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Interference
Interference is the phenomenon in which two waves superpose to
form the resultant wave of the lower,higher or same amplitude .The
most commonly seen interference is the optical interference is the
optical interference or light interference.This is because light waves are
randomly generated every which way by most sources.This mean that
light waves coming out of a source do not have a constant
amplitude,frequency or phase.One of the best examples of the
interference is demonstrated by the light reflected from a film of oil
floating on water.
Interference
Types of interference:
• There are two types of interference
1. Constructive interference of light: When the amplitude of the waves
increases because of the wave amplitudes reinforcing each other is
known as constructive interference.
2. Destructive interference of light:When the amplitude of the waves
of the waves reduces because of the wave amplitudes opposing
each other is known as destructive interference.
Types of interference
Newton Ring
• Newtons ring is a phenomenon in which an interference
pattern is created by the reflection of light between two
surfaces;a spherical surface and an adjacent touching flat
surface.Circular interference fringe is produced by enclosing
a vert thin film of air or any other transparent medium of
varying thickness between a plane glass plate and a convex
lens of a large radius of curvature.Such fringe were first
obtained by Newton and are known as Newtons rings.
Newtons Rings
Newtons ring due to reflected light
• For the bright rings due to reflected light,
• 2ut cosθ =(2n-1)λ/2
• 2ut =(2n-1)λ/2, when u =1 and cosθ =1
• n=1,2,3……
• For the dark rings due to reflected light,
• 2ut cosθ =nλ,n=0,1,2….
• 2t =nλ
Newtons ring due to reflected light
• Therefore,the radius of the bright ring
• r^2 =(2n-1)λR/2
• Therefore,the radius of the dark ring
• r^2 =nλR
• Therefore ,Dn^2 =4nλR
• Measure the diameter of the (n+m)th dark ring .Let it be Dn+m
•
(Dn+m )^2
4
=(n+m) λR………. …. …(i)
• (Dn+m)^2 =4(n+m) λR…….. …. ……(ii)
Newtons ring due to reflected light
• Subtracting (i) from (ii)
(Dn+m )^2
4
- C =4m λR
• λ =
Dn+m 2
−
(Dn )^2
4
4𝑚𝑅
• Hence the wavelength of a given monochromatic source of
light can be determined.
Fresnel bi-prism
Fresnel used a biprism to show interference phenomenon.A Fresnel
Biprism is a thin double prism placed base to base and have very small
refracting angle (0.5˚).This is equivalent to a single prism with one of its
angle nearly 179˚ and other two of 0.5˚ each.It can be used to determine
the wavelength of a light source (monochromatic),thickness of a thin
transparent sheet,thin film,refractive index of medium etc.
Fresnel Biprism
Fresnel bi-prism
• Let us get the image of virtual sources by placing a convex lens as
shown above.
• We get from lens equation the relation between the distance u and v
• (1/v ) + (1/u ) = 1/f ......................(2)
• where v is lens-to-image distance and u is lens-to-object distance
• magnification of image = ( d1 / d ) = v/u ......................(3)
• When v > u , we get magnified image.
Fresnel bi-prism
• Let us get the image of virtual sources by placing a convex lens as
shown above.
• We get from lens equation the relation between the distance u and v
• (1/v ) + (1/u ) = 1/f ......................(2)
• where v is lens-to-image distance and u is lens-to-object distance
• magnification of image = ( d1 / d ) = v/u ......................(3)
Fresnel bi-prism
• When v > u , we get magnified image.
• In lens equation (2) , u and v are interchangeable .
• Hence if we position the convex lens at a distance v from object , we
get a diminshed image at a distance v
•
• new magnification of image = ( d2 / d ) = u / v ....................(4)
Fresnel bi-prism
• By multiplying eqn.(3) and eqn.(4) , we get , ( d1 / d ) ( d2 / d ) = ( v/u
) ( u / v ) = 1
•
• From above expression , we get , d =√(d1 d2 )
• Hence , using the distance d between virtual sources in eqn.(1) and
from the measured values
• of fringewidth β it is possibe to detrmine the wavelength λ of
monochromatic light .
