Download CHAPTER 1 PHYSICAL OPTICS: INTERFERENCE • Introduction

CHAPTER 1
PHYSICAL OPTICS: INTERFERENCE
What is “physical optics”?
• Introduction
• Waves
• Principle of superposition
• Wave packets
• Phasors
• Interference
• Reflection of waves
• Young’s double-slit experiment
• Interference in thin films and air gaps
The methods of physical optics are used when the
wavelength of light and dimensions of the system are of
a comparable order of magnitude, when the simple ray
approximation of geometric optics is not valid. So, it is
intermediate between geometric optics, which ignores
wave effects, and full wave electromagnetism, which is a
precise theory.
In General Physics II you studied some aspects of
geometrical optics. Geometrical optics rests on the
assumption that light propagates along straight lines and
is reflected and refracted according to definite laws, such
Or the use of a convex lens as a magnifying lens:
as Fermat’s principle and Snell’s Law. As a result the
positions of images in mirrors and through lenses, etc.
can be determined by scaled drawings. For example, the
s!
s
production of an image in a concave mirror.
s
Object
object
•F
y
Image
•C
f
y!
image
2
s!
1
f
But many optical phenomena cannot be adequately
explained by geometrical optics. For example, the
iridescence that makes the colors of a hummingbird so
brilliant are not due to pigment but to an interference
effect caused by structures in the feathers.
The colors you see in a soap bubble are also due to an
interference effect between light rays reflected from the
front and back surfaces of the thin film of soap making
the bubble. The color depends on the thickness of film,
ranging from black, where the film is thinnest, to
magenta, where the film is thickest.
Likewise, the colors you see in a thin film of oil floating
on water are due to an interference effect between light
rays reflected from the front and back surfaces of the oil
Another example is the “spectrum of colors” you see
film.
when you reflect light from the active side of a CD. The
effect that produces these colors is closely related to
interference; it is called diffraction.
Here is an example of diffraction caused by the edges of
the razor blade when viewed from behind with
Waves
monochromatic blue light.
In physics,we come across three main types of waves:
Mechanical waves: water waves, sound waves, seismic
waves, waves on a string ...
Electromagnetic waves: visible and ultraviolet light,
radio and television waves, microwaves, x-rays and radar
waves. All e-m waves travel in vacuum with a speed of
The pattern of “fringes” is easily observable with
monochromatic light. With white light, fringes due to
the different wavelengths overlap making them more
difficult to observe.
Diffraction and interference cannot be expained by
geometrical optics; instead, light has to be treated as
waves.
c = 2.99792458 ! 108 m s.
Matter waves: the waves associated with electrons,
protons and other fundamental particles, atoms and
molecules.
However, all waves have features in common.
Take the ripples (waves) in a pond caused by a small
stone being dropped into the water ...
y
$
A
x
x!
"A
Shown is a snapshot of the wave at some time t. It is
described by the general equation:
y(t, x) = A cos(!t " kx)
At a position, x ! say, the disturbance y(t) varies
sinusoidally with time (i.e., simple harmonic motion).
x
Similarly, at a time t, as shown here, the disturbance
y
!
A
x
"A
Vertical and horizontal illustrations of the wave
To describe the “disturbance” of such a wave we need
two variables, t and x.
varies sinusoidally with position, x. The parameter
k = 2# $ is called the wavevector, where $ is the
wavelength. The product kx is called the phase angle
(= %). A negative (positive) sign indicates the wave is
traveling to the right (left).
Also, ! is the angular frequency of the wave given by
! = 2#f , and T = 1 f is the periodic time of the wave.
y
y
!
!x
A
x
x
wave at t
$A
wave at t + !t
To find the speed of a wave, we take two snapshots at a
Note that the disturbance at some fixed time t !,
y(t ! , x) = y(t ! , x ± n!),
time interval !t apart. If the wave (i.e., the red dot)
i.e., the wave is reproduced at displacements of n!,
travels a distance !x in that time, then the speed of the
where n is an integer.
wave is
y
v = !x !t .
