Optics Lecture PPTS Prof.Kiran Madhale.pdf

Optics in Engineering Physics

1. Welcome to Physics
Lecture.
Mr. Kiran Madhale
Associate Professor in Physics
Walchand College of Engineering, Sangli.

2. Welcome to Walchand College of Engg.Sangli.
•Very happy in welcoming to this institute.
•Congratulations for selection of this institute as your dream institute.
•We are celebrating platinum jubilee year.
•You are so lucky to be a student of this college.
•Consider this is a happy moment in your life.
•In your group, new friends, new faces, new subjects, new teachers, all
new new……
•You may feel perplexed but it is transitory.
•Very soon you feel more comfortable and this comfort will takes you
to confidence.

3. Welcome to Walchand College of Engg.Sangli.
•You assume that, here in Walchand, you are
representing your family or your village or your
city or your district or your state…….
•You have to do something in Walchand so that
your family or village or your city or your
district will feel more proud.

4. Looking back a little…… ( थोडेमागेवळूनपाहताना)
•So far you had school life.
•You are entering college life.
•No matter system is online or
offline you should study regularly.
•Every moment is teaching
something.
•The days do not stop.
•Surely good atmosphere will be
created.

5. Some important instructions for joining the lecture….
•Before each lecture ,you are given a lecture link on
your WhatsApp group and you have to join the lecture
in time using that link.
•Avoid asking the questions in between the lecture.
•It is better to ask the questions at the end of lecture.
•Do remember the above instructions and act
accordingly and cooperate with us.

6. Let us now turn to our subject….
Our Subject is……

7. What are the Modules in Syllabus of this subject
Engineering Physics (5PH101)?
There are 6 modules.
Module No.1: Optics.
Module No.2: Quantum Physics.
Module No.3: Ultrasonics.
Module No.4: Solid State Physics.
Module No.5: Gravitation and Central Force Motion.
Module No.6: Computer Instrumentation.
Module No.1,3 and 5 will teach by Mr.Kiran V Madhale.
Module No.2,4 and 6 will teach by Dr. Bashir N Jamadar.

8. What are the contents in these modules ?
Module No.1: Optics.
Introduction, types of optics, diffraction, types of diffraction, Fresnel’s diffraction: Fresnel’s half period
zones, zone plate, diffraction at straight edge. Fraunhofer’s diffraction: diffraction due to single slit,
double slits, plane diffraction grating. Polarization: optical activity, specific rotation of optical active
substances, Laurent’s half shade polarimeter.
Module No.2: Quantum Physics.
Introduction, black body radiation, Planck’s quantum theory, Wien’s displacement law and Rayleigh –
Jeans law, phase velocity, group velocity and particle velocity, de-Broglie’s hypothesis, Compton effect:
theory and experimental verification, Heisenberg’s uncertainty principle and its applications, wave
function and its physical significance, Schrödinger’s wave equation: time independent and time
dependent, applications of Schrödinger’s wave equation.
Module No.3: Ultrasonics.
Introduction, classification of sound, ultrasonic waves, generation of ultrasonic waves
(Magnetostriction and Piezoelectric method), detection of ultrasonic waves by Kundt’s tube, thermal
detection and sensitive flame method, velocity of ultrasonic waves in liquid, applications of ultrasonic
waves in scientific and engineering field.

9. What are the contents in these modules ?
Module No.4: Solid State Physics.
Introduction, formation of energy bands in solid, classification of solid on the basis of band
theory, number levels in band, density of states, Fermi-Dirac statistics, Fermi level, variation
of Fermi level with change in temperature for semiconductor, electrical conductivity of
metal and semiconductor, Hall effect, basic concept of p-n junction.
Module No.5: Gravitation and Central Force Motion.
Law of gravitation, Gravitational potential energy, Inertial and gravitational mass,
Potential and field due to spherical shell and solid sphere, Motion of a particle under a
central force field, Two body problem and its reduction to one-body problem and its
solution, The energy equation and energy diagram, Kepler’s Laws, Satellite in circular orbit
and applications, Geosynchronous orbits.
Module No.6: Computer Instrumentation.
Introduction, instrumentations, measurement system, control system, Transducer and
Sensor: transducers, sensors, classification of transducers, characteristics of transducers,
selection criterion for transducers, temperature transducers, strain gauge, pressure
transducers, force transducers, optical transducers, actuators.

