Double Slit Experiment and diffrac5on gra5ngs Diffraction
Diffraction is normally taken to refer to various
phenomena which occur when a wave encounters an
obstacle. It is described as the apparent bending of
waves around small obstacles and the spreading out of
waves past small openings
Superposition, or Interference
Another characteristic of a WAVE is the ability to
undergo INTERFERENCE. There are TWO types.
We call these waves IN PHASE.
We call these waves OUT OF PHASE.
Interference
applet
Double-Slit Experiment
to illustrate wave nature of light
Thomas Young’s Double Slit
Interference Experiment
•  Gave credibility to the theory
that light is a wave
•  Showed an interference
pattern
•  Measured the wavelength of
the light
http://galileo.phys.virginia.edu/classes/USEM/SciImg/home_files/introduction_files/doubleslit.jpg
Young’s Double Slit Experiment
Young’s Double Slit
Interference Pattern
Interference Fringes on a Screen
applet
Conditions
for Interference of Light Waves
Chapter 15
Diffraction – The Central Maximum
Suppose you had TWO sources
each being allowed to emit a
wave through a small opening or
slit.
The distance between the slits
denoted by, d.
The distance from the slit spacing
to the screen is denoted by the
letter, L.
If two waves go through the slit and then proceed straight ahead to the
screen, they both cover the SAME DISTANCE and thus will have constructive
interference. Their amplitudes will build and leave a very bright intense spot
on the screen. We call this the CENTRAL MAXIMUM.
Diffraction – The Central Maximum
Figure 2
Figure 1
Here is the pattern you will see. Notice in figure 2 that there are several
bright spots and dark areas in between. The spot in the middle is the
BRIGHTEST and thus the CENTRAL MAXIMUM. We call these spots
FRINGES.
Notice we have additional bright spots, yet the intensity is a bit less. We
denote these additional bright spots as ORDERS. So the first bright spot
on either side of the central maximum is called the FIRST ORDER
BRIGHT FRINGE. Figure 1 represents the intensity of the orders as we
move farther from the bright central maximum.
Diffraction – Orders, m
We use the letter, m, to
represent the ORDER of
the fringe from the bright
central.
Second Order
Bright Fringe
First Order
m=2
Bright Fringe
m=1
First Order
Central
Bright Fringe
Maximum m = 1
m=0
Second Order
Bright Fringe
m=2
It is important to understand that we see these bright fringes as a result of
CONSTRUCTIVE INTERFERENCE.
So bright fringes are ordered m = 0,1,2,3….
Diffraction – Bright Fringes
The reason you see additional
bright fringes is because the
waves CONSTRUCTIVELY build.
There is a difference, however, in the
intensity as you saw in the previous slide.
As you can see in the picture, the BLUE
WAVE has to travel farther than the RED
WAVE to reach the screen at the position
shown. For the BLUE WAVE and the RED
WAVE to build constructively they MUST
be IN PHASE.
Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO
TRAVEL SO THAT THEY BOTH HIT THE SCREEN IN PHASE?
Diffraction – Path difference
Notice that these 2 waves are
IN PHASE. When they hit they
screen they both hit at the
same relative position, at the
bottom of a crest.
λ
How much farther did the red
wave have to travel?
Diffraction – Path difference
Notice that these 2 waves are
IN PHASE. When they hit they
screen they both hit at the
same relative position, at the
bottom of a crest.
λ
How much farther did the red
wave have to travel?
ONE WAVELENGTH
Diffraction – Path difference
Notice that these 2 waves are
IN PHASE. When they hit they
screen they both hit at the
same relative position, at the
bottom of a crest.
λ
How much farther did the red
wave have to travel?
ONE WAVELENGTH
We call this extra distance the PATH DIFFERENCE. The path
difference and the ORDER of a fringe help to form a pattern.
Diffraction – Path Difference
2λ
2λ
1λ
1λ
The bright fringes you see
on either side of the central
maximum are multiple
wavelengths from the bright
central. And it just so
happens that the multiple is
the ORDER.
Therefore, the PATH DIFFERENCE
is equal to the ORDER times the
WAVELENGTH
P.D. = mλ
(Constructive)
Diffraction – Dark Fringes
We see a definite DECREASE in intensity between the
bright fringes. In the pattern we visibly notice the DARK
REGION. These are areas where DESTRUCTIVE
INTERFERENCE has occurred. We call these areas
DARK FRINGES or MINIMUMS (minima).
Diffraction – Dark Fringes
First Order
Dark Fringe
m=1
ZERO Order
Dark Fringe
m=0
ZERO Order
Dark Fringe
m=0
First Order
Dark Fringe
m=1
It is important to understand that we see these dark fringes as a result of
DESTRUCTIVE INTERFERENCE.
So dark fringes are ordered m = 0,1,2,3….
Diffraction – Dark Fringes
On either side of the bright
central maximum we see areas
that are dark, or minimum
intensity.
