This PowerPoint summarizes double-slit interference by transforming the interference patterns formed by Young's experiment into a right triangle. It is an attempt to clarify the significance of all of its components in order to determine different angles of which bright fringes appear on a screen L distance away from the two slits and eventually to conceptualize the distance between bright fringes by knowing the value of the wavelength of light.

1. LearningObject – CamilleAlvarez (LC2)

2. PhET Simulation
• The points of
constructive
interference appear
as bright bands
(green) on the
screen.
• The points of
destructive
interference appear
as dark bands.
These bands are
referred to as “fringes”.
Let’s go into more
detail.

3.  We can determine the angles where
maximum light intensity (bright
fringes) occurs by thinking about a
right triangle.
 Fraunhofer Condition:
The distance from A1 to B is
approximately the same as the
distance from C to B ONLY IF the
distance from the screen L is much
greater than the slit separation d.
 The extra distance that light travels
from A2 to the screen at B is
Dsinθ
This extra distance coincides to the
path difference (Hence the fact that
light travels different lengths from both
slits to a point on the screen).
http://theory.uwinnipeg.ca/
physics/light/img30.gif
C
A1
A2
BThese angles
are the same
dsinθ

4. Assuming the light waves from the source are
in phase, the path difference for
 Constructive Interference:
Δx = mλ
 Destructive Interference:
Δx = (m+1/2)λ

5. Red = bright fringe (constructive interference)
Blue = dark fringe (destructive interference)

6.  When the path difference corresponds to
exactly a whole number of λ, constructive
interference (bright fringes) occurs.
 So:
Dsinθ = mλ
 For all values of m other than zero that is an
integer, the maxima for different
wavelengths have different angles.
 An m value of zero means that θ = 0 ° and the light would appear in the middle
of the screen.
Path difference Δx !!

7.  Accordingly, when the path difference
corresponds to an odd number of half λ,
destructive interference (the dark fringes)
occurs.
 So:
Dsinθ = (m+1/2)λ

8. http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/imgpho/diffgrat.gif

9.  Let’s assume that y is much less
than L so we can use small angle
approximation for
sinθ ≈ y/L
Therefore
dsinθ = mλ
becomes
d(y/L) = mλ
y = (mλL)/d
This equation allows us to calculate
the distance of a bright fringe to the
point on the screen opposite of the
center of the slits.

10. Red has the largest wavelength in
the visible spectrum.
Using
y = (mλ)L/d
we can see mathematically that the
distance between bright fringes is a
lot bigger than purple (smallest
wavelength).