In 1799, Young set out to prove his theory with
the double slit experiment.
Thomas Young

4. The Double Slit Experiment
Young’s experiment was simple in design. A white light was shone towards a
screen with two narrow and closely spaced slits. The pattern that appeared
on a wall behind the screen with slits was then noted. If the light acted
strictly as particles, Young would have seen two points on the back wall (two
maxima) which corresponds to the straight movement of the particles through
the two slits. Instead, he found that the back screen possessed alternating
bars of light and dark.
The alternating bars were the result of the light acting as waves. As the light
passed through the two slits, the slits acted as production points of two new
waves from the original light wave. These new two waves then propagated
outwards, causing constructive and destructive interference with each other
to create the alternating light and dark pattern seen on the back screen.
As a result of these findings, Young demonstrated that light could not only
behave like a particle, but like a wave as well.

5. The Math Behind the Pattern
For the math of the Double-Slit Interface, we have to look at the situation
from a geometric perspective. This allows us to derive angles for which
the points of maximum light intensity appears on the back screen:
• Light travels through slits D1 and D2, where the light from D1 reaches
P first.
• Because of this, the light from D2 has an extra distance to travel to
reach P.
• By dividing the D2 to P distance into two parts, we can see how much
further the light from that slit has to travel. The right triangle formed
has an angle q, which by using trigonometry can be used to find the
extra distance. This extra distance is dsinq.
• So: D1 to P = (D2 to P) – dsinq
• This restriction is referred to as the Fraunhofer condition (which
works in most important cases).
D1
D2
P
O

6. The Math Behind the Pattern cont.
The an important consequence of this mathematical derivation is we can
now mathematically explain how constructive and deconstructive
interference arises from the Double-Slit experiment.
• The region where constructive interference takes place is easy to
identify: its where the bright patches appear on the back screen.
• These bright regions appear when the extra distance corresponds to an
exactly whole number of wavelengths (1,2,3,…,etc.). As such, the
situation can be shown as:
dsinq = ml
• “m” is an integer that is referred to as the spectral order, which
denotes how many wavelengths fit in the extra distance.
• Likewise, the areas of deconstructive interference (the dark patches) is
represented similarly, with the only difference being the extra distance
is equal to an odd number of half wavelengths:
dsinq = (m+1/2)l
D1
D2
P
O

7. The Math Behind the Pattern cont.
As a result of these conditions, different values of wavelength and “m”
will effect the angle of reflection, and subsequently the interference
pattern.
• Special situation: when m = 0, angle of 0 degrees results from any
wavelength, so shining a white light at the slits will result in the white
beam in the centre of the interference pattern.
• For other “m” values, the maximum for different wavelengths occur at
many different angles. Due to dsinq = ml, the longer the wavelength,
the larger the deflection angles.
• The direct central maximum (the point of maximum diffraction) is
called the zero order, while the spectral orders above are labelled m=
+1, +2,… and below labelled m = -1, -2, …
D1
D2
P
O

8. Sample Question
A laser beam illuminates a double slit screen that is 15 cm. away from the back screen. The m = 1 fringe is
5 cm away from the m = 0 (the zero order) fringe. The distance between the slits is 2 cm.
a) What is the wavelength of the laser beam?
b) The wavelength of the beam is increased by 1 cm. What is the maximum order that can occur?

9. Sample Question Part A
A laser beam illuminates a double slit screen that is 15 cm. away from the back screen. The m = 1 fringe is 3 cm away
from the m = 0 (the zero order) fringe. The distance between the slits is 2 cm.
a) What is the wavelength of the laser beam?
First, we have to determine the angle q. By looking at the previous mathematical diagram, you can see that a right
triangle forms if you draw a line from d to P if you use L as a base. Assuming P is m = 1 and the point parallel to d is m
=0, then you can use the inverse of tangent and the given distances to determine the angle in radians. This given angle
is the same as the rest of the q seen on the diagram.
Arctan(3 cm/15 cm) = q = 0.197 rad.
Then use the formula dsinq = ml to determine the wavelength. Remember d = 2 and m = 1.
(2 cm.)sin(0.197) = (1)l = l
l = 0.39 cm.
So the wavelength is equal to 0.39 cm.

10. Sample Question Part B
A laser beam illuminates a double slit screen that is 15 cm. away from the back screen. The m = 1 fringe is 5 cm away
from the m = 0 (the zero order) fringe. The distance between the slits is 2 cm.
a) What is the wavelength of the laser beam? (Solved)
b) The wavelength of the beam is increased by 1 cm. What is the maximum order that can occur?
So, add 1 cm. to the previously found wavelength and work backwards through the dsinq = ml formula to find “m”. To
find the maximum order, the sinq must be set equal to 1. This is because the maximum possible angle that can occur
where the interference still takes place is just below 90 degrees.
0.39 cm. + 1 cm. = 1.39 cm.
(2 cm.)(1) = m(1.39 cm.) ((2 cm.)/1.39 cm) = m = 1.44.
When you get a decimal answer for the spectral order, you truncate the value down (round down) to the nearest integer.
So in this case the maximum order is m = 1.

11. Applications
So now that the mathematics behind the Double-slit Interference is known, what can they tell us about light
and how can they be used to help us better understand the world?
• The Double-Slit experiment helped to establish the wave-particle duality theory of light, which we take
today to be the best way to explain the behavior of light.
• We can determine the spectral order of any point on an interference pattern, provided we know the
angle and the wavelength or they can be calculated.
• Additionally, the wavelength or the angle can be determined if the other two necessary variables are
known or can be calculated.
• By determining the wavelength, other characteristics of the original light can be determined. This
principle will eventually lead to Diffraction Grating, which can be used to determine the characteristic
of an element based on the colour (wavelength) of the light it produces when white light is shone
through it.

12. Conclusion
In the end, Thomas Young’s experiment had a profound effect on how we consider the behavior of light
today. The experiment not only cemented the idea of the wave-particle duality of light in the scientific
community, but also provided the means for the analysis of different colours (wavelengths) of light in the
later developed Diffraction Gratings (which shot light through not two, but thousands of slits to determine
its properties).