1. I &II Semester -2020-21 Engineering Physics Unit-1. Wave OpticsBy
Hasan Ziauddin
Assistant Professor
Department of Physics
RIET , Jaipur
Rajasthan

2. Introduction
Self Introduction, Interaction with students with the discussion of following:- Science: ??? Engineering: ???? Technology: ?????

4. Basics for Optics Path difference and Phase difference.
Coherent Source ( Temporal , Spatial).
Young’s Double Slit Experiment.
Interference in Thin Films.
Interference in Wedge –Shaped Film.

5. Path difference and Phase difference

6. Coherence

7. Coherent Source

8. Temporal & Spatial Coherence
Temporal Coherence
Spatial Coherence
The type of coherence related with time
The type of coherence related with position
It is known as longitudinal coherence.
It is known as transverse coherence.
The temporal coherence of light is related to frequency bandwidth of the source. monocromaticity
Spatial coherence is related to size of light source

9. Interference Of Light
Interference is the phenomenon in which two waves superpose to form the resultant wave of the lower, higher or same amplitude.
I = A^2

10. Young’s Double Slit Experiment (Not in syllabus)
3rd order
2nd order
1st order

12. Thin Film Interference (Not in syllabus)
When the light is made incident on this thin film partial reflection and partial refraction occur from the top surface of the film. The refracted beam travels in the medium and again suffers partial reflection and partial refraction at the bottom surface of the film. In this way several reflected and refracted rays are produces by a single incident ray. As they moves are superimposed on each other and produces interference pattern. 2 1 5
Stokes treatment : When a beam is reflected from a denser medium, a path change of λ /2 (or phase π ) occur for the ray. 3 4 6

13. Interference in Parallel Film ( Reflected Rays) (Not in syllabus)
Consider a thin film of uniform thickness ‘t’ and refractive index  bounded between air. Let us consider monochromatic ray AB is made incident on the film, at B part of ray is reflected (R1) and a part is refracted along BC. At C The beam BC again suffer partial reflection and partial refraction,  the reflected beam CD moves again suffer partial reflection and partial refraction at D. The refracted beam R2 moves in air. These two reflected rays R1 and R2 interfere to produce interference pattern.

14. R1 & R2
Assignment

15. Stokes treatment : when a beam is reflected
from a denser medium (ray R1 at B), a path
change of λ /2 (or phase π ) occur for the ray.
Assignment
Shape of the fringes when
t= λ/4 , λ/2 and λ
Hint: Haidinger fringes

16. Special Case
When angle of incident is 90 degree then
Angle of refraction r =0
Assignment:
Condition for constructive and destructive interference in case of reflected.
What will be the pattern of fringes when observed from transmitted side.

17. Interference in Wedge Shaped Film (Reflected Rays) (Not in syllabus)
The wedge shaped film has a thin film of varying thickness, having thickness zero at one end and increases at the other. The angle of wedge is θ.

18. Stokes treatment : When a beam is reflected from a denser medium (ray R1 at B), a path change of λ /2 (or phase π ) occur for the ray.

19. Special Case
When angle of incident is 90 degree and θ or small angles.
Angle of refraction r=0 and θ= 0 (approx.), (r+ θ)
Cos (r+ θ)= cos 0 =1
Condition for constructive and destructive interference : Assignment

20. Thin film neither parallel nor wedge (Not in syllabus)
Assignment:
What will be the shape of the fringe
when the slit is neither parallel nor
wedge shaped
Hint: Fizeau fringes

21. Unit – 1 Wave Optics Newton’s Ring’s Michelson’s Interferometer
Fraunhofer Diffraction ( Single slit)
Diffraction Grating
Rayleigh criterion for limit of resolution

22. Newton’s Ring’s ( History)
It is named after Isaac Newton, who investigated the effect in his 1704 treatise Opticks.

23. Newton’s Ring’s Introduction
The formation of Newton’s Ring’s is an important
application of interference of
Light wave from opposite
faces of a thin film of variable
thickness.
Newton’s Ring’s
Reflected light
Transmitted light.