Once again, we notice that the
BLUE WAVE had to travel
farther than the RED WAVE to
reach the screen. At this point ,
however, they are said to
destructively build or that they
are OUT OF PHASE.
Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO
TRAVEL SO THAT THEY BOTH HIT THE SCREEN OUT OF PHASE?
Diffraction – Path difference
Notice that these 2 waves are OUT OF
PHASE. When they hit they screen
they both hit at the different relative
positions, one at the bottom of a crest
and the other coming out of a trough.
Thus their amplitudes SUBTRACT.
0.5 λ
How much farther did the red wave
have to travel?
Diffraction – Path difference
Notice that these 2 waves are OUT OF
PHASE. When they hit they screen
they both hit at the different relative
positions, one at the bottom of a crest
and the other coming out of a trough.
Thus their amplitudes SUBTRACT.
0.5 λ
How much farther did the red wave
have to travel?
ONE HALF OF A WAVELENGTH
Diffraction – Path difference
Notice that these 2 waves are OUT OF
PHASE. When they hit they screen
they both hit at the different relative
positions, one at the bottom of a crest
and the other coming out of a trough.
Thus their amplitudes SUBTRACT.
0.5 λ
How much farther did the red wave
have to travel?
ONE HALF OF A WAVELENGTH
The call this extra distance the PATH DIFFERENCE. The path
difference and the ORDER of a fringe help to form another pattern.
Diffraction – Path Difference
1.5λ
1.5λ
0.5λ
0.5λ
The dark fringes you see on
either side of the central
maximum are multiple
wavelengths from the bright
central. And it just so
happens that the multiple is
the ORDER.
Therefore, the PATH DIFFERENCE is equal to the
ORDER plus a HALF, times A WAVELENGTH
P.D. = (m+1/2)λ
(Destructive)
Path Difference - Summary
For CONSTRUCTIVE INTERFERENCE or
MAXIMA use:
P.D. = mλ
Where m = 0, 1, 2, 3…
For DESTRUCTIVE INTERFERENCE or
MINIMA use:
P.D. = (m + 1/2) λ
Where m = 0, 1, 2, 3…
Where “m”, is the ORDER.
Young’s Experiment
In 1801, Thomas Young successfully showed that light does produce an
interference pattern. This famous experiment PROVES that light has WAVE
PROPERTIES.
Suppose we have 2 slits separated by a distance, d and a
distance L from a screen. Let point P be a bright fringe.
P
y
d
θ
central max
L
We see in the figure that we
can make a right triangle
using, L, and y, which is the
distance a fringe is from the
bright central. We will use
an angle, θ, from the point
in the middle of the two
slits.
We can find this angle using
tangent!
Similar Triangles
(or more often, y )
Diffraction – Another way to look at
This right triangle is
“path difference”
SIMILAR to the one
made by y & L.
P
θ
d
θ
central max
d
P.D.
P.D. = dsinθ
P.D.
Notice the blue wave travels farther. The
difference in distance is the path difference.
(Note that for a large distance, L, from the screen, ray 1 and
ray 2 are approximately parallel).
Similar Triangles lead to a path difference
y
θ
These angles
Are EQUAL.
d
L
dsinθ
θ
sinθ = opp/hyp = path length difference / d
So path length difference = d sinθ
Diffraction – Putting it all together
Path difference is equal to the following:
  mλ for constructive interference
  (m+1/2)λ for destructive interference
  dsinθ for both
Therefore, we can say:
Constructive Interference
d sin θ = mλ
Will be used to find the
angle!
€
Similar Triangles
(or more often, y )
Notice: tan θ = y / L
Example #1
A viewing screen is separated from a double slit source by 1.2
m. The distance between the two slits is 0.030 mm. The
second -order bright fringe ( m=2) is 4.5 cm from the central
maximum. Determine the wavelength of light.
Example #1
A viewing screen is separated from a double slit source by 1.2
m. The distance between the two slits is 0.030 mm. The
second -order bright fringe ( m=2) is 4.5 cm from the central
maximum. Determine the wavelength of light.
2.15 degrees
d sin θ = mλ
(3x10 −5 )sin(2.15) = 2 λ
λ = 5.62x10-7 m
€
Small Angle Approximation
When L is large, the
hypotenuse is very close in
value to L
If θ is small (less
than a few deg), we
can say that
sin θ ≈ tan θ
L
d sin θ = mλ m = 0,1,2,3... Maximum
dy
d sin θ ≈ d tan θ =
= mλ
L
€
also,
tan θ =
ym
L
or
y m = L tan θ ≈ Lsinθ
mλL
€ so for maxima, y m =
d
Maxima
€
€
Minima
m
ym (+/-)
m
ym (+/-)
0
1
2
3
0
1
2
3
0
Lλ/d
2Lλ/d
3Lλ/d
1
2
Lλ/2d
3Lλ/2d
5Lλ/2d
7Lλ/2d
& for minima, d sin θ = (m + ) λ m = 0,1,2,3... Minimum
(or P.D.)
