Transcript 2.4 Interference

Properties of a stationary wave (2)

All particles between two adjacent nodes
(within one vibrating loop) are in phase.
Video
1. Stationary waves (string)
2. Stationary waves (sound)
4 Interference




When two waves meet, they
interfere.
Superposition occurs to give
constructive and destructive
interferences.
To produce a permanent
interference pattern, the
sources must be coherent.
The waves from coherent
sources have
(1) the same frequency,
(2) the same wavelength,
(3) constant phase difference.

If their phase difference is not constant, at a
certain point, there may be reinforcement at
one instant and cancellation at the next. If
these variations follow one another rapidly,
the interference pattern will change quickly.
 The wave causing interference should have
roughly the same amplitude. This is to ensure
the wave cancel each other to produce a
minima (zero amplitude).
Constructive interference:
The waves arrive at a point in phase
S1
P
S2
resultant
Destructive interference:
The waves arrive at a point exactly out of
phase.
S1
P
resultant
S2
½l
1
1
1
Path difference from 2 sources equals l , 1 l , 2 l
2
2
2
and so on.
Factors affecting the interference pattern

(1) Source separation
When source separation increases, the
separation between antinodal (or nodal) lines
decreases.
Increase the
separation of two
sources

(2) Wavelength
When wavelength decreases, the separation
between antinodal (or nodal) lines decreases.
Decrease the
wavelength
Young’s experiment
http://www.fed.cuhk.edu.hk/sci_lab/download/project/interferen
ce/interference.html
Young’s double-slit experiment

It is very important to use a single light source
and a double slit, rather than two light
sources. It is because the two sets of light
waves passing through the double-slit are
coherent.
 Since the wavelengths of light waves are very
small, the separation between the slits must
be very small.
 The screen should be placed at an
appreciable distance from the slits so that the
separation of fringes is observable.
Interference pattern of light

Explanation:
 Diffraction of light occurs at each slit. Since
the two diffracted waves overlap, interference
occurs.
 Bright fringes are where constructive
interference occurs while dark fringes are
where destructive interference occurs.
Path difference for Young’s
double-slit experiment
P
X
a
Q a
R
Y
Since a << D, PX and PY are almost parallel ⇒ q ≈ a
and  PQY ≈ 90o
By geometry q’ = a ⇒ q ≈ q’
Path difference = PY – PX = QY ≈ a sin q.
Path difference for Young’s double-slit
experiment
P
X
a
Q a
R
Y
Path difference = a sin q.
Constructive interference:
nl
If the
bright fringe is at P, a sin qn = nl ⇒ sin q n 
a
Destructive interference:
nth
If the mth bright fringe is at P, a sin qm = (m – ½) l ⇒
m 1
2l
sin q m 
a
Fringe position
P
X
a
Q a
R
Y
Path difference = a sin q.
nth bright fringe
Let yn be the distance between the nth bright fringe and
the central bright fringe.
nl
D
yn = D tan qn ≈ D sin qn =
a
Fringe position
P
X
a
Q a
R
Y
yn = nl D
a
Dy = distance between 2 successive bright fringes
n  1l D  l D
nl
D

= yn – yn-1 =
a
a
a
1.
2.
The fringes are evenly spaced ( Dy 
lD
)
a
The fringe spacing for red light is greater than for blue
light.
l
aDy
D
∴
lred > lblue
3.
The interference is incomplete because for all fringes
except the central bright one, the amplitudes of the
two wave-trains are not exactly equal.
Appearance of Young’s interference
Fringes
http://micro.magnet.fsu.edu/primer/java/doubleslit/index.html

If white light is used the central fringe is
white and the fringes on either side are
coloured.
Interference Fringe Pattern
Measuring wavelength of light
l can be measured by using the formula
l
aDy
D
http://www.matter.org.uk/schools/Content/Interference/
doubleslits_1.html
Interference by Thin Films

Thin film interference patterns seen in
Thin film of soapy water
Seashell
A thin layer of oil on the
Water of a street puddle
Parallel-sided Thin Film (1)

Consider a film of soap with uniform thickness
in air
When a beam of light is incident
on to the surface of the film, part
of incident light is reflected on
the top surface and part of that
transmitted is reflected on the
lower surface.
air
t
If the film is not too thick, the two
Soap film
reflected beams are close
together to produce an
interference effect.
http://webphysics.davidson.edu/physlet_resources/bu_semester2/c26_thinfilm.html
Phase change of p