24. Experimental Set Up
Ref
Trans

25. Experimental setup

26. Theory Explained
When a Plano convex lens of long focal length is placed in contact on a plane glass plate, a thin air film is enclosed between the upper surface of the glass plate and the lower surface of the lens. The thickness of the air film is almost zero at the point of contact O and gradually increases as one proceeds towards the periphery of the lens. Thus points where the thickness of air film is constant, will lie on a circle with O as center. By means of a sheet of glass G, a parallel beam of monochromatic light is reflected towards the lens L. Consider a ray of monochromatic light that strikes the upper surface of the air film nearly along normal. The ray is partly reflected and partly refracted as shown . The ray refracted in the air film is also reflected partly at the lower surface of the film. The two reflected rays, i.e. produced at the upper and lower surface of the film, are coherent and interfere constructively or destructively. When the light reflected upwards is observed through microscope M which is focused on the glass plate, series of dark and bright rings are seen with center as O. These concentric rings are known as " Newton's Rings ". At the point of contact of the lens and the glass plate, the thickness of the film is effectively zero but due to reflection at the lower surface of air film from denser medium, an additional path of λ/2 is introduced or phase π (Stokes treatment). Consequently, (In reflected) the center of Newton rings is dark due to destructive interference.

27. Formation of Newton’s Ring’s
Newton’s Ring’s In reflected light
Optical path difference
between two successive
reflected waves QS1R1
and NS2R2
= 𝟐μ𝒅 ± 𝝀/𝟐….(1)
(refer: Interference in Wedge –Shaped Film
Note : In wedge thickness is t )
d= thickness of the air film at N and 𝜆/2 is the additional path difference due to reflection at G.

28. 𝟐μ𝒅=(𝟐𝒎 + 𝟏)𝝀/𝟐 {or (𝟐𝒎 - 𝟏)𝝀/𝟐 }…. (2) , Where 𝒎 = 𝟎, 𝟏, 𝟐,
Condition for constructive interference:
𝟐μ𝒅 = odd multiple of 𝝀/𝟐
𝟐μ𝒅=(𝟐𝒎 + 𝟏)𝝀/𝟐 {or (𝟐𝒎 - 𝟏)𝝀/𝟐 }…. (2) , Where 𝒎 = 𝟎, 𝟏, 𝟐,
Condition for destructive interference :
𝟐μ𝒅= even multiple of 𝝀/𝟐
𝟐μ𝒅 =𝟐𝒎.𝝀/𝟐 …… (3) , Where 𝒎=𝟎,𝟏,𝟐,𝟑
A fringe of a given order (m) will be along the loci of points of equal film thickness (d) and hence the fringe will be circular.

29. QQ1 is the radius of mth order bright or dark ring
From fig
QQ1 is the radius of mth order bright or dark ring
𝐐𝐐𝟏=𝐫𝐦
R= radius of curvature of the convex surface.
Since 𝑹≫𝒅 , we can write,

30. From eqn. 5 and 6 we can conclude that the radius of bright and dark rings is proportional to the square root of odd natural numbers and natural numbers respectively.

31. At the point of contact of lens and glass plate d=0. So from equation
As D= 2 r
Central Fringe:
At the point of contact of lens and glass plate d=0. So from equation
the condition for destructive interference will be satisfied with m=0. This indicates that the central fringe is dark and appears as dark spot.

32. Application of Newtons Ring
(Also try from dark fringes)
Where 𝒎=𝟎,𝟏,𝟐,𝟑 , and
p = Any fixed (+) integer..1,2,3…… ( Discuss in Experiment in detail) V.Imp

33. Determination of refractive index of liquid:

34. Newton’s ring’s with transmitted light:
Newton’s ring can also be observed with transmitted light. There are two differences in the reflected and transmitted systems of rings-
The rings in transmitted light are exactly
complementary to those seen in the reflected
light, so that the central spot is now bright.
The rings in transmitted light are much
poorer in contrast than those in reflected
light.

