UNIT-1
INTERFERENCE
Interference:
When two or more waves are superimposed then there is a modification of intensity or
amplitude in the region of superposition. This modification of intensity or amplitude in the
region of super position is called Interference.
When the resultant amplitude is the sum of the amplitudes due to two waves, the
interference is known as Constructive interference and when the resultant amplitude is equal to
the difference of two amplitudes, the interference is known as destructive interference.
PRINCIPLE OF SUPERPOSITION:
This principle states that the resultant displacement of particle in a medium acted upon by
two or more waves simultaneously is the algebraic sum of displacements of the same particle due
to individual waves in the absence of the others.
Consider two waves traveling simultaneously in a medium. At any point let y1 be the
displacement due to one wave and y 2 be the displacement of the other wave at the same instant.
Then the resultant displacement due to the presence of both the waves is given by
y = y1 ± y2
+ve Sign has to be taken when both the displacements y1 & y 2 are in the same direction
−ve Sign’ has to be taken when both the displacements y1 & y2 are in the opposite direction.
INTERFERENCE IN THIN FILMS
Consider a thin film of thickness t and refractive index µ . A ray of light OA incident on the
surface at an angle i is partly reflected along AB and partly refracted into medium along AC,
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,making an angle of refraction r .at C it is again partly reflected along CD. Similar refractions
occur at E.
To find the path difference between the rays, draw DB perpendicular to AB
Then the path difference = µ ( AC + CD ) − AB …………….(1)
From triangle ACE
CE
cos r =
AC
CE t
AC = = ………………..(2)
cos r cos r
From triangle CDE
CE
cos r =
CD
CE t
CD = = ……………….(3)
cos r cos r
From triangle ABD
AB
cos(90 − i ) =
AD
AB = AD cos(90 − i ) = 2 AE sin i ………..(4) ( ∵ AD = 2 AE )
FROM triangle ACE
AE
sin r = ⇒ AE = AC sin r
AC
t sin r t
AE = ( ∵ AC = )
cos r cos r
From Eq (4)
2t sin r
AB = × sin i
cos r
2t sin r sin i sin r
AB = ×
cos r sin r
2
2µt sin r sin i
AB = ………………………(5) (∵ µ = )
cos r sin r
On substituting the values of AC , CD & AB from Eq(2),(3)&(5) in Eq(1) ,we get
t t 2µt sin 2 r
The path difference = µ ( + )−
cos r cos r cos r
2µt 2µt cos 2 r
= (1 − sin 2 r ) = = 2µ t cos r
cos r cos r
∴The path difference = 2 µ t cos r
According to the theory of reversibility, when the light ray reflected at rarer-denser interface, it
λ
introduces an extra phase difference π (or) path difference of
2
∴The actual path difference = 2 µ t cos r − λ
2
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,Case.1: condition for maximum intensity
We know that the intensity is maximum when path difference= nλ
λ
∴ From Eq.(6) 2µt cos r − = nλ
2
λ
2 µ t c o s r = ( 2 n + 1)
2
Case.2: condition for minimum intensity
λ
We know that the intensity is minimum when path difference = (2n + 1)
2
λ λ
∴ from Eq.(6) 2µt cos r − = (2n + 1)
2 2
2 µ t cos r = ( n + 1)λ
NEWTON’S RING EXPERIMENT
A Plano convex lens(L) having large focal length is placed with its convex surface on the
glass plate(G2).a gradually increasing air film will be formed between the plane glass plate and
convex surface of Plano convex lens. The thickness of the air film will be zero at the point of
contact and symmetrically increases as we go radially from the point of contact.
A monochromatic light of wavelength ‘λ’ is allowed to fall normally on the lens with the
help of glass plate (G1) kept at 450 to the incident monochromatic beam. A part of the incident
light rays are reflected up at the convex surface of the lens and the remaining light is transmitted
through the air film. Again a part of this transmitted light is reflected at on the top surface of the
glass plate (G1).both the reflected rays combine to produce an interference pattern in the form of
alternate bright and dark concentric circular rings, known as Newton rings. The rings are circular
because the air film has circular symmetry. These rings can be seen through the travelling
microscope.
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, THEORY
Consider a Plano convex lens is placed on a glass plate. Let R be the radius of curvature and r
be the radius of NEWTON ring, corresponding to constant film thickness.
As one of the rays suffers reflection at denser medium, so a further phase changes of π or path
λ
difference of takes place.
2
λ
The path difference between the rays =2µt cos r + ----------------------------- (i)
2
For air µ = 1 , and normal incidence r = 0
λ
∴ Path difference = 2t +
2
AT THE POINT OF CONTACT
The thickness of the air film t=0, µ=1 & for normal incidence r = 0.
λ
Then the path difference = .
2
λ
If the Then the path difference = then the corresponding phase difference is π.so that
2
gives a dark spot is formed at the centre.
For bright ring
λ
2t + = nλ
2
λ
2t = (2n − 1) ----------------------------------- (ii)
2
For Dark ring
λ λ
2t + = (2n + 1)
2 2
2t = nλ ------------------------------------------- (iii)
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INTERFERENCE
Interference:
When two or more waves are superimposed then there is a modification of intensity or
amplitude in the region of superposition. This modification of intensity or amplitude in the
region of super position is called Interference.
