light waves

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35.1 Interference
and coherent
sources of light
Waves, Optics & Modern Physics
203-NYC-05
Greg Mulcair

Light as learned
in High School
n  In high school you learned about light and
how it behaves when reflecting off
surfaces and passing through lenses.
n  This branch of physics (geometric optics)
treats light as “rays” (straight lines).
n  Although this works very accurately, light
is actually a wave when you look close
enough. Wave optics is the topic of Unit 2.

Electromagnetic (EM) waves
n  Light is the term used for the visible portion
of a wide range of waves around us known
as electromagnetic (EM) radiation.
n  Light is EM radiation with a wavelength
between approx. λ = 400nm and 700nm.

EM waves are transverse waves.

Electromagnetic (EM) waves
n  Just like for mechanical waves, EM waves
transmit energy at a certain wave speed.
¨ But, EM waves don’t need a medium! They can
transmit this energy through vacuum.
¨ EM waves travel at speed v = 3E8 m/s (which
we will call the speed of light, “c”)
n  We will say “light” during most of our discussions, but what
we will learn applies to all EM waves.

Note for the curious: The nature of EM radiation is explained in chapter 32 but is not part of the NYC curriculum. The energy transmitted in EM
radiation is both electrical and magnetic. The waves alternate rapidly, from positive to negative in electrical terms, and from North to South pole in
magnetic terms. A nice visualization of this can be seen here: http://www.ephysics.ca/Abbott/ActiveFigures/Ch.23-46/SWFs/AF_3403.html

Interference of light waves
n  Light
waves interfere with each other the
same way we’ve seen: by superposition.
¨ Recall:This is a fancy way of saying that the
waves add up. The resultant displacement is
the displacement of all the waves combined.

n  Note: We use the term “displacement” in a
general sense here.
¨  With waves on the surface of a liquid, we mean
the actual displacement of the surface above or
below its normal level, as we learned.
¨  With sound waves, the term refers to the excess
or deficiency of pressure, as we learned.
¨  For electromagnetic waves, we usually mean a
specific component of electric or magnetic field.

Note: Waves in 2D and 3D
n  Waves on a string travel in one dimension
along the length of the string) and ripples
on a pond go in two dimensions (along the
plane of the surface).
n  For light we will work with waves traveling
in two or three dimensions.

Visualizing waves as wavefronts
n  A wavefront is just a line
joining common points on
the wave. E.g.: All the
peaks (crests) of the wave.
n  These move outward with
time as the wave travels,
and the distance between
each is the wavelength
(which makes sense).

Note: Waves in 2D and 3D
n  For example, the figure
shows a succession of
wave crests (the peaks)
from a sound source.
¨  If the waves are in 2D, like
waves on the surface of a
liquid, the circles represent
circular wave fronts.
¨  If the waves are in 3D, the
circles represent spherical
wave fronts spreading
away from the source.

Note: Monochromatic light
n  Most light sources (light bulbs, fire…) emit a
continuous distribution of wavelengths.
n  But for our study, it will be easier to
consider light that is made to shine at only
one frequency and wavelength (colour):
monochromatic light.
¨ Wewill assume monochromatic light for this
chapter (interference) and the next (diffraction).

n  An excellent example of this is the laser

Note: Coherent sources
n  Furthermore, sources of monochromatic
waves that have a constant phase
difference (not necessarily zero) are
coherent.
n  We often refer to waves from such
sources as coherent waves.
n  If the waves are light, we usually just say
coherent light.

Note: Polarization
n  Andthe last assumption we will make is that
the wave sources have the same
polarization (the transverse wave
disturbances lie along the same line).
¨ Youdon’t need to know more than that,
although it’s an interesting topic!

n  Ok,
with all those assumptions and
simplifications to our scenario, we can now
take a look at interference of light waves!

n  Goodnews, it’s the same rule as we
learned for sound wave interference.

Example of two ideal sources
n  Considertwo
sources (S1 and S2)
of equal amplitude,
equal wavelength
(monochromatic)
and same
polarization (in the
case of transverse
waves such as light).