T
A
Since the disturbances (y) are equal
t
$A
T = 1 f = 2" #
Similarly, the disturbance at some fixed position x !,
y(t, x ! ) = y(t ± T, x ! ) = y(t ± m 2" # , x ! ),
i.e., the wave is reproduced after time intervals of
m 2" # , where m is an integer.
y(t, x) = y(t + !t, x + !x),
so "t # kx = "(t + !t ) # k(x + !x),
i.e., "!t = k!x,
$v = " k .
Note also v = " k = 2%f 2% & = f&.
The velocity of a fixed point on a wave (such as the red
dot) is called the phase velocity.
Principle of superposition
“If two or more waves are traveling through a medium,
y1 = 6sin x
the resultant disturbance at any point is the algebraic
sum of the individual disturbances.”
y 2 = 5sin 2x
Waves that obey this principle are called linear waves.
y 3 = 4sin 3x
One consequence is that two waves can “pass” through
y 4 = 3sin 4x
each other!
y2
y1
y2
y1
y 5 = 2sin 5x
y 6 = sin 6x
y1 + y 2
y1 + y 2
y1 + y 2
y2
y2
y1
y1
(Waves that do not obey this principle are called nonlinear waves.)
y = y1 + y 2 + y 3 + y 4 + y 5 + y 6
So, the superposition of several waves of differing
wavelengths and amplitudes produces complex
waveforms. For example, to produce a square wave ...
Consider the superposition of two waves of equal
amplitude but slightly different frequencies and
wavelengths. Then, the resultant is
y(x, t) = A sin (!1t " k1x ) + A sin (! 2 t " k 2 x )
$ #!
#k '
= 2A cos&
t"
x) sin ( !t " kx ),
% 2
2 (
where #! = !1 " !2 , #k = k1 " k 2 , ! =
and k =
WBX06VD1.MOV
A square wave can be expressed as a so-called Fourier
(! 1 + ! 2 )
2
( k1 + k 2 ) . When plotted, at some time t, we
2
get
y(x)
A
series:
x
!
1 $ n# '
f (x) = "
sin & x ) ,
%L (
n
n=1,3,5…
where L = * 2 , i.e., one-half of the wavelength.
y(x)
2A
x
A Fourier series decomposes a periodic function into a
sum of simple oscillating functions, i.e., sines and/or
cosines.
i.e., the waves are separated into “groups”.
y(x)
2A
y(x)
2A
x1
x2
x
x1
!x
$ !"
!k '
y = (x,t) = 2A cos&
t#
x) sin ( "t # kx )
% 2
2 (
The first term is the envelope, i.e., the green curve. The
waves.
x
!x
So, successive minima (at time t) occur at x 2 and x1 where
$ !"
!k ' $ !"
!k '
&
t#
x2 ) # &
t#
x ) = *,
% 2
( % 2
2
2 1(
2*
i.e, x 2 # x1 = !x =
.
!k
second term is the wave within the envelope. Both the
envelope and wave within the envelope are traveling
x2
Thus, the spatial extent of the group is !x + !k #1. If we
plot the resultant as a function of t at a fixed point, we get
The envelope moves with velocity v g = !" !k , called
the group velocity. The wave inside the envelope moves
with velocity v p = " , called the phase velocity.
k
The amplitude of the envelope is zero when
$ !"
!k '
*
&
t#
x n ) = (2n +1) ,
% 2
(
2
2
where n = 0, ±1, ±2 !.
y(t)
2A
t
In the case of sound waves, this waveform produces the
phenomenon of “beats”.
y(t)
2A
t2
t1
t
The group velocity (the velocity of the envelope) is
" d kv
%
"
%
dv
p
d!
$
' = $v p + k p ' .
vg =
=
dk k $
dk '
dk &
#
&k #
k
[
]
(
)
If the phase velocity is the same at all frequencies and
!t
Successive minima (at point x) occur at t 2 and t1 where
$ !"
!k ' $ !"
