The document provides an overview of interference and diffraction concepts. It discusses interference due to thin films of uniform and non-uniform thickness. For thin films of uniform thickness, the conditions for constructive and destructive interference are derived based on the total path difference being equal to integral or half-integral multiples of wavelength. For wedge-shaped thin films of non-uniform thickness, the optical path difference expression is derived and the conditions for bright and dark fringes are obtained. Key features of interference patterns from wedge-shaped films like equidistant, straight and parallel fringes are also summarized.

1. UNIT I
INTERFERENCE
AND
DIFFRACTION

2. Syllabus contents
• Interference and Diffraction
Interference due to thin films of uniform thickness and non-uniform thickness
(with derivation), Newton’s rings, Applications of interference.
Fraunhoffer diffraction at a single slit; condition of maxima and minima, Plane
Diffraction grating (diffraction at multiple slits).

3. Outline
• Introduction
• Definition of interference
• Young’s double slit experiment and principle of superposition
• Interference due to thin films of uniform thickness
• Interference due to thin films of non uniform thickness (wedge shaped film)

4. Techniques of Interference
Division of wavefront Division of amplitude

5. THIN FILM INTERFERENCE
Thin film: A film is said to be thin when its
thickness is of the order of one wavelength of
visible light ~ 5500 Å (0.55μm).
↑ Iridescence caused by interference : Colours as seen due to
interference phenomena on oil layer, soap bubbles, beetle body,
oil film on rock and morph butterfly wings, Colours on CD and
DVD.

6. INTERFERENCE
PRINCIPLE OF SUPERPOSITION:
The resultant displacement of a particle of the
medium acted upon by two or more waves
simultaneously is the algebraic sum of the
displacements of the same particle due to
individual waves.
y = y1 + y2
y’= y1 - y2

7. INTERFERENCE
Definition: Interference is the physical effect caused by
superimposition of two or more wave trains traveling in the same
direction at the same time.
When two or more waves superimpose, the resultant amplitude
(intensity) in the region of superposition is different than the
amplitude (or intensity) of individual waves. This modification in
the distribution of intensity in the region of superposition is
called interference.

8. INTERFERENCE OF WAVES
The equation of two waves are
The wave resulting from their superimposition is
y=y1+y2





 



x
T
t
a
y 2
sin
1





 





x
T
t
a
y 2
sin
2

9. INTERFERENCE OF WAVES
The wave resulting from their superimposition is





 



x
T
t
a
y 2
sin 




 




x
T
t
a 2
sin





 







 2
/
2
sin
.
cos
2
x
T
t
a
y
+

10. INTERFERENCE OF WAVES
Constructive Interference: Sum of amplitudes due
to two waves when they meet in phase
Destructive Interference: Difference of amplitudes
due to two waves when they meet at a point in
opposite phase

11. INTERFERENCE OF WAVES
Constructive Interference: Sum of amplitudes due to two
waves
Path difference:
=n=(2n) /2 where n=0,1,2,3….
=0, , 2, 3, ……..n
Phase difference:
=(2/) 
=(2n ) where n=0, 1, 2, 3….
=0, , 2, 3,….n

12. INTERFERENCE OF WAVES
Destructive Interference: Difference of amplitudes due to
the two waves
Path difference:
=(2n+1) /2, where, n=0,1,2,3…..
=/2, 3/2, 5/2 …..(2n+1)/2
Phase difference:
=(2/ )  = (2n+1) where n=0,1,2,3….
=, 3, 5,………

13. Techniques of obtaining interference
These techniques can be divided into two broad classes:
[1] Wave front splitting: A wave front of light emerging from a
narrow slit can be divided it into two by passing it through two closely
spaced slits.
Young’s double slit expt., Fresenl’s expt. of double mirror,
Fresenel’s biprism, Lloyd’s mirror, etc employ this technique.
[2] Amplitude splitting: The amplitude (intensity) of a light wave is
divided into two parts, namely reflected and transmitted components,
by partial reflection at a surface.
Interference in thin films (parallel, wedge shaped, Newton’s
ring, etc), Michelson’s interferometer etc utilize this method. Also,
optical elements such as beam splitters, mirrors are for achieving
amplitude divison

14. INTERFERENCE DUE TO THIN FILMS
OF UNIFORM THICKNESS
A transparent thin film of uniform thickness bounded by
two parallel surfaces is known as a plane parallel thin film.
ASSUMPTIONS:
[1]According to Stokes, when a light wave is reflected at the
surface of optically denser medium, it suffers a phase change
of  i.e. a path difference of /2.
[2]A distance ‘l’ traversed by light wave in a medium of
refractive index ‘’ has it’s equivalent optical path ‘l’.