Interference occurs for rays 1 and 2
ray 1 ray 2
A
phase change
of p
C
air
film
B
d
air
Suppose the thickness of the film is d and its refractive index
is n.
Let l be the wavelength of light in air.
Consider almost normal incidence (angle of incidence ≈ 0o)
Interference due to reflected rays
(Optical) Path difference for rays 1 and 2 = 2nd
Phase change of p
ray 1 ray 2
A
phase change
of p
C
air
film
B
d
air
If light travelling in a less dense medium is reflected by a dense medium,
the reflected wave is phase-shifted by π.
No phase change will be experienced by transmitted rays.
(Optical) Path difference for rays 1 and 2 = 2nd
Conditions for constructive interference and destructive interference
Bright fringes:
Dark fringes:
2nd = (m + ½)l
where m = 0, 1, 2, 3…..
(a phase change of p occurs at A)
2nd = ml
where m = 0, 1, 2, 3…..
ray 1 ray 2
A
phase change
of p
C
air
film
B
D
d
air
ray 3 ray 4

Interference due to transmitted rays (ray 3 and ray 4)
Bright fringes:
Dark fringes:
2nd = ml
where m = 0, 1, 2, 3…..
(no phase change of p occurs at B)
2nd = (m + ½)l
where m = 0, 1, 2, 3…..
Blooming of Lenses (1)



The process of coating a
film on the lens is called
blooming.
A very thin coating on the
lens surface can reduce
reflections of light
considerably.
This makes use of
destructive interference
of light to reduce the
reflection.
http://users.erols.com/renau/thinfilm.html
air (n = 1)
phase change of p
Coating film MgF2 (n = 1.38)
Glass (n = 1.72)


Path difference of the rays = 2nd
For destructive interference between rays 1 and 2
2nd = l/2 (both rays undergo a phase change of p)
d = l/(4n)
Thickness of coating
Put l = 5.5 x 10-7 m, n = 1.38 (refractive index of
coating)
d = 5.5 x 10-7 / (4 x 1.38) = 9.97 x 10-8 m
air (n = 1)
phase change of p
Coating film MgF2 (n = 1.38)
Glass (n = 1.72)
Note:
 1
The thickness of the film (coating) should be of ¼
wavelength of light in the film.
 2. With suitable blooming, the reflectance can be reduced
from 4% to less than 1%.
 3
The interference is complete for one wavelength only.
An average value of l (i.e. green – yellow) is chosen.
For red and blue light, the reflection is weakened but
not eliminated and bloomed lens appears purple.
no phase change
air (n = 1)
Coating film MgF2 (n = 1.38)
phase change of p


Glass (n = 1.72)
While destructive interference occurs between
reflected rays, constructive interference occurs
between transmitted rays.
If there is constructive interference on one side of the
film, there will be destructive interference on the other
side (energy conservation).
Brilliant colours of oil film on water
Investigating oil film on water
Brilliant colours of oil film
Constructive
interference
Air
qred
na = 1.0
Constructive
interference
qblue
oil film
no = 1.4
water
nw = 1.33

Interference occurs between two wave-trains – one
reflected from the surface of the oil and the other from the
oil-water interface.

When the path difference gives constructive interference
for light of one wavelength, the corresponding colour is
seen in the film.
Constructive
interference
Air
qred
na = 1.0
Constructive
interference
qblue
oil film
no = 1.4
water
nw = 1.33

The path difference varies with the thickness of the film
and the angle of viewing, both of which affects the colour
produced.

If the film is not thin, for a particular angle of viewing,
constructive interference between reflected rays occurs for
more than one colour. Therefore, many colours are present
in the reflected light. This gives the appearance of white
light and no specific colour is seen.
Soap film
phase change of p

A soap film mounted on a ring is held vertically. At first
the film appears uniformly bright. As the soap drains to the
bottom, a series of interference fringes are seen.
Soap film

For normal incidence, bright fringes
are observed if

2nd = (m – ½) l ,
phase change
where n is the refractive index of soap,
of p
l is the wavelength of light in air, and
m = 1, 2, 3, …

Minimum thickness of the film for
bright fringe

dmin = l /(4n)

Hence, when the upper part of the film
becomes extremely thin < l /(4n),
constructive interference does not take
place and a black area or black fringe
is observed.

As time goes by, the film drains downwards further and
does not break, the fringe pattern changes:
1
2
3
4



Dark area at the top increases and moves downwards.
The number of fringes increases.
Fringes are more closely spaced towards the bottom.