!k '
&
t2 #
x) # &
t1 #
x ) = *,
% 2
2 ( % 2
2 (
wavelengths, i.e., there is no dispersion then
dv p
dk = 0 i.e., v g = v p .
2*
i.e, t 2 # t1 = !t =
.
!"
Hence, the temporal extent !t + !" #1.
So, we find that
!x.!k , constant (2* )
and
!t.!" , constant (2* ) .
The phase velocity of the individual harmonic waves is
v p = " k , " = v p k.
Case A.mpg
A medium in which
dv p
dk = 0 is said to be non-
dispersive. (An example is an electromagnetic waves in
vacuum.) Glass, for instance, is a dispersive medium.
Shown here is a plot of the phase velocity in flint glass.
(
Wave Packets
A “wave packet” can be created by superposing many
waves spanning a wavevector range k ! ± !k 2 .
a(k n )
)
v p !10 8 m/s
1.98
y
k! " !k 2
Red
dv p
1.97
dk
k! + !k 2
# "39.3 m2 s
To generate the wave packet shown above, we put
Violet
y=
1.95
(
#
In this case,
If
dv p
kn
k!
!k
1.96
0
x
k !10 7 m"1
1.0
dv p
1.2
1.4
dk < 0, so v p > v g .
dk > 0, then v g > v p .
1.6
)
(
25
# a(k n )cos( k n x )
n="25
)
where k n = k ! + n !k 50 and the amplitudes a(k n ) are
a Gaussian distribution, i.e.,
2 2
a(k n ) = e " n $ .
We set k ! = 5, !k = 3 and $ = 10. Thus, there are 51
equally spaced values of k n in the sum.
In constructing a wave packet in this way we find the
For example, if we use the full-width at half maximum
interesting result that the spatial extent or “length” of the
(FWHM) for !x and !k we find that !x " !k # 5.54.
wave packet, !x, is inversely proportional to !k, viz:
!k = 1
!k = 0.333
x
x
a(k)
!x
!k = 2
x
!x
!k = 3
x
!x
!k = 4
x
!k
!x
!x " !k
1
$ 45 ± 1
$ 45
2
$ 22 ± 1
$ 44
3
$ 15 ± 1
$ 45
4
$ 11 ± 1
$ 44
arbitrary units
!x
!k
!k = 0.666
!x
!k = 1.00
x
!x
Although it is difficult to locate exactly the start and end
points of the wave packets, it is clear that
!x
a result we found earlier. The actual value of the
constant depends on the amplitude function a(k n ) and
the precise definitions of !x and !k.
!k
!x
!x" !k
0.333
16.65
5.54
0.666
8.33
5.55
1.00
5.52
5.52
1.33
4.16
5.53
arbitrary units
!k = 1.33
x
!x " !k # constant,
k
k!
x
The important point is that
there is a relationship
!x
between !x and !k, i.e.,
the range of k values and the width of the packet. We
will see the significance of this result in chapter 6.
Again, if we had we summed originally over a frequency
range, !f , instead of !k, we would have found similar
condition also occurs between the frequency range and
the temporal extent, !t ,
i.e., !f " !t # !$ " !t # constant.
vg
x
!x
Furthermore, a wave packet is composed of many
Question 1:
individual component waves of differing wavevector
(hence wavelength and frequency). So, a wave packet
For the wave packet shown, (a) estimate the range in
traveling through a dispersive medium, i.e., one in which
wavevectors !k. (b) If !t is the time it takes the wave
the phase velocity depends on frequency (and
packet to pass a point in space, what is the range of
wavevector) - such as glass, water, etc. - will broaden
frequencies !f in the wave packet?
because the component waves travel with different
velocities.
(b) The frequency is given by the number of oscillations
!x
(a) First, we determine how many oscillations there are
in the wave packet; but it is difficult to say definitively
passing the point in time !t , i.e.,
f = N !t .
"!f = !N !t = 1 !t .