Constructive interference
n  Consider a point a on
the x-axis.
n  From symmetry the
two distances from S1
to a and from S2 to a
are equal.
n  So waves from the two
sources take equal
times to travel to a.

n  So for any point on
the x-axis, waves that
leave in phase arrive
in phase and there is
constructive
interference.
n  The total amplitude is
the sum of the two
source amplitudes.

n  Nowconsider point b,
where the distance
from S2 to b is
exactly two
wavelengths greater
than the distance
from S1 to b.

n  Soa wave crest from
S1 arrives at b
exactly two cycles
earlier than a crest
emitted at the same
time from S2 and
again the two waves
arrive in phase.

n  So the waves arrive
in phase and again
there is constructive
interference.
n  The total amplitude is
the sum of the two
source amplitudes.

n  In general, waves interfere constructively
if their path lengths differ by an integral
number of wavelengths:

r2 − r1 = mλ
m = 0, ±1, ±2, ±3,…

Destructive interference
n  Now consider point c,
where the path
difference for the
distance from S2 to c
compared to distance
from S1 to c is exactly
2.5 wavelengths.

n  So the wave crests
from the two sources
arrive exactly one
half cycle out of
phase.
n  In this case, there is
destructive
interference.

n  Waves
interfere destructively if their path
lengths differ by a half-integral number of
wavelengths

" 1%
r2 − r1 = $ m + ' λ
# 2&
m = 0, ±1, ±2, ±3,…

Summary: interference of light
n  Waves interfere constructively if
their path lengths differ by an
integral number of wavelengths:
r2 − r1 = mλ m = 0, ±1, ±2, ±3,…

n  Waves interfere destructively if
their path lengths differ by a
half-integral number of
wavelengths:
" 1%
r2 − r1 = $ m + ' λ m = 0, ±1, ±2, ±3,…
# 2&

Question
n  Two sources S1 and S2 oscillating in phase emit
sinusoidal waves. Point P is 7.3 wavelengths
from source S1 and 4.3 wavelengths from source
S2. As a result, at point P there is
¨  A. constructive interference.
¨  B. destructive interference.
¨  C. neither constructive nor destructive interference.
¨  D. not enough information given to decide.

We just subtract the two and find that the path difference is:

Answer r2 – r1 = 7.3λ – 4.3λ = 3λ
As seen, this gives constructive interference.


Question

We just subtract the two and find that the path difference is:

Answer r2 – r1 = 7.3λ – 4.6λ = 2.7λ
This gives neither constructive or destructive interference.


Interference
of light
n  Ifwe draw a curve
through all points
where constructive
interference occurs,
we get the shapes
shown.
n  We call these
antinodal curves.

Interference
of light
n  In ch.15 and ch.16,
antinodes were points of
maximum amplitude in
n  But be careful: There
was no net flow of
energy in those cases,
but in this case there is
(it is just channeled such
that it’s greatest along
the antinodal curves).

Interference
of light
n  For 3D sources, try to
picture rotating the red
antinodal curves about
the y-axis.
n  The result would be
antinodal surfaces that
follow the same idea.

35.2 Two source
interference of light
Waves, Optics & Modern Physics
203-NYC-05
Greg Mulcair

Two-source interference of light
n  Studying the interference of two sources of
light is not as easy two wave ripples since
we can’t see the light as easily.
n  In 1800, the English scientist Thomas
Young devised an experiment that will be
important for us now and in later chapters:
Young’s double-slit experiment

n  First,
monochromatic light is sent through
a small (1µm) slit S0 to obtain a source
that is an idealized point source.
n  Reason: Emissions from different
parts of an ordinary source are
not synchronized, whereas if they
all pass through a small slit they
are very nearly synchronized.

Recall: 1µm = 1E-6m

n  Next,
this light encounters two other slits,
S1 and S2 (each ≈1µm wide, ≈1mm apart)

n  Note: Because the cylindrical
wave fronts travel an equal
distance from the source S0 to the
slits S1 and S2 , they reach the
slits in phase with one another
(i.e. when a crest is at S1 a crest
is at S2 , and so on).

n  The
waves emerging from S1 and S2 are
therefore always in phase, so S1 and S2 are
coherent sources of monochromatic light.
n  Recall: Monochromatic
light is light that is made
to shine at only one
frequency and
wavelength (colour)
n  Recall: Coherent
sources emit waves
which have a constant
phase difference

n  Lastly,
we observe how these two
monochromatic coherent sources of light
interfere with each other on a screen.
Questions:

What do you think is
happening at these
bright bands?
And what’s happening
at these dark bands?

n  We see an interference pattern on the screen:
¨  Brightbands: there was constructive interference
¨  Dark bands: there was destructive interference

Question
n  What did we learn earlier was the way to
know if there is constructive or destructive
interference between two waves?