The figure above shows an air
wedge formed by a thin film
and a glass block. They are
separated by a thin piece of
paper so that the wedge angle
q is very small.

In the arrangement,
monochromatic light from a
source is partially reflected
vertically downwards by a
glass plate G.

When a microscope is focused
on the wedge, bright and dark
equally-spaced fringes are
seen.

This is because the reflected
rays interfere with each other
to form an interference pattern.
Thin Film of Air

Light rays reflected from the upper and lower
surfaces of a thin wedge of air interfere to
produce bright and dark fringes.
 The fringes are equally spaced and parallel
to the thin end of the wedge.
http://www.gg.caltech.edu/~zhukov/applets/film/applet.html
Thin Film of Air




Consider almost normal incidence.
Path difference of two rays = 2d
For dark fringes, 2d = nλ.
For bright fringes, 2d = (n+½)λ.
θ
Phase change of p
d
Thin Film of Air, Wedged-shaped (2)
θ
d
Phase change of p
Fringe separation
For two adjacent dark fringes, Dd = ½[ml – (m – 1)l]  ½l
Dx 
Dd
l
l


tan q
2 tan q 2q
since q is very small.
Note:
 1. If the path difference > coherent length, no fringe
is formed.
 2
In order to have a clear fringe pattern, the fringe
separation should be increased. This can be done by
making the air wedge as thin as possible.
 3
At the practical level, every film absorbs some of
the light going through it. Thick films absorb
proportionately more than thin ones, thereby reducing
the dark and light bands in an interference pattern.
Applications of air wedge
1.Measuring diameter of a metal wire
q
5 cm
d
l
l
Dx 
 q 
2q
2Dx
Suppose the distance between the 1st fringe and the 91st fringe
observed is 16.2 mm and the wavelength of light emitted from
the light source is 690 nm.
Fringe separation Dx = 16.2 mm / 90 = 0.18 mm
690 109
3
o
radian
=
0.110

1
.
91

10
Angle of the wedge =
2  0.18  103
If the length of the air wedge is 5 cm, the diameter of the
metal wire d
≈ 5 cm x 1.91 x 10-3 = 9.58 x 10-3 cm = 9.58 x 10-5 m
2. Testing the flatness of surface
Concave upward


Concave downward
In making of optical ‘flats’, the plate under test is
made to form an air wedge with a standard plane
glass surface.
Any uneven parts of the surface will show up as
irregularities in what should be a parallel, equallyspaced, straight set of fringes.
Find the thickness of the air wedge at P.
Wavelength of white light l ≈ 5.5 x 10-7 m
xP
At P, destructive interference occurs between the reflected rays.
Path difference = l
2t = l
t = l / 2 = 2.75 x 10-7 m
Find the thickness of the air wedge at Q.
Wavelength of white light l ≈ 5.5 x 10-7 m
xQ
What is the thickness of the air
wedge at R?
xR
At Q, constructive interference occurs between the reflected rays.
Path difference = 1.5 l
2t = 1.5l
t = 1.5l / 2 = 4.125 x 10-7 m
Thickness of air wedge around the
ring is equal.
Flatness of Surfaces


Observed fringes for a wedged-shaped air film
between two glass plates that are not flat.
Each dark fringe
corresponds to a region of
equal thickness in the film.
 Between two adjacent
fringes the change in
thickness is λ/2μ.
where μ is the refractive
index of the film.
Newton’s Rings (AL only)

When a curved glass surface is placed in
contact with a flat glass surface, a series of
concentric rings is seen when illuminated from
above by monochromatic light. These are
called Newton’s rings.
Diffraction pattern through an obstacle
Diffraction Patterns
Interference by multiple slits

The following shows interference fringes by a single
slit, double-slit and 3 slits, 7 slits and 15 slits with the
use of monochromatic light and white light.
1 slit
2 slits
7 slits
3 slits
15 slits
Computer simulations

The locations of the
principle maximum are
the same for any number
of slits.
 As the number of slits
increases,
 the width of maximum
becomes narrower
(sharper),
 intensity of maximum
increases (brighter), and
 the sub peak between any
two maxima falls.
Relative intensity
N=1
N=2
N=3
N=4
N=5
N = 10
Diffraction grating

A large number of equally
spaced parallel slits is called a
diffraction grating.
A diffraction grating is made by making many parallel
scratches on the surface of a flat piece of transparent material.
The scratches are opaque but the areas between the
scratches can transmit light. Thus, a diffraction grating
becomes a multitude of parallel slit sources when light falls
upon it.