But # = 2$f , i.e., !# = 2$!f ,
" !# 2$ = 1 !t ,
i.e., !# % !t = 2$ (a constant).
where the waves start and stop. We find there are
between 12 and 13 oscillations, so, let us take N = 12 but
with an uncertainty !N = 1.
"# = !x N = 1612 = 1.33 units.
2$ 2$N
But k =
=
.
#
!x
2$
2$
"!k =
!N =
% 1 = 0.393 inverse units.
!x
16
Note that !x & !k = 2$, i.e., a constant whose value
depends on the uncertainty !N.
Earlier we found that !x % !k = 2$ ( = !#.!t )
" !# !k = !x !t ,
which the speed of the wave packet; it is called the
group velocity ( v g). We note that
!x % !k = 2$ and !# % !t = 2$
represent the minimum values, with !N = 1, i.e., the
uncertainty in the number of oscillations.
A monchromatic source is defined as one with a constant
wavelength (or frequency). However, if a source
produces a single wave train of finite length, i.e., for a
finite length of time, the disturbance cannot be
monchromatic.
y
Phasors
Returning to single frequency (i.e., monochromatic)
waves. Suppose we add two sine waves. What is the
resultant?
r1
t
!t
As the wave train shown above is not a simple harmonic
wave - it has a start and an end - and it is non-repeating,
it must be represented as an infinite sum of harmonic
waves of a range of different frequencies, centered on
the principal frequency. Thus, strictly speaking, unless
the wave train produced by a source is infinitely long, it
cannot be truly monchromatic.
However, in what follows, we will ignore such effects.
S1
P
Consider two separated
r2
sources, S1 and S2, with
S2
identical frequency,
wavelength and amplitude.
Such sources are described as
coherent. Then at P we have:
y1 = A cos(!t " kr1 ) and y 2 = A cos(!t " kr2 )
Thus, the resultant disturbance at P is:
y1 + y 2 = A cos(!t " kr1 ) + A cos(!t " kr2 )
How can we work this out ??
... remember phasors ??
You might think that we could use trigonometric
identities but, in general, that is not the best approach.
y(t) = A cos!t = A cos"(t)
Let’s solve this problem using phasors ...
(" + # $ %)
A
"
y(t)
R
!
#
A
%
Put y1 = A cos(/t $ kr1 ) = A cos #
and y 2 = A cos(/t $ kr2 ) = A cos %
i.e., # = /t $ kr1 and % = /t $ kr2 .
A
y1
y(t)
!
y2
The length of the resultant phasor is R = 2A cos ! .
But, 2! + (" + # $ %) = " & ! = $ (# $ %) 2 .
(
+
'R = 2A cos*) $ (# $ %) 2 -, .
But cos. is an even function, i.e., cos! = cos($! ).
(
+
'R = 2A cos* (# $ %) 2 - .
)
,
t
A “phasor” is a rotating vector. Phasors can be added just
like ordinary vectors.
http://www.animations.physics.unsw.edu.au/jw/phasor-addition.html
Thus, the projection of this phasor onto the x-axis is
y1 + y 2 = R cos(# + ! ),
(
+
= 2A cos*) (# $ %) 2 -, cos(# + ! ) .
amplitude
Substituting for !, " and # we get:
&$ %$ )
&
$ + $1 )
y1 + y 2 = 2A cos( 2 1 + cos( ,t % 2
+,
' 2 *
'
2 *
where ($2 % $1 ) = (kr2 % kr1 ) = -, i.e., the phase difference
between the two waves.
We can also get the result trignometrically, viz:
y1 + y 2 = A cos(,t % kr1 ) + A cos(,t % kr2 )
= A cos(,t % $1 ) + A cos(,t % $2 ).
Using the trignometric relationship:
& X % Y)
& X + Y)
cosX + cosY = 2cos(
+ cos(
+,
' 2 *
' 2 *
we find, as before, (see top of page),
& $ % $1 )
&
$ + $1 )
y1 + y 2 = 2A cos( 2
+ cos( ,t % 2
+.
' 2 *
'
2 *
Although this seems like much less work, it is only
applicable if the amplitudes of y1 and y 2 are equal. The
method using phasors is more generally applicable ... see
the next example ...