Answer: Their path difference
n  Waves interfere constructively if
their path lengths differ by an
integral number of wavelengths:
r2 − r1 = mλ m = 0, ±1, ±2, ±3,…

n  Waves interfere destructively if
their path lengths differ by a
half-integral number of
wavelengths:
" 1%
r2 − r1 = $ m + ' λ m = 0, ±1, ±2, ±3,…
# 2&

n  Consider a point P on the screen. We see that
the wave coming from source S2 travels a bit
further than the wave coming from source S1.

n  But how do we
calculate the
exact difference
in the distances
the waves travel
(r2 – r1) ?

n  First we simplify: Since the distance R is much
greater than the distance d between the two slits,
we can consider the rays as being parallel.

n  And so the path difference is approximately equal
to the extra distance travelled from S2 (which is
dsinθ from trigonometry and similar triangles).
r2 − r1 = d sin θ
Similar triangles, so
this θ equals this θ n  This is a very good
approximation when R is
θ large. Make sure that
you understand the
trigonometry (it is
important now and will
be important later too).

n  So the difference in path length is r2 − r1 = d sin θ
¨  Where θ is the angle between a line from slits to screen
(the thick blue lines in the figure) and the normal to the
plane of the slits (the thin black line in the figure).

n  And we’ve learned that we have:
¨  Constructive interference when r2 − r1 = mλ m = 0, ±1, ±2,…

" 1%
¨  Destructive interference when r2 − r1 = $ m + ' λ m = 0, ±1, ±2,…
# 2&

n  So we can conclude that we have:
¨  Constructive interference when d sin θ = mλ m = 0, ±1, ±2,…

! 1$
¨  Destructive interference when d sin θ = # m + & λ m = 0, ±1, ±2,…
" 2%

n  So we get a succession of
bright and dark bands on the
screen as the angle changes
(i.e. for different locations on
the screen). We call these
interference fringes.
n  The central band (which
corresponds to m = 0) is the
brightest

Let’s be sure we understand
n  We describe locations on the screen in
terms of the angle from horizontal.
n  For any θ, if the value of dsinθ works
out to be an integer multiple
of wavelengths
( d sinθ = mλ )…
Light from two very close
coherent sources. We
+θ
see interference fringes
on the screen at different
angles, θ, from the
horizontal
-θ
…then you will
have constructive
interference (a bright band).
! $
n  And if d sinθ = # m + 1 & λ you get a dark band.
" 2%

Measuring the y-position of fringes
n  If the angle θ is small (< 20º) we can
describe each fringe’s position in terms
of it’s distance from the middle fringe.
n  For example, for the m=3
fringe the angle
θm = θ 3 ym = y3
θm = θ 3

n  Corresponds to a
height above horizontal of
ym = y3. From trig, what is this?
Note that the screen is a distance R from the slits

n  From trig, the y-position of the third
bright fringe (m=3) is: y3 = Rtanθ3
n  R is actually very large compared to y3
so this equation is very nearly
equal to: y3 = Rsinθ3
ym = y3
θm = θ 3

R

n  And since we also
know that d sin θ = mλ we
can say dsinθ3 = 3λ è sinθ3 = 3λ/d
which gives us: y3 = R3λ/d

n  In general, the y-position of any bright
(constructive interference) fringe is:
mλ
ym = R
d
ym

θm

n  So the distance
between adjacent bright
bands in the pattern is inversely
proportional to the distance d between
the slits. The closer together the slits are, the
more the pattern spreads out. When the slits are far
apart, the bands in the pattern are closer together.

Measuring the wavelength of light
n  This equation can also be used to find
the wavelength of incoming light since
we can measure R, d and ym.
mλ
ym = R
d
ym

θm

n  Young's experiment
was actually the first direct
measurement of wavelengths of light.

This is a great simulation of this scenario. Start by watching it with max slit separation and minimum wavelength.
The red/green dots where the wave crests (maximus) meet follow antinodal curves on their path to the screen.
When they reach the screen, this constructive interference (large amplitude) gives a bright band on the screen.

mλ
ym = R
d

http://www.awakensolutions.com/Abbott/ActiveFigures/NYB/SWFs/AF_3702.html

Now try decreasing the slit separation, and see that ym increases (the pattern spreads out) as expected.
And you can change the wavelength as well and see that ym increases (the pattern spreads out) as well.