15. INTERFERENCE DUE TO THIN FILMS OF
UNIFORM THICKNESS
When the film is observed in reflected light

16. INTERFERENCE DUE TO THIN FILMS
OF UNIFORM THICKNESS
The effective path difference
between waves BR and DR1 is,
=2tcosr + (/2)

17. When the film is observed in reflected light
Condition for darkness: The total path difference should be odd multiple of
/2
 = (2n + 1)/2 where n=0,1,2,3,…..
2tcosr + (/2)= (2n + 1)/2
2tcosr = n
Condition for brightness: The total path difference should be an
integral multiple of 
 = n where n=0,1,2,3,…..
2tcosr + (/2)= n
2tcosr = (2n+1)/2
INTERFERENCE DUE TO THIN FILMS OF
UNIFORM THICKNESS

18. When the film is observed in transmitted light
INTERFERENCE DUE TO THIN FILMS OF
UNIFORM THICKNESS

19. When the film is observed in transmitted light
Condition for brightness: The total path difference should be odd multiple
of /2
 = (2n + 1)/2 where n=0,1,2,3,…..
2tcosr + (/2)= (2n + 1)/2
2tcosr = n
Condition for darkness: The total path difference should be an
integral multiple of 
 = n where n=0,1,2,3,…..
2tcosr + (/2)= n
2tcosr = (2n+1)/2
INTERFERENCE DUE TO THIN FILMS OF
UNIFORM THICKNESS

20. INTERFERENCE DUE TO THIN FILMS OF
UNIFORM THICKNESS
Some important points
[a] The conditions of interference depend on three parameters i.e. t,  and r.
In case of parallel film, t and r remain constant.
The conditions of interference solely depend on the wavelength ‘’.
[b] When parallel beam of monochromatic light is incident normal to the film, the
whole film will appear uniformly dark or uniformly bright.
The film will appear bright in reflected light when the thickness of the film is
/4, 3/4, 5/4…..
The film will appear dark in reflected light when the thickness of the film is /2,
/, 3/2…..
[c] A change in the angle of incidence of the rays leads to a change in the path
difference. Consequently, if the inclination of the film with respect to the light
beam is changed gradually, the film appear dark and bright or bright and dark in
succession.
[d] If a parallel beam of white light is incident on the film, those wavelengths for
which the path difference is ‘n’, will be absent from the reflected light. The other
colours will be reflected

21. INTERFERENCE DUE TO WEDGE-
SHAPED FILMS OF NON-UNIFORM
THICKNESS
ASSUMPTIONS:
[1]According to Stokes, when a light waves is reflected at the
surface of optically denser medium (i.e) surface backed by a
denser medium), it suffers a phase change of  i.e. a path
difference of /2.
[2]A distance ‘l’ traversed by light wave in a medium of
refractive index ‘’ has it’s equivalent optical path ‘l’.

22. INTERFERENCE DUE TO WEDGE-SHAPED FILMS OF
NON-UNIFORM THICKNESS

23. INTERFERENCE DUE TO WEDGE-SHAPED
FILMS OF NON-UNIFORM THICKNESS

24. INTERFERENCE DUE TO WEDGE-SHAPED FILMS OF
NON-UNIFORM THICKNESS
From the complete geometry of above figure, BPQ and
BPC have a common side BP,
From BPQ, BQP=α
QBP=90
BPQ=90-α
From BPC, CBP=90-r
CPB=90-α
Hence BCP=r+α
PG is parallel to DP and ray BCP is assumed to be
straight and intersecting at PQ and DP at C and P,
hence BPD=r+α
DS=SP=t, CS is common side of both triangle CSD and
CSP, i. e. CP=CD,
Hence, CDS= r+α

25. The optical path difference  is given by,
 =  (BC + CD) – BF
=  (BE + EC + CD) –  (BE)
=  (EC + CP)
=  (EP)
From EPD,
= 2t cos(r + )
Due to reflection from a surface backed by a
denser medium suffers an abrupt phase change
of  which is equivalent to a path difference of
/2.
eff = = 2t cos(r + )  /2
INTERFERENCE DUE TO WEDGE-SHAPED
FILMS OF NON-UNIFORM THICKNESS