Question 2:
Find the resultant of the two waves whose disturbances
at a given location vary with time as follows:
y1 = 4 cos!t and y 2 = 3cos !t + " 3 .
(
)
Interference
Draw the two waves as phasors.
3
R
$
#t
r1
( )
3sin ! 3 = 2.60
!
3
3cos ! 3 = 1.50
( )
4
( )) (
Earlier, we found the resultant
r2
at P for two identical sources S1
S2
and S2 is:
#! "! &
#
! + !1 &
y1 + y 2 = 2A cos% 2 1 ( cos% )t " 2
(,
$ 2 '
$
2 '
where (!2 " !1 ) = k(r2 " r1 ) = k*r =
The length of the resultant phasor is given by
(
S1
P
( ))
2
2
R 2 = 4 + 3cos ! 3 + 3sin ! 3 = 37,
"R = 6.08.
The resultant wave (y1 + y 2 ) is given by the projection
of R onto the horizontal axis, i.e.,
R cos(#t + $),
& 2.60 )
!
where $ = tan %1 (
+ = 25.3 , 0.44 rad.
' (4 + 1.5) *
"y1 + y 2 = 6.08cos(#t + 0.44 rad).
2+
*r. Here,
,
• (r2 " r1 ) = *r is called the path difference, and
• (!2 " !1 ) = - is the phase difference
between the two waves. Note, the resultant disturbance
(y1 + y 2 ) is a maximum when
(!2 " !1 )
- k
+
= n+ = = *r = *r,
2
2 2
,
EVEN
i.e., when *r = n, or - = 2n+.
Thus, when the path difference is an integral number (n)
of wavelengths or the phase difference is an even number
of +, (y1 + y 2 ) is a maximum. This condition is known as
constructive interference.
r1
S1
P
r2
S2
The resultant at P is:
#! "! &
#
! + !1 &
y1 + y 2 = 2A cos% 2 1 ( cos% )t " 2
(.
$ 2 '
$
2 '
But, if
(!2 " !1 ) * 3* 5*
(2n + 1)
= ,
,
, etc =
*,
2
2 2 2
2
the resultant disturbance (y1 + y 2 ) = 0. That occurs
when
(!2 " !1 ) + k
*
(2n + 1)
= = ,r = ,r =
*
2
2 2
2 ODD
#
1&
i.e., ,r = % n + ( - or + = (2n + 1)*
$
2'
So, when the path difference is a half-integer number of
wavelengths or the phase difference is an odd number
of *, (y1 + y 2 ) = 0. This is destructive interference.
Question 3:
If a transparent sheet of plastic of thickness 1.00mm and
refractive index µ = 1.25 is placed in the path of a ray,
what is the optical path difference ( !r) it introduces?
1.00mm
L
(a)
The general wave equation
Earlier we wrote:
y(t, x) = A cos(!t " kx)
(b)
t
Let the overall distance be L, the thickness of the plastic
be t and the wavelength of the light be !. Then, the
number of wavelengths in case (a) is
L
Na = ,
!
and the number of wavelengths in case (b) is
L " t t L " t µt L + (µ " 1) t
Nb =
+ =
+ =
.
!
!#
!
!
!
So, Nb > Na , and the optical path difference is
% L + (µ "1)t L(
$r = ( Nb " Na )! = '
" * ! = (µ "1)t.
&
!
!)
Using t = 1.00 + 10"3 m and µ = 1.25, we get
$r = 0.25 + 1.00 +10 "3 m = 0.25mm.
to describe the wave. However, this is not the only
expression we could have chosen. For example,
y(t, x) = Bsin(!t " kx) and y(t, x) = Ce "i(!t" kx)
are also waves. They may look different but they are all
solutions of the “general wave equation”
#2 y 1 #2 y
=
.
#x 2 v 2 #t 2
Example: if y(t, x) = A cos(!t " kx),
#y
#2y
=
kA
sin(!t
"
kx)
and
#x
#y
#2y
=
"!A
sin(!t
"
kx)
and
= "! 2 A cos(!t " kx).