35. Numerical’s
Question: In a Newton's ring experiment the diameters of 4th and 12th dark rings are 0.4 cm and 0.7 cm respectively. Deduce the diameter of 20th dark ring. Ans: In Newton's ring experiment. Given that: m= 4; (m+p)=12, p=8 Dm = 0.4 cm and D m+p =0.7 cm. The wavelength of sodium light using Newton's ring is λ = D²m+p - Dm²/4pR 4λR = D²m+p - Dm²/p 4λR =(0.7)²-(0.4)²/p……(1) We know that the diameter of the dark ring in presence of air is Dm² = 4mλR D20² = 20 X (4λR)……(2) Putting the value of 4λR from Eq (1) in Eq (2) D20² = 20 X [(0.7)²- (0.4)²]/8 D20= 0.91 cm.

36. Question: In a Newton's ring set up, diameter of 20th dark ring is found to be 7.25mm. The space between spherical surface and the flat slab is then filled with water (μ= 1.33). Calculate the diameter of the 16th dark ring in new set up.
Ans. Given that : D20= 7.25 mm
We know that the diameter of mth ring in presence of air is
Dm² = 4mλR
D20² =4 X 20 X λR
4λR= (7.25)²/20……………………………(1)
New set up:
Now liquid is introduced, then diameter of the ring is
D`m² = 4mλR/μ
D`16²= 4 X 16 X λR/1.33
= 16 X (4λR)/1.33…………………………..(2)
Putting the value of 4λR from equation (1) in (2)
We get,
D`16²= 16 X (7.25)²/20 X 1.33
D`16 = 5.62 mm.

37. Luminiferous aether or ether

38. Michelson –Morley Experiment (Not in syllabus)

40. Conclusion
Einstein : "If the Michelson–Morley experiment had not brought us into serious embarrassment, no one would have regarded the relativity theory as a (halfway) redemption

41. Interferometer
Interferometers are investigative tools used in many fields of science and engineering. They are called interferometers because they work by merging two or more sources of light to create an interference pattern, which can be measured and analyzed; hence 'Interfere-o-meter', or interferometer.

42. MICHELSON’S INTERFERROMETER
Principle:- The MI works on the principle of division of amplitude. When the incident beam of light falls on a beam splitter which divided light wave in two part in different directions. These two light beams after traveling different optical paths, are superimposed to each other
and due to superposition interferences fringes formed.
(Image of M2)

43. Construction:- It consists of two highly polished plane mirror M1 and M2, with
two optically plane glass plate G1 and G2 which are of same material and same
thickness. The mirror M1 and M2 are adjusted in such a way that they are
mutually perpendicular to each other. The plate G1 and G2 are exactly parallel to
each other and placed at 45° to mirror M1 and M2. Plate G1 is half silvered from
its back while G2 is plane and act as compensating plate. Plate G1 is known as
beam-splitter plate.
The mirror M2 with screw on its back can slightly titled about vertical and
horizontal direction to make it exactly perpendicular to mirror M1. The mirror M1
can be moved forward or backward with the help of micrometer screw and this
movement can be measured very accurately.
Working: Light from a broad source is made parallel wavefront by using a convex lens L.
Light from lens L is made to fall on glass plate G1 which is half silver polished
from its back. This plate divides the incident beam into two light rays by the
partial reflection and partial transmission, known as Beam splitter plate. The
reflected ray travels towards mirror M1 and transmitted ray towards mirror M2.
These rays after reflection from their respective mirrors meet again at 'O' and
superpose to each other to produce interference fringes. This firings pattern is
observed by using telescope.
Functioning of Compensating Plate: In absence of plate G2 the reflected ray passes
the plate G1 twice, whereas the transmitted ray does not passes even once.
Therefore, the optical paths of the two rays are not equal. To equalize this path the
plate G2 which is exactly same as the plate G1 is introduced in path of the ray
proceeding towards mirror M2 that is why this plate is called compensating plate
because it compensate the additional path difference.