26. INTERFERENCE DUE TO WEDGE-SHAPED
FILMS OF NON-UNIFORM THICKNESS
Condition for brightness: The total path difference should be an integral multiple
of 
=n where n=0,1,2,3,…..
2tcos(r+) + (/2)= n
2tcos(r+) = (2n+1)/2
Condition for darkness: The total path difference should be odd multiple of /2
=(2n + 1)/2 where n=0,1,2,3,…..
2tcos(r+) + (/2)= (2n + 1)/2
2tcos(r+) = n

27. WEDGE-SHAPED FILMS
Salient features of the Interference Pattern due to wedge
shaped thin film
(1) Fringes are equidistant
Consider two consecutive dark
fringes at point A and C as shown
in figure.
The nth dark band be
formed at point A at a distance x1
from the edge of contact ‘O’ and t1
be the thickness of film at A.
The (n+1) th dark band is formed at point C at a distance
x2 from ‘O’ and t2 is the thickness of film at point C.

28. WEDGE-SHAPED FILMS
For destructive interference
condition,
2 μ t cos r = n λ (normal incidence,
cos r =1)
or 2 μ t1= nλ ------- (1)
For (n+1)th dark fringe,
2 μ t2 = (n+1) λ --------- (2)
Subtracting (1) from (2) we get,
2 μ (t2 – t1) = λ -------- (3)
From figure, in right angled triangle AEC,
tan θ = 𝐶𝐸/𝐴𝐸=t2 − t1/x2−x1
Since θ is small, tan θ ~ θ,t2 − t1= (x2−x1 ) θ ---- (4)

29. WEDGE-SHAPED FILMS
Substituting value of t2-t1 in equation (3) we get
2 μ (x2-x1) θ = λ
Since x2-x1 = β = Fringe width (ie. distance between two consecutive dark
fringes).
Therefore, 2 μ β θ = λ or Fringe width β = 𝜆/2μ𝜃
For air film, refractive index μ =1,
β = 𝜆/2𝜃
Since 𝜆, 𝜃 are constant, β is constant. Hence fringes are equidistant .
Interference
pattern

30. WEDGE-SHAPED FILMS
[2] The fringe at apex is dark
At the apex, the thickness of the wedge is very small compared to
λ, i.e. 𝑡≪𝜆. Therefore thickness of film at apex is zero.
The optical path difference becomes, Δ = 2μt−λ/2 = λ/2
For path difference of 𝜆/2, the interfering rays will always be 180°
out of phase and interfere destructively . Therefore, the fringe at
the apex of the wedge is always dark.

31. WEDGE-SHAPED FILMS
[3] Fringes are straight and parallel
The locus of points having the same thickness lie along lines parallel
to the apex line of the wedge.Thus, the fringes are straight. Since
the fringes are equidistant and straight, they are parallel.
[4] Fringes of equal thickness
Each maximum or minimum is a locus of constant film thickness, the
fringes are called fringes of equal thickness.

32. WEDGE-SHAPED FILMS
[5] Fringes are localized
The fringes are located at the top surface of the
wedge film.
[6] Wedge Angle ()
Experimentally, we can find the
wedge angle ‘θ’ using a travelling
microscope.
As shown in the fig consider two dark
fringes formed at points A and B at a
distance x1and x2 respectively from
apex ‘O’.

33. WEDGE-SHAPED FILMS
Let the thickness of the film be
t1 and t2 at A and B respectively.
 At point A, 2 μ t1 = n λ
From the figure, t1 = x1tan θ ~
x1 θ (as θ is very small)
∴2μ x1 θ = nλ ------(1)
Similarly, at point B,
2μ x2 θ = (n+N) λ ------- (2)
where N is the number of fringes between A and B.
Subtracting (1) from (2) we get, 2 μ(x2-x1) θ = Nλ
Hence 𝜃 = 𝑁λ/2 μ(x2-x1)
for air film, 𝜃 = 𝑁λ/2 (x2-x1) ----- (3)

34. WEDGE-SHAPED FILMS
[7] Spacer thickness
To find the thickness of spacer/sheet or diameter of
wire ‘t’
From figure , t = 𝑙 tan θ
where 𝑙 is the length of air wedge
Substituting the value of θ from eqn.3 we get,
𝑡 = 𝑙 𝜃 =λ/2μ (x2-x1) -----(usingeqn.3)