LASER :
LASER APPLICATION
LASER APPLICATION
Polarization of light
Polarized light waves are light waves in which the vibrations
occur in a single plane.The phenomenon of polarization has
helped to establish beyond doubt that light waves are
transverse waves.The process by which light waves vibrating
in a particular plane is called polarization of light.
Types of Polarization
• Depending upon the longitudinal and transverse wave movement,the
classification of polarization is three types.
1. Linear polatization
2. Circular polarization
3. Eliptical polarization
Types of Polarization
Methods of achieving Polarization
1. Reflection
2. Scattering
3. Refraction
4. Double refrection
Application of Polarization
1. Polarization is used in sunglasses to reduce the glare.
2. Three dimensional movies are produced and shown with the help of
polarization.
3. Polarization is used for differentiating between transverse and
longitudinal waves.
4. Infrared spectroscopy uses polarization.
Application of Polarization
Application of Polarization
Diffraction of Light
The rhythmic variations in intensity and the bending of light around the
corners of an obstacle or the encroachment of light into the region of
geometrical shadow constitute a class of phenomena known as the
diffraction of light.The diffracting object or aperture effectively becomes
a secondary source of the propagating wave.Italian scientist Francesco
Maria Grimadi coined the word diffraction and was the first to record
accurate observations of the phenomenon in 1660.
Diffraction of Light
Application
1. CD reflecting rainbow colours.
2. Holograms
3. From the shadow of an object.
4. Bending of light at the corner of the door.
5. Spectrometer.
6. X-ray diffraction
7. To separate white light.
Application
Application
Application
Application
Nicol Prism
Construction
• It consist of a rhombohedral crystal of Iceland spar that has been cut at
an angle of 68˚ with respect to the crystal axis ,cut again
diagonally,and then rejoined as shown,using a layer of transparent
medium Canada balsam as a glue.Unpoiarized light ray enters through
the left face of the crystal ,as shown in the diagram ,and is split into
two orthogonally polarized ,differently directed rays by the
birefringence property of calcite.

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Interference of Light Waves

  • 1. Interference Interference is the phenomenon in which two waves superpose to form the resultant wave of the lower,higher or same amplitude .The most commonly seen interference is the optical interference is the optical interference or light interference.This is because light waves are randomly generated every which way by most sources.This mean that light waves coming out of a source do not have a constant amplitude,frequency or phase.One of the best examples of the interference is demonstrated by the light reflected from a film of oil floating on water.
  • 3. Types of interference: • There are two types of interference 1. Constructive interference of light: When the amplitude of the waves increases because of the wave amplitudes reinforcing each other is known as constructive interference. 2. Destructive interference of light:When the amplitude of the waves of the waves reduces because of the wave amplitudes opposing each other is known as destructive interference.
  • 5. Newton Ring • Newtons ring is a phenomenon in which an interference pattern is created by the reflection of light between two surfaces;a spherical surface and an adjacent touching flat surface.Circular interference fringe is produced by enclosing a vert thin film of air or any other transparent medium of varying thickness between a plane glass plate and a convex lens of a large radius of curvature.Such fringe were first obtained by Newton and are known as Newtons rings.
  • 7. Newtons ring due to reflected light • For the bright rings due to reflected light, • 2ut cosθ =(2n-1)λ/2 • 2ut =(2n-1)λ/2, when u =1 and cosθ =1 • n=1,2,3…… • For the dark rings due to reflected light, • 2ut cosθ =nλ,n=0,1,2…. • 2t =nλ
  • 8. Newtons ring due to reflected light • Therefore,the radius of the bright ring • r^2 =(2n-1)λR/2 • Therefore,the radius of the dark ring • r^2 =nλR • Therefore ,Dn^2 =4nλR • Measure the diameter of the (n+m)th dark ring .Let it be Dn+m • (Dn+m )^2 4 =(n+m) λR………. …. …(i) • (Dn+m)^2 =4(n+m) λR…….. …. ……(ii)
  • 9. Newtons ring due to reflected light • Subtracting (i) from (ii) (Dn+m )^2 4 - C =4m λR • λ = Dn+m 2 − (Dn )^2 4 4𝑚𝑅 • Hence the wavelength of a given monochromatic source of light can be determined.