44. Formation of fringes in MI
When the mirror M1 and
the virtual image M2ꞌ of M2
are not exactly parallel
localized fringes are produced.
Assignment
Shape of the fringes when
d= λ/4 ,λ/2 and λ

45. Formation of Circular Fringes:
The shape of fringes in MI depends on inclination of mirror M1 and M2. Circular fringes are produced with monochromatic light, if the mirror M1 and M2 are perfectly perpendicular to each other. The virtual image of mirror M2 and the mirror M1 must be parallel. Therefore it is assumed that an imaginary air film is formed in between mirror M1 and virtual image mirror M'2. Therefore, the interference pattern will be obtained due to imaginary air film enclosed between M1 and M’2.From Fig. if the distance M1 and M2 and M'2 is 'd', the distance between S'1 and S'2 will be 2D.
If the light ray coming from two virtual sources
making an angle θ with the normal then the path
difference between the two beams from S1 and S2
will becomes
As one of the ray is reflecting from denser medium
mirror M1, a path change of λ/2 occurs in it (Stokes treatment).
Hence the effective path difference between them will be
Formation of Circular Fringes:

46. Where d and λ are constants, so θ will be constant for given order number (m). Hence maxima will be in the form of concentric circles about the foot of the perpendicular from the eyes to the mirror as a common center.
For Small angle θ
 This type of fringes are called fringe of equal inclination.
 Fringes are non-localized and situated at infinity.

47. Formation of Localized fringes:
When the mirror M1 and the virtual image M2ꞌ of M2 are not exactly parallel localized fringes are produced.

48. Application of Michelson Interferometer
Determination of Wavelength of a monochromatic light:
For this purpose the interferometer is adjusted to obtain circular fringes in the field of view of the observing telescope. Then the mirror M1 is through a distance λ/2. The path difference will be changed by 2× λ/2= λ and hence the position of a bright fringe is taken by the next bright one.
Let, position of M1 is shifted by a distance x until N bright fringes cross the cross-wire of the observing telescope.
Therefore, 𝒙=N𝝀/𝟐
𝝀=𝟐𝒙/N
Now, x can be measured with the help of a micrometer screw. Thus by counting m we can find out λ.

49. Q.1. In MI 200 fringes cross the field of view when the movable mirror is displaced through mm. Calculate the wavelength of the monochromatic light used.
Solution:- Given
N=200
x= mm = X 10-3 m
So the wavelength

50. Determination of difference in Wavelength:
If the source emits the light of two wavelengths λ1 and λ2 (λ1> λ2) then each wave will produce an interference system of its own. In this situation if M1 is displaced, the field will be alternately distinct and indistinct. The fringes will be in consonance when the bright rings of one wave coincide with the bright ring of another. Similarly the fringes will be in dissonance bright ring of one wave coincide with the dark ring of another.
Let mirror M1 is displaced by a distance d so that the fringes pass from one consonance to next consonance through the intermediate state of dissonance. This will happen when value of d is such that.
Where m and (m+1) represents the number of fringe shift for the light of wavelength 𝜆 1and 𝜆2.
If 𝜆1 is known, we can find 𝜆2 from the above relation. Then difference in wavelength can be determined.
λ =λ1- λ2 = 2 d (λ1. λ2) check

52. Determination of refractive index of a material:
To determine the refractive index of a material (μ), the interferometer is first to be adjusted for white light fringes when the optical path for two interfering beam are made equal. A thin wire is attached to the middle of the mirror M1 and the central achromatic fringe with white light is to be made coincident with the wire.
Now a thin plate (𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑖𝑛𝑑𝑒𝑥=μ 𝑎𝑛𝑑 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠=𝑡) is introduced in the path of one of the interfering rays. An extra optical path (μ−1)𝑡 is introduced in the side of the plate. Since the ray travels twice though the plate, the path difference introduced is 2(μ−1)𝑡 between the two interfering beam.
Due to this extra path central fringe will be displaced from the wire. The mirror M1 is then to be displaced 𝑑 until the central fringe again coincides with the wire. In that case
If thickness of the plate t is known, we can find μ from the above relation.