10. Which books we have to use?
M. N. Avadhanulu and P. G. Kshirsagar, “A Text book of Engineering Physics” S.Chand and
Company
R. K. Gaur and S. L. Gupta “Engineering Physics”, Dhanpat Rai Publications
For more readings,
1. Halliday, Resnic and Walker, “Fundamentals of Physics”, John Wiley, 9th edition 2011.
2. A. Beiser, “Concepts of Modern Physics”, McGraw Hill International, 5th edition, 2003.
3. Ajoy Ghatak, “Optics”, Tata McGraw Hill 5th edition, 2012.
4. P. M. Mathews, K. Venkatesan, “Text Book of Quantum Mechanics”, Tata McGraw Hill
2nd Edition, 2010
5. M.K Harbola, “Engineering Mechanics”, Cengage 2nd edition, 2013.
6. D. Kleppner & R. Kolenkow, “An Introduction to Mechanics”, McGraw Hill Education, 1st
edition 2017

11. Course Objectives are……….
1.To provide basic concepts to solve many
engineering and technical issues.
2.To give deep insights into the understanding of
engineering courses.
3.To encourage them to understand engineering
and technical development.

12. Why we have to study physics in engineering?
•If you are interested in building your career in Engineering, then studying
physics is obvious.
•We cannot imagine our life without Physics.
•The future technology directly or indirectly is related to physicists who have
made a huge contribution to the world.
•Engineering is a profession in which scientific knowledge and mathematics
are used for innovations, to develop new things that benefit mankind, which
is important to society and nature, making everything around us easier.
•Basic sciences have their own importance in the broad spectra of the
engineering education system.
•Knowledge of science helps theengineerunderstand the constraints inherent
in a problem and helps theengineerdevelop possible approaches for a
solution.

13. Role of physics in engineering
•Now let’s throw some light on the role of physics in engineering……..
•Physics is one of the most fundamental scientific disciplines, and its
main goal is to understand how the universe behaves.
•Physicsgenerates fundamental knowledge needed for the future
technological advances.
•Study of physics develops the ability of Problem Solving, Logical
Thinking and also the Ability to think intellectually.
•Physics concepts, such asClassical mechanics, Thermodynamics and
Statistical mechanics, Electromagnetism, Quantum mechanics, Atomic
physics, Molecular physics, Optics, Condensed Matter Physics, Nuclear
Physics etc., play a vital role in the process of innovation, which is,
crucial in the development of engineering branches.

14. Role of physics in engineering
•Engineering is basically physics applied to create something more practical.
•It can be mechanical, electrical, civil, etc., but they’re all basically governed
by physics.
•There’s no way you would solve complex engineering problems without
understanding the physics behind it.
•In Civil Engineering, the laws of physics can tell you about forces, tension,
harmonic vibrations and oscillations, tensile strength, elasticity, and all kinds
of other concepts that you can use to make calculations about your designing
and construction work .
•In Mechanical Engineering, you need the help of physics in dealing with
aircraft, watercraft, engines, robotics, weapons, cars, pneumatics, hydraulics
and others by using core areas including mechanics, dynamics,
thermodynamics, materials science, structural analysis, and electricity.

15. Role of physics in engineering
•Electrical engineering involves designing electrical circuits
including motors, electronic appliances, optical fiber
networks, computers, and communication links.
•Electrical engineers often need to convert electrical
energy to other forms of energy, with the understanding
of mechanics and thermodynamics.
•Knowing the fundamentals of Electrical Engineering, in
addition to, how small-scale components like integrated
circuits and various types of transistor logic, all functions
require at least an intermediate understanding of
Electromagnetism, which you learn from Physics.