2
#t
#t
#x 2
# 2y
= "k 2 A cos(!t " kx),
$ k ' 2 #2y
# 2y 1 #2y
=& )
, i.e.,
=
,
#x 2 % ! ( #t 2
#x 2 v 2 #t 2
where v is the phase velocity of the wave.
Preliminary observations prior to discussing interference
Standing waves
and diffraction effects.
Consider two waves with the same amplitude traveling in
Look at what happens
opposite directions along the x-axis,
when plane water waves,
i.e., y R = a cos(!t " kx) and y L = a cos(!t + kx).
Then applying the superposition principle, the resultant is
traveling from the left,
encounter a barrier with a
y R + y L = a cos(!t " kx) + a cos(!t + kx)
small opening. The waves
Using the trignometric relationship:
# A + B&
# A " B&
cos A + cosB = 2cos%
( cos%
(,
$ 2 '
$ 2 '
on the right are circular and centered on the opening just
as if there was a point source of waves at the opening.
we get
y(x, t) = y R + y L = 2a cos(!t)cos("kx )
d
= 2a cos(!t)cos( kx ) .
This is a standing wave; because the crests and troughs
“stand in-place” but the amplitude varies sinusiodally
with a maximum of 2a!
(a) ! << d
(b) ! " d
(c) ! >> d
If we vary the wavelength and the size of the opening we
find that when ! << d, as in (a), the ray approximation is
See animation ...
http://www.walter-fendt.de/ph14e/stwaverefl.htm
valid but when ! " d the waves spread out. The
spreading effect in (b) and (c) is called diffraction.
P
Two-slit interference pattern (Young, 1801)
r1
P
S1
r1
S1
yn
r2
r2
d
S2
d
S2
*
*
L
L
d sin *
L >> d
If the path difference to P between the two rays is
!r = (r2 " r1 ) = n#,
where n is an integer, we have constructive interference,
i.e., maximum disturbance at the point P on the screen.
However, if
$
1'
!r = (r2 " r1 ) = & n + ) #,
%
2(
we have destructive interference, i.e., zero disturbance
at the point P.
yn
!
!
d sin !
L >> d
Thus, we have interference maxima at angle !, when
"r = dsin ! n = n# ,
and interference minima when
$
1'
"r = dsin ! n = & n + ) #.
%
2(
The phase difference at point P is
2+
* = k"r = d sin !.
#
If L >> d, y n = L tan !n . For small angles, tan ! , sin !,
so the condition for maximum disturbance at P is
$ #L '
y n = n& ) .
( n = 0, ± 1, ... ).
% d (
Let E1 be the electric field at P due to the waves from S1
For zero disturbance at P,
!
1 $ 'L
yn = # n + & .
"
2% d
and E2 be the electric field at P due to the waves from S2.
Since both electric fields result from the same single source
( n = 0, ± 1, ... ).
So, providing ( ) small, the positions of the maxima and
minima are separated by a distance
1 'L
*y =
,
2 d
i.e., they are equally spaced.
If light is used as the source, what is the intensity
distribution across the screen?
With all e-m radiation (like light) the “disturbance” or
“wave function” is the electric field vector, i.e.,
! !
E = A" sin +t.
that illuminates the two slits, they have the same frequency
!
and amplitude, and the E-vectors will be parallel. When
they reach P, they will have a phase difference ! = k"r.
So, we can represent the individual wave functions at P as
E1 = A" sin #t and E2 = A" sin(#t + !).
The resultant is
E = E1 + E2 = A" sin #t + A" sin(#t + !).
Using the identity
($ & %) ($ + %)
sin $ + sin % = 2cos
sin
,
2
2
we find
'
!
!*
E = 2A" cos .sin ) #t + , .
(
2
2+
!
Thus the amplitude of the wave is 2A " cos .
2
The intensity of e-m radiation is proportional to the
square of the amplitude, i.e.,
!