53. Numerical’s
Question: A thin transparent sheet of refractive index μ =1.6 is introduced in one of the beams of Michelson interferometer and a shift of 24 fringes for λ= 6000 A° is obtained. Calculate the thickness of the sheet. Ans: In Michelson interferometer, Given that :λ=6000 Å, μ=1.6, We know that, 2t(μ-1)= 2d …….(1) 2d=mλ , …… (2) From (1) and (2) 2t(μ-1)=mλ t= mλ/2(μ-1) t= 1.2 X 10⁻⁵ m.

54. Diffraction of Light
Diffraction refers to various phenomena that occur when a wave encounters an obstacle or a slit. It is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle.
Assignment:
Fresnel Distance

55. Types of diffraction

58. Fraunhofer Diffraction at a Single Slit

61. Intensity distribution by single slit diffraction
Central Maxima
Principal Minima

62. Secondary Maxima (Not in syllabus)

63. Intensity distribution by single slit diffraction

64. Width of the central maximum

65. Diffraction Grating

66. Formation of Spectra with Diffraction Grating
With Monochromatic Light
With White Light

67. Plane Wavefront Passing through Grating

68. Prism and Grating

69. Theory for transmission grating
(resultant intensity and amplitude) (Assignment)

73. Intensity distribution by diffraction Grating

74. Theory due to diffraction from each slit (Same as single slit)
Secondary Maxima
Resultant Intensity
Central Maxima
Principal Minima

75. Theory due to Interference of N slits (Not in syllabus)
Resultant Intensity
Principal Maxima’s

76. Manima’s
Secondary Maxima’s

77. Intensity of Secondary Maxima’s (Not in syllabus)

78. Intensity distribution by Diffraction Gratings

79. Characteristics of Grating Spectra
1. ABSENT SPECTRA ( Assignment)

80. 2. Maximum Number of Order Observed by Grating

81. 3. Width of principal maxima (Assignment)

82. 4. Dispersive Power of Diffraction Gratings

83. Experimental Set up to determine wavelength

84. Resolving Power
Resolution: When two objects or their images are very close to each other they appeared as a one and it not be possible for the eye to seen them separate. Thus to see two close objects just as separate is called resolution.
Limit of resolution: The smallest distance between two object, when images are seen just as separate is known as limit of resolution.
Resolving Power: The ability of an optical instrument to produce two distinct separate images of two objects located very close to each other is called the resolution power.

85. Rayleigh Criterion for Resolution
Lord Rayleigh ( ) a British Physicist proposed a criterion which can manifest when two object are seen just separate this criterion is called Rayleigh’s Criterion for Resolution
Well Resolved
Just resolved
Not resolved

86. Resolving power of a telescope (Not in syllabus)
Resolving power of telescope is defined as the reciprocal of the smallest angle sustained at the object by two distinct closely spaced object points which can be just seen as separate ones through telescope. Let a is the diameter of objective telescope as shown in fig and P1 and P2 are the positions of the central maximum of two images. According to Rayleigh criterion these two images are said to be separated if the position of central maximum of the second images coincides with the first minimum or vice versa.
The path difference between AP2 and
BP2 is zero and the path difference
between AP1 and BP1 is given by
…….. (1)
If dθ is very small sin dθ = dθ
…….. (2)
For rectangular
aperture

87. Resolving power of a Diffraction Grating(Not in syllabus)