  • 10. Fresnel bi-prism Fresnel used a biprism to show interference phenomenon.A Fresnel Biprism is a thin double prism placed base to base and have very small refracting angle (0.5˚).This is equivalent to a single prism with one of its angle nearly 179˚ and other two of 0.5˚ each.It can be used to determine the wavelength of a light source (monochromatic),thickness of a thin transparent sheet,thin film,refractive index of medium etc.
  • 12. Fresnel bi-prism • Let us get the image of virtual sources by placing a convex lens as shown above. • We get from lens equation the relation between the distance u and v • (1/v ) + (1/u ) = 1/f ......................(2) • where v is lens-to-image distance and u is lens-to-object distance • magnification of image = ( d1 / d ) = v/u ......................(3) • When v > u , we get magnified image.
  • 13. Fresnel bi-prism • Let us get the image of virtual sources by placing a convex lens as shown above. • We get from lens equation the relation between the distance u and v • (1/v ) + (1/u ) = 1/f ......................(2) • where v is lens-to-image distance and u is lens-to-object distance • magnification of image = ( d1 / d ) = v/u ......................(3)
  • 14. Fresnel bi-prism • When v > u , we get magnified image. • In lens equation (2) , u and v are interchangeable . • Hence if we position the convex lens at a distance v from object , we get a diminshed image at a distance v • • new magnification of image = ( d2 / d ) = u / v ....................(4)
  • 15. Fresnel bi-prism • By multiplying eqn.(3) and eqn.(4) , we get , ( d1 / d ) ( d2 / d ) = ( v/u ) ( u / v ) = 1 • • From above expression , we get , d =√(d1 d2 ) • Hence , using the distance d between virtual sources in eqn.(1) and from the measured values • of fringewidth β it is possibe to detrmine the wavelength λ of monochromatic light .
  • 19. Polarization of light Polarized light waves are light waves in which the vibrations occur in a single plane.The phenomenon of polarization has helped to establish beyond doubt that light waves are transverse waves.The process by which light waves vibrating in a particular plane is called polarization of light.
  • 20. Types of Polarization • Depending upon the longitudinal and transverse wave movement,the classification of polarization is three types. 1. Linear polatization 2. Circular polarization 3. Eliptical polarization
  • 22. Methods of achieving Polarization 1. Reflection 2. Scattering 3. Refraction 4. Double refrection
  • 23. Application of Polarization 1. Polarization is used in sunglasses to reduce the glare. 2. Three dimensional movies are produced and shown with the help of polarization. 3. Polarization is used for differentiating between transverse and longitudinal waves. 4. Infrared spectroscopy uses polarization.
  • 26. Diffraction of Light The rhythmic variations in intensity and the bending of light around the corners of an obstacle or the encroachment of light into the region of geometrical shadow constitute a class of phenomena known as the diffraction of light.The diffracting object or aperture effectively becomes a secondary source of the propagating wave.Italian scientist Francesco Maria Grimadi coined the word diffraction and was the first to record accurate observations of the phenomenon in 1660.
  • 28. Application 1. CD reflecting rainbow colours. 2. Holograms 3. From the shadow of an object. 4. Bending of light at the corner of the door. 5. Spectrometer. 6. X-ray diffraction 7. To separate white light.
  • 34. Construction • It consist of a rhombohedral crystal of Iceland spar that has been cut at an angle of 68˚ with respect to the crystal axis ,cut again diagonally,and then rejoined as shown,using a layer of transparent medium Canada balsam as a glue.Unpoiarized light ray enters through the left face of the crystal ,as shown in the diagram ,and is split into two orthogonally polarized ,differently directed rays by the birefringence property of calcite.