= d sin θ
€
(m +1/2) λL
ym =
d
EXAMPLE #2
Suppose that Young’s experiment is
performed with blue-green light of 500 nm.
The slits are 1.2 mm apart, and the viewing
screen is 5.4 m from the slits. How far
apart are the bright fringes?
EXAMPLE #2
Suppose that Young’s experiment is
performed with blue-green light of 500 nm.
The slits are 1.2 mm apart, and the viewing
screen is 5.4 m from the slits. How far
apart are the bright fringes?
From the table on the previous slide we see that the separation between bright fringes is Δy =
EXAMPLE #2
Suppose that Young’s experiment is
performed with blue-green light of 500 nm.
The slits are 1.2 mm apart, and the viewing
screen is 5.4 m from the slits. How far
apart are the bright fringes?
From the table on the previous slide we see that the separation between bright fringes is Δy =
Lλ / d
€
EXAMPLE #2
Suppose that Young’s experiment is
performed with blue-green light of 500 nm.
The slits are 1.2 mm apart, and the viewing
screen is 5.4 m from the slits. How far
apart are the bright fringes?
From the table on the previous slide we see that the separation between bright fringes is Δy =
Lλ / d
−9
Lλ /d = (5.4m)(500 × 10 m) /0.0012m
€
EXAMPLE #2
Suppose that Young’s experiment is
performed with blue-green light of 500 nm.
The slits are 1.2 mm apart, and the viewing
screen is 5.4 m from the slits. How far
apart are the bright fringes?
From the table on the previous slide we see that the separation between bright fringes is Δy =
Lλ / d
−9
Lλ /d = (5.4m)(500 × 10 m) /0.0012m
= 0.00225m = 2.25mm
€
A diffraction grating that consists of a large number of
parallel slits overcomes both of these difficulties.
A diffraction grating uses interference to disperse light. It is
often an important component in optical instrumentation for
wavelength determinations.
For a diffraction grating, the intensity falls away from these
maxima much more rapidly than that for a double slit. Because
there are so many slits to act as sources, any angle other than those
for maxima will be dark or nearly dark.
Diffraction gratings
  A diffraction grating is a precise array of tiny engraved
lines, each of which allows light through.
  The spectrum produced is a mixture of many different
wavelengths of light.
How a Diffraction Grating Works
When you look at a diffracted
light you see:
—  the light straight ahead as
if the grating were
transparent.
—  a "central bright spot".
—  the interference of all other
light waves from many
different grooves produces
a scattered pattern called a
spectrum.
Note: We cannot use the small
angle approximation for
diffraction gratings.
For gratings, since d is very small, to
observe some of the higher order fringes,
we usually have to place the screen very
close to the grating; therefore, the angle θ is
no longer small.
The Diffraction Grating
Diffraction grating is an arrangement consisting of a large
number of parallel, closely spaced slits.
Gratings with as many as 40 000 slits per centimeter can be made, depending
on the production method. In one method a diamond-tipped cutting tool is used
to inscribe closely spaced parallel lines on a glass plate, the spaces between
the lines serving as the slits.
Double Slit & Grating
The bright fringes produced
by a diffraction grating are
much narrower than those
produced by a double slit.
Note the three small
secondary bright fringes
between the principal bright
fringes of the grating. For a
large number of slits, these
secondary fringes become
very small.
Grating Formula
wavelength
of light (nm)
distance between grating lines (m)
mλ = d sinθ
d sin θ = (m + 1 2 ) λ
constructive
destructive
Note: sometimes the given data for slit separation is
given as “lines per _ meters” instead of “d”
Set 1/d = number of lines per m
(NOTE: a “line” is the solid material separating each slit).
€
Double Slit & Grating
The bright fringes produced
by a diffraction grating are
much narrower than those
produced by a double slit.
Note the three small
secondary bright fringes
between the principal bright
fringes of the grating. For a
large number of slits, these
secondary fringes become
very small.
Example #3
A light with wavelength, 450 nm, falls on a diffraction
grating (multiple slits). On a screen 1.80 m away the
distance between dark fringes on either side of the
bright central is 4.20 mm. a) What is the separation
between a set of slits? b) How many lines per meter
are on the grating?
Example #3
A light with wavelength, 450 nm, falls on a diffraction
grating (multiple slits). On a screen 1.80 m away the
distance between dark fringes on either side of the
bright central is 4.20 mm. a) What is the separation
between a set of slits? b) How many lines per meter
are on the grating? NOTE: a “line” is the solid
material separating each slit.
Example #3
A light with wavelength, 450 nm, falls on a diffraction grating (multiple slits).
On a screen 1.80 m away, the distance between dark fringes on either side
of the bright central is 4.20 mm. a) What is the separation between a set of
slits? b) How many lines per meter are on the grating? NOTE: a “line” is the
solid material separating each slit.
λ = 450x10 −9 m L = 1.80m
y = 4.2mm /2 = 0.0021m
θ =?
d =?
0.067 degrees
5197.2 lines/m
€
0.0001924 m