I = 4I! cos2 ,
2
where I! is the intensity of the light on the screen from
2#
either slit separately and ! = k"r = d sin %.
$
Intensity
Question 4:
4I!
Iav = 2I!
y
Using a conventional two-slit apparatus and light that has
a 589nm wavelength, 28 bright fringes per centimeter are
observed near the center of a screen 3.00m away. What
is the slit separation?
n=
,4
,3
,2
,1
0
1
2
3
4
An alternating intensity pattern is observed on the screen,
called an interference pattern. Maximum intensity occurs
& $L )
where y = n ( + , with n = 0, ± 1, ± 2, etc.
' d *
From earlier, when ! is small, the spacing between bright
1 $ #L'
and dark fringes is "y = & ) . Since bright and dark
2% d (
fringes are equally spaced, the spacing between
neighboring bright fringes is
"y n,n+1 =
,d =
1 * 10+2
#L
m/cm
=
= 3.57 * 10 +4 m .
d
28 fringes/cm
#L
(589 * 10 +9 m)(3 m)
=
= 4.95 *10 +3 m
+4
"y n,n+1
3.57 *10 m
= 4.95mm.
Class discussion problem: At the dark fringes there is
zero light intensity and so no energy is arriving. When
light waves interfere and produce an interference pattern
what happens to the energy in the light waves?
At the dark fringes there is zero light intensity and so no
Conditions for interference
energy is arriving. When light waves interfere and
produce an interference pattern what happens to the
To observe interference,
energy in the light waves?
the sources must be coherent, i.e., they must maintain a
The energy is re-distributed non-uniformly. The energy
constant phase difference with respect to each other.
in the dark regions is less than the average; the energy in
the bright regions is above the average.
Class discussion problem: Which of the following are
coherent sources:
Intensity
4I!
Iav = 2I!
y
So, the energy “missing” in the dark fringes is
“transferred” to the bright fringes.
(a) two candles,
(b) a point source and its image in a plane mirror,
(c) two pinholes illuminated by the same source,
(d) two headlights of a car,
(e) two images of a point source reflecting from the
top and bottom surfaces of a glass block.
Class discussion problem: If you were to blow smoke
into the region between the slits and the viewing screen,
would the smoke show evidence of the interference?
Class discussion problem: What would the interference
P
pattern look like if the light was composed of two
different wavelengths; say, red light with ! R = 750nm
and blue light with ! B = 430nm?
r1
S1
S2
r2
Note: we cannot use the
small angle approximation
400m
300m
Question 5:
since d ! L ! y.
1000m
(a) From earlier the condition for a maximum is
"r = dsin # = n$ . At the n = 2 maximum,
Two radio antennas, separated by 300m, simultaneously
broadcast identical coherent signals. A radio in a car
traveling north receives the
signals. (a) If the car is at
400m
300m
1000m
the position of the second
maximum, what is the
wavelength of the signals?
(b) How much farther north does the car have to travel
to reach the next minimum?
# = tan %1 (400 m 1000 m ) = 21.8! .
dsin # (300 m) sin 21.8!
&$ =
=
= 55.7 m.
n
2
(b) After the n = 2 maximum, the next minimum occurs
when the path length is one-half wavelength greater,
i.e., when
'
1*
5
dsin # = ) n + , $ = $.
(
2+
2
' 5$ *
' 5 - 55.7 m*
&# = sin %1 ) , = sin %1 )
, = 27.7! .
( 2 - 300 m +
( 2d +
So at the minimum: y = (1000 m) tan 27.7! = 525 m.
Therefore, the car must travel an additional
"y = (525 % 400) m = 125 m.
Interference in thin films and air gaps
Consider a thin film (of water or oil) that is viewed at a
small angle to the normal. Some of the light from the
Reflection of waves
source is (directly) reflected from the top surface (#1).
Most light enters the film and is refracted. Some of that
Incident pulse
When a traveling wave or
light is reflected from the
pulse encounters a change in
lower surface and refracted at
the medium, part of the wave
#1
#2
the upper surface (#2). At the
(or all of the wave, if it is a
hard medium) is reflected.