Summary for Young’s
double-slit experiment
n  For
any angle, the position of the fringes
can be described using these equations:
d sin θ = mλ m = 0, ±1, ±2,…

! 1$
d sin θ = # m + & λ m = 0, ±1, ±2,…
" 2%

n  And
if the angle θ is small (< 20º) this
equation works too:
mλ Recall from a few slides back, this equation
ym = R came from saying the height ym = Rtanθm is very
close to ym = Rsinθm . However this isn’t the
d case if θ cannot be considered small.

Example 35.1 (p.1213)
n  In a two-slit interference experiment, the slits are
0.200 mm apart, and the screen is at a distance of
1.00 m. The third bright fringe (not counting the
central bright fringe straight ahead from the slits) is
found to be displaced 9.49 mm from the central
fringe. Find the wavelength of the light used.
d sin θ = mλ

! 1$
d sin θ = # m + & λ
" 2%
mλ
ym = R
d

mλ
ym = R
d
Solution using the ‘approximate’ equation
n  The third bright fringe corresponds to m = 3. To
mλ
determine the wavelength we may can use y =R
d
m

since R = 1.00 m is much greater than d = 0.200
mm or y3 = 9.49 mm
mλ
ym = R
d
y d (9.49E − 3)(0.2E − 3)
λ= m =
mR (3)(1)
λ = 633E − 9m = 633nm

Note that this fringe could also have corresponded to m = -3.
You would obtain the same result for the wavelength since y and m would be negative.

d sin θ = mλ
Solution using the ‘pure’ equation
n  Using the pure equation we need to first find the
angle to the screen: tan θ = opp = 9.49 E − 3 ⇒ θ = 0.544o
adj 1

n  And we use this to find the exact wavelength:
d sin θ = mλ
(0.2 E − 3) sin( 0.544) = 3λ
λ = 6.326E − 7 m
λ = 633E − 9m = 633nm

Note that this fringe could also have corresponded to m = -3.
You would obtain the same result for the wavelength since y and m would be negative.

Question 35.11 (p.1230)
n  Coherent light from a sodium-vapor lamp is
passed through a filter that blocks everything
except light of a single wavelength. It then falls
on two slits separated by 0.460 mm. In the
resulting interference pattern on a screen 2.20 m
away, adjacent bright fringes are separated by
2.82 mm. What is the wavelength?

Answer 35.11 (p.1230)

mλ
ym = R
d
(m + 1)λ
ym +1 = R
d
(m + 1)λ mλ λ λ
Δy = ym +1 − ym = R −R = R ( (m + 1) − m ) = R
d d d d
d Δy (0.46 E − 3)(2.82 E − 3)
λ= = = 5.9E − 7m = 590nm
R 2.2

Question 35.13 (p.1230)
n  Two very narrow slits are spaced 1.80 µm apart
and are placed 35.0 cm from a screen. What is
the distance between the first and second dark
lines of the interference pattern when the slits are
illuminated with coherent light with λ = 550 nm?
n  For this problem, do not use the approximation
that tanθ ≅ sinθ (consider that the angle is too
large for this approximation since the distance
from the screen, R, is less than 1m).

Two very narrow slits are spaced 1.80 µm apart and are placed 35.0 cm from a
screen. What is the distance between the first and second dark lines of the
interference pattern when the slits are illuminated with coherent light with λ = 550 nm?

Answer 35.13 (p.1230)
n  Inthe previous problem we were told that
the angle wasn’t small enough for us to use
the small angle approximation (which lets
mλ
us say sinθ ≅ tanθ, thus giving us ym = R d ).
n  So we had to use the main equation from
which the above is derived: d sinθ = ! m + 1 $ λ
# &
" 2%

Answer 35.13 (p.1230)
n  Let’s
see how the small angle
approximation (sinθ ≅ tanθ) breaks down
as the angle gets bigger:
¨ For θ1 = 8.8º we get:
n  sinθ1 = 0.1528 tanθ1 = 0.1546
¨ For θ2 = 27.3º we have a larger angle and get:
n  sinθ2 = 0.4583 tanθ2 = 0.5157
n  As
you can see, for larger angles the small
angle approximation is less and less valid

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