This document discusses the principles of interference and diffraction of light waves. It defines constructive and destructive interference based on the path difference between light waves. The conditions for constructive and destructive interference are outlined. The document also discusses Young's experiment to demonstrate light interference and derive an expression for the bandwidth of interference fringes. It explains how the biprism experiment can be used to measure the wavelength of light using interference principles.

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Interference and Diffraction
Principle of Superposition of Waves
a. Statement :
When two or more waves arrive at a point in the
medium simultaneously then each wave produces
its own displacement independent of the other
waves. The resultant displacement at that point is
given by the vector sum of individual displacements
produced by each wave.
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Interference of Light
Redistribution of light intensity in the region of medium
is due to the physical process called as “Interference of
light”
Constructive Interference
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Destructive Interference
Path Length and Path Difference
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Condition
for
Destructive
Constructive
Interference
Interference
(Or
Conditions
and
for
Brightness & Darkness of a point)
a. Condition
for
Constructive
(brightness):1. Overlapping
2. Path difference = n λ = (2n) λ /2
[n = 0, 1, 2, ...........]
3. Phase difference
4. 4. Intensity
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Interference

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b. Condition
for
Destructive
Interference
(darkness):1. Overlapping
2. Path difference =(n – 1/2)λ
= (2n – 1) λ/2
[n = 1, 2, 3, 4..........]
3. Phase difference
4.
Intensity
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M.C.Q
Q.1
In interference of light
(a.1) Light energy is created
(c.1) Light energy is redistributed
Q.2
(b.1) Light energy is destroyed
(d.1) Light energy is doubled
Select the correct statement
(a.2) Only longitudinal waves produce interference.
(b.2) Only transverse waves
produce interference.
(c.2) Only standing waves produce interference. (d.2) Both longitudinal and
transverse waves produce interference.
Q.3
Two sources of light are said to be coherent if they emit light waves of the same
(a.3) Frequency and speed
(b.3) Wavelength and constant
phase difference
(c.3) Frequency and amplitude
Q.4
(d.3) Intensity and frequency
Which one of the following quantities is conserved in the interference of light waves?
(a.4) Phase difference
(b.4) Amplitude
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(c.4) Intensity
(d.4) Path difference

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Young’s Experiments to Demonstrate the
Phenomenon of Interference
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Theory of Interference Band
(Expression for bandwidth or fringe width of an
interference band.)
Expression for path difference S2P – S1P :In ∆PNS2 and ∆PMS. By Pythagoras theorem.
S2P2 = S2N2 + PN2
= D2 + (x + d/2)2
..................... (1)
S1P2 = S1M2 + PM2
= D2 + (x - d/2)2
...................... (2)
S2P2 – S1P2 = (x + d/2)2 – (x – d/2)2
= x2 + 2x d/2 + d2/4 – x2 + 2x d/2
– d2/4
= 2xd
..................... (3)
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If x, d < < D, then S1P ≈ S2P ≈ D
Equation (3) becomes.
2D (S2P – S1P) = 2xd
S2P – S1P = xd/D
..................... (4)
For a point P to be bright.
S2P – S1P = (2n) λ/2
.................... (5)
From (3) and (4), xd/D = n λ
x = n λ D/d
Let xn, xn – 1 = distances of nth and n + 1th bright
bands from O.
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Using these in (6),
xn = nλ D/d and xn - 1 = (n + 1)λ D/d
X = Band width of bright band
= xn+1 – xn = (nλ D/d) + (λ D/d) – (n λ D/d)
X = λD/d
..................... (7)
e. For a point P to be dark
S2P – S1P = (2n – 1) λ /2
..................... (8)
From (3) and (8), xd/D = (2n – 1) λ/2
x = (2n – 1) λD/2d
...................... (9)
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Let xn, xn -1 = distances of nth and n - 1th dark
bands from O. Using these in (9).
Xn = (2n – 1) λ D/2d
Xn – 1 = [2(n + 1) – 1] λD/2d = (2n + 1) λD/2d
X = Band width of dark band
= Xn-1 – Xn
= (2nλD/2d) + (λ D/2d) λ – (2n D/2d) + (λ D/2d)
X = λ D/d
...................... (10)
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Assuming the Expression for Path Difference
Obtain an Expression for Band Width of An
Interference Band
a. From above diagram, bright and dark bands are
placed alternately. Let, n = 0, 1, 2, 3....................
λ = Wavelength of the light used.
D = Distance between the sources and the screen.
d = Distance between the two sources.
For a bright band.
xn = n λ D/d and xn+1 = (n + 1) λ D/d
X = λ D/d
b. For a dark band.
Xn = (2n -1) λ D/2d
Xn-1 = [2 (n + 1) - 1] λ D/d = (2n + 1) λ D/2d
X = λ D/d
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Conditions for Obtaining Well Defined Steady
Interference Pattern
a. Statement :Two sources must be
i.
equally bright
ii. monochromatic
iii. coherent
iv. as narrow as possible
v. as closed as possible
b. Explanation :i.
This will not give a well-defined interference
pattern.
ii. This will not give a well defined interference
pattern.
iii. This produces unstable interference pattern.
iv. These will not a well defined interference pattern.
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v. Since X = λ D/d. For λ, D = constant
X α 1/d. Therefore smaller the distance between
the sources higher the bond width. This gives a well
defined interference pattern.
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Biprism
Two thin prisms of extremely small refracting angle
are connected bases to base. Such a prism formed
is called as “Biprism”. Biprism is used for producing
two coherent sources from a single source.
Biprism Experiment
This
experiment
is performed
to
find
wavelength of monochromatic source of light.
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unknown

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a. Measurement of D :- Distance between slit and the
eyepiece is measured directly from the scale on the
optical bench.
d. Measurement of X :- For this purpose vertical wire
of the cross wire is made coincide with one edge of
the white band and corresponding reading (xn) on
the micrometer scale is recorded using slow motion
screw. The cross wire is moved through known
number of bands (n). Vertical wire is again made
coincide with one edge of the bright band and
corresponding heading (xn) on the micrometer is
recorded.
Mean band width X is obtained by the formula.
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X = Xn – X0/n
e. Measurement of d (method of conjugate foci) :S1, S2 are virtual coherent sources. Distance
between these sources (d) cannot be measured
directly.
1. A convex lens is interposed in between biprism
is so adjusted that real, bright and enlarged
images S1’ and S2’ of S1, S2 are observed in the
focal plane of the eyepiece. Using slow motion
screw distance (d1) between S1’ and S2’ is
measured.
u = object distance
v = Image distance
object size / Image size
= object distance / Image distance
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d/d1 = u/v
......................... (1)
2. Convex lens is displaced towards eyepiece and
its position is so adjusted that real, diminished
and reduced images S1’’ and S2’’ of S1, S2 are
observed in the focal plane of the eyepiece.
Using slow motion screw distance (d2) between
S1” and S2” is measured.
d/d2 = v/u
.........................(1)
Multiplying (1) by (2),
d/d2 = u/v × v/u
d2/d1d2 = 1
d2 = d1d2
d=
d1d 2
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f.
Unknown wavelength l is calculated by :λ = X d/D = X d1d 2 / D
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