When the resultant amplitude is the sum of the amplitudes due to two waves, the
interference is known as Constructive interference and when the resultant amplitude is equal to
the difference of two amplitudes, the interference is known as destructive interference.
PRINCIPLE OF SUPERPOSITION:
This principle states that the resultant displacement of particle in a medium acted upon by
two or more waves simultaneously is the algebraic sum of displacements of the same particle due
to individual waves in the absence of the others.
Consider two waves traveling simultaneously in a medium. At any point let y1 be the
displacement due to one wave and y 2 be the displacement of the other wave at the same instant.
Then the resultant displacement due to the presence of both the waves is given by
y = y1 ± y2
+ve Sign has to be taken when both the displacements y1 & y 2 are in the same direction
−ve Sign’ has to be taken when both the displacements y1 & y2 are in the opposite direction.
INTERFERENCE IN THIN FILMS
Consider a thin film of thickness t and refractive index µ . A ray of light OA incident on the
surface at an angle i is partly reflected along AB and partly refracted into medium along AC,
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,making an angle of refraction r .at C it is again partly reflected along CD. Similar refractions
occur at E.
To find the path difference between the rays, draw DB perpendicular to AB
Then the path difference = µ ( AC + CD ) − AB …………….(1)
From triangle ACE
CE
cos r =
AC
CE t
AC = = ………………..(2)
cos r cos r
From triangle CDE
CE
cos r =
CD
CE t
CD = = ……………….(3)
cos r cos r
From triangle ABD
AB
cos(90 − i ) =
AD
AB = AD cos(90 − i ) = 2 AE sin i ………..(4) ( ∵ AD = 2 AE )
FROM triangle ACE
AE
sin r = ⇒ AE = AC sin r
AC
t sin r t
AE = ( ∵ AC = )
cos r cos r
From Eq (4)
2t sin r
AB = × sin i
cos r
2t sin r sin i sin r
AB = ×
cos r sin r
2
2µt sin r sin i
AB = ………………………(5) (∵ µ = )
cos r sin r
On substituting the values of AC , CD & AB from Eq(2),(3)&(5) in Eq(1) ,we get
t t 2µt sin 2 r
The path difference = µ ( + )−
cos r cos r cos r
2µt 2µt cos 2 r
= (1 − sin 2 r ) = = 2µ t cos r
cos r cos r
∴The path difference = 2 µ t cos r
According to the theory of reversibility, when the light ray reflected at rarer-denser interface, it
λ
introduces an extra phase difference π (or) path difference of
2
∴The actual path difference = 2 µ t cos r − λ
2
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,Case.1: condition for maximum intensity
We know that the intensity is maximum when path difference= nλ
λ
∴ From Eq.(6) 2µt cos r − = nλ
2
λ
2 µ t c o s r = ( 2 n + 1)
2
Case.2: condition for minimum intensity
λ
We know that the intensity is minimum when path difference = (2n + 1)
2
λ λ
∴ from Eq.(6) 2µt cos r − = (2n + 1)
2 2
2 µ t cos r = ( n + 1)λ
NEWTON’S RING EXPERIMENT
A Plano convex lens(L) having large focal length is placed with its convex surface on the
glass plate(G2).a gradually increasing air film will be formed between the plane glass plate and
convex surface of Plano convex lens. The thickness of the air film will be zero at the point of
contact and symmetrically increases as we go radially from the point of contact.
A monochromatic light of wavelength ‘λ’ is allowed to fall normally on the lens with the
help of glass plate (G1) kept at 450 to the incident monochromatic beam. A part of the incident
light rays are reflected up at the convex surface of the lens and the remaining light is transmitted
through the air film. Again a part of this transmitted light is reflected at on the top surface of the
glass plate (G1).both the reflected rays combine to produce an interference pattern in the form of
alternate bright and dark concentric circular rings, known as Newton rings. The rings are circular
because the air film has circular symmetry. These rings can be seen through the travelling
microscope.
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, THEORY
Consider a Plano convex lens is placed on a glass plate. Let R be the radius of curvature and r
be the radius of NEWTON ring, corresponding to constant film thickness.
As one of the rays suffers reflection at denser medium, so a further phase changes of π or path
λ
difference of takes place.
2
λ
The path difference between the rays =2µt cos r + ----------------------------- (i)
2
For air µ = 1 , and normal incidence r = 0
λ
∴ Path difference = 2t +
2
AT THE POINT OF CONTACT
The thickness of the air film t=0, µ=1 & for normal incidence r = 0.
λ
Then the path difference = .
2
λ
If the Then the path difference = then the corresponding phase difference is π.so that
2
gives a dark spot is formed at the centre.
For bright ring
λ
2t + = nλ
2
λ
2t = (2n − 1) ----------------------------------- (ii)
2
For Dark ring
λ λ
2t + = (2n + 1)
2 2
2t = nλ ------------------------------------------- (iii)
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