However, the reflected pulse
Reflected pulse
upper surface, a change of
t
phase of ! occurs on
reflection. At the lower
or wave is inverted, i.e., it
surface, there is no change of phase on reflection. If the
undergoes a phase change of
rays are nearly parallel, the path difference is "r = 2t so
! (180!).
the phase difference is
(
)
k #"r = 2! $ # 2t,
http://www.kettering.edu/~drussell/Demos/reflect/reflect.html
where $ # = $ n is the wavelength of light in the medium
(of refractive index n).
n1 = 1
n 2 " 1.33
n 3 " 1.5
Because of the phase change on reflection at the upper
surface, the total phase difference between #1 and #2 is
! = 2" $ # 2t + ".
(
)
• if 2t = m$ # , i.e., a whole number of wavelengths,
So, if a film of water is on a glass slide, say, rays
reflected at the air-water interface and at the waterglass interface both undergo a phase change of !, so the
then
! = " + 2"m = (2m + 1)"
(= ", 3", 5" …),
phase difference between these rays when they emerge
in air is determined solely by the thickness of the water
and we get destructive interference.
film, i.e., 2t.
• if 2t = (2m + 1) $ # 2 , i.e., an odd number of halfwavelengths, then
! = " + (2m + 1)" = 2(m + 1)"
Air wedge
(= 2", 4", 6" …)
and we get constructive interference.
1
Consider a wedge between
2
t
Note: there is only a change in phase of " if the
reflection occurs at an interface where the incident and
reflected rays are in the less dense medium; a so-called
hard reflection.
x
# = tx
two glass slides. Ray #1 is
reflected from the upper
glass- air interface, where
there is no change of phase. Ray #2 is reflected from
the lower air-glass interface, where there is a change of
phase of !.
Therefore, the total phase difference between rays #1
and #2 is
(
Newton’s rings
)
! = k"r + # = 2# $ 2t + #.
When 2t = m$, then
! = 2#m + # = (2m + 1)#
(= #, 3#, 5# …),
i.e., destructive interference occurs.
But, since % = t x , that condition occurs when
m = 2t $ = 2x% $ , i.e., x = m$ 2% .
So, dark fringes appear that are equally spaced in x. The
linear density of dark fringes, i.e., the number per unit
length is
m = 2% .
x
$
They are caused by interference
between light reflected from the
bottom of the curved surface and
the light reflected from an optically
air gap
optical flat
Shown alongside are the
flat surface. If the curved surface
fringes produced by two very
is spherical the fringes are circular. Since the thickness
flat glass plates inclined at a
of the air gap increases away from the point of contact,
very small angle.
the fringe spacing decreases.
2r
L
From the notes, dark fringes appear when 2t = m!.
Therefore, if dark fringes appear when 2t = m!, bright
Question 6:
fringes occur when
"
1%
2t = $ m + ' !,
#
2&
(m =
Light, with wavelength 600nm, is used to illuminate two
glass plates at normal incidence. The plates are 22.0cm
in length. They touch at one end but are separated at the
other end by a wire of radius 0.025mm. How many
bright fringes appear along the length of the plates?
( m = 0, 1, 2, 3 …).
2t 1
) .
! 2
The maximum value of m occurs when t = 2r. Then
2(2r) 1 4 * 0.025 * 10 )3 1
m=
) =
) = 166.2.
!
2
2
600 * 10 )9
Hence, mmax = 166,
since m must be a whole number.
Note, the spacing of bright fringes is
22.0 * 10 )2
=
= 1.33 * 10 )3 m (1.33mm)
mmax
166
L
Class discussion question:
Class discussion question:
Consider the thin film we looked at previously. What
difference, if any, would it make to the appearance of
Consider the air-wedge shown below, where the air gap
the fringes if white light was used instead of
increases linearly. Is the first band (fringe) at the right
monochromatic light?
hand edge - where the gap is zero - bright or dark?
#1
#2
t