Diffraction grating VS double slit

For measuring wavelength of light accurately, a diffraction
grating is used

Diffraction grating can achieve a sharper and brighter fringe
pattern.

Angular separation of fringes can be made larger by using a
diffraction grating as slit separation can be made many times
smaller.
Types of grating
1. Transmission grating ― Light passes the spaces
between lines.
2. Reflection grating ― Light reflects on the unruled
parts.
 Fine and coarse grating
 A fine grating (e.g. 600 lines / mm.)
 A coarse grating (e.g. 100 lines /mm)
Typical pattern using white light
Colour band (1st order)
Central band
(zero-th order)
Colour band (2nd order)


A typical pattern consists of
1.
a central bright band, called the zero order image, is white
2.
on either side of the central band, there are brilliant bands
of colours, called first – and second order spectra.
Note: Dispersion increases with order.
Theory of diffraction grating
Monochromatic light
A
B
Diffraction grating

X
1st
order maximum
(m = 1)
Zero th order
maximum
(m = 0)
1st order maximum
(m = -1)
θ
d
θ
Y
Path difference
= d sin θ
Consider wavelets coming from points A and B on
two successive slits and traveling at an angle q to the
direction of the incident beam.
 Path difference between the ray X and Y = d sin q.
 For constructive interference (mth order maximum)
 d sin q = ml
Turntable
Diffraction grating
Collimator C
θ
Light
source
Telescope T
Eyepiece
Achromatic
lenses
Cross-wire





Example
Find the wavelength of light if the angle turned for 1st
order maximum is 20o when a diffraction grating of 500
line / mm is used.
Solution:
For 1st order maximum,
d sin q = l
where d = 1mm / 500 = 2 x 10-6 m
The wavelength of light l = 2 x 10-6 sin 20o
= 6.84 x 10-7 m
Turntable
Diffraction grating
Collimator C
θ
Light
source
Telescope T
Eyepiece
Achromatic
lenses
Cross-wire



For constructive interference (mth order maximum)
d sin q = ml
Find all the values of q for all red fringes.
(l = 7 x 10-7 m, diffraction grating = 200 lines per mm)
Solution:
Maximum number of colour fringes



The number of orders of maximum is limited by the
grating spacing d and the wavelength l.
For a diffraction grating of 5000 lines / cm, slit
spacing d = 1 cm / 5000 = 2 x 10-6 m
Maximum for red light is achieved when
ml red
sin q 
d
∵ sin q  1
mlred
d
2  106
1 m 
m
 m  2.86
∴
7
d
lred
7  10
We can at most observe the 2nd order red fringe.

Similarly, for the highest order of violet fringe observed,
mlviolet
d
2  106
1 m 
m
m5
7
d
lviolet
4  10
We can should be able to observe the 5th order violet fringe.

The highest order maximum is the greatest integer 
.
d
l
Overlapping of colour bands
sin q m 
ml1
represents the condition for the mth order maximum
d
of wavelength l.
nl2
is for the nth order maximum of wavelength l2
sin q n 
d
If q m  q n
, the two colour band overlap.
Example 3


Show that 2nd order orange fringe will overlap with 3rd
order violet fringe.
It is given that the wavelengths for orange light and
violet light are 6 x 10-7 m and 4 x 10-7 m respectively.
Solution:
For 2nd order orange fringe,
sin q orange 2
2  6  107 1.2  106


d
d
For 3rd order violet fringe,
3  4  107 1.2  106
sin q violet3 

 sin q orange 2
d
d
∴ qorange2 = qviolet3 ⇒ 2nd order orange fringe overlaps with 3rd order
violet fringe.
Note:
*For a diffraction grating of 5000 lines / cm (d =
sin q red 2
2  7  107

 q red 2  44.4
6
2  10
*For 3nd order violet fringe,
sin q violet3
3  4  107

 q vilet3  36.9  q red 2
6
2  10
*Hence, 2nd spectrum overlaps with 3rd spectrum.
*For 2nd order red fringe,
1 cm
= 2 x 10-6 m)
5000
Application of diffraction grating


A diffraction grating can be used to determine the
wavelength of waves emitted by a source.
A diffraction grating is placed in front of a methane air
flame (left) and methane oxygen flame (right).
A methane air flame
A methane oxygen flame
By comparing the spectra produced, we can decide which
flame is hotter. Why?