16. Role of physics in engineering
•Electronicsinclude the workings of transistors,
diodes and semiconductors.
•Integrated circuituses physics to study how various
tiny transistors are connected in circuits.
•Electromagnetismis used for antennae design, RF
signals, wireless communications, etc.
•The field of roboticsrelies on a lot of physics things
such asdynamics, chaos, mechanics, motors, etc. as
well asoptics(for cameras for computer vision).

17. Role of physics in engineering
•Since Electrical engineering leads to Electronics engineering
and finally to Computer engineering & Information
Technology and Artificial Intelligence.
•It can be concluded that the mother of all engineering
branches is Physics.
•An engineer might design the product itself, or just figure out
a way to build it.
•But either way, success is impossible without an
understanding of the physics behind each of them.
•Thus, it is true that Physics has a significant role in
Engineering.

18. Let us start…….
Module No.1
OPTICS

19. Module No.1: OPTICS
Lecture Plan:
1.Introduction to OPTICS
2.Types of OPTICS
3.Diffraction
4.Diffraction of sound waves
5.Diffraction of light
6.Types of diffraction
7.Difference between diffraction and interference
8.Examples of diffraction of light observed in daily life.

20. Module No.1 OPTICS
Introduction
Physics is a base of all engineering branches.
There are 36 branches of Physics.
Some important branches are..
1. Astrophysics
2. Atomic and Molecular Physics
3. Bio-Physics
4. Geo-Physics
5. Mechanics (Classical and Quantum)
6. Nuclear Physics
7. Thermodynamics
8. Electronics
9. Nanotechnology
10. Optics

21. Module No.1 OPTICSIntroduction
•What is optics?
•One of the important field of Physics.
•Science of light.
•What is light?
•Type of energy
•It is e.m radiation within certain portion of the e.m
spectrum.
•The word usually refers to visible light which is
visible to human eye and is responsible for the
sense of light.

22. Module No.1 OPTICS Introduction
•Why we have to learn optics ?
•Optics is also the study of scenes or the study of the behaviour of
light and how other radiations are transmitted and reflected.
•It studies facts such as how light is reflected from objects or gives
energy in other forms of electromagnetic waves.
•According to the study of optics, there are large number of
applications to technology and they are extremely useful in many
ways.
•In fact, without optics, none of you see or read.
•Whatever I type on my computer, sent to the internet and
identified by this computer, This is only possible with the help of
optics.
•That is why optics has importance in life.

23. Module No.1 OPTICS Introduction
•Types of Optics:
Optics
Geometrical
Optics
Physical
Optics
Quantum
Optics

24. Geometrical optics
•It describes the propagation of light in terms of
rayswhich travels in straight lines whose paths
are governed by the laws of reflection
,refraction at interfaces between two media.
•When the ray of light hits the boundary
between two transparent materials, it is divided
into reflected and refracted ray.
•It deals with the formation of images by using
optical devices like mirrors, lenses and prisms.
•In Geometrical optics light is considered as a ray.

25. Physical optics
Light is considered as a Wave

26. QuantumOptics
•Light is considered as a Particle
•It is also considered as interaction of light with matter.

27. In brief…
•Geometrical Optics:
Light considered as a RAY
•Physical Optics:
Light considered as a WAVE
•Quantum Optics:
Light considered as a PARTICLE

28. DIFFRACTION
•Wave -Property-Bending
•Bending-noticeable-when size of obstacle is comparable with
wavelength of wave
•If λ= d ----Diffraction
•If d > λ ----Reflection
•If λ>d ----Only geometrical shadow will form but not
diffraction

29. Definition of Diffraction
•Bending of waves round the obstacle or
aperture whose size is comparable with
wavelength of waves is called diffraction.
•Diffraction occurs with all types of waves:
Sound waves
Water waves
Electromagnetic waves: Visible light, X-rays etc.
•Where there are waves diffraction takes place.

30. Diffraction of sound waves

31. Diffraction of sound waves
•Wavelength of Ultrasonic waves : 1 cm
……..Not diffracted by ordinary obstacles
•Wavelength of radio waves : 2.5 cm to 250 m
……… diffraction takes place due to large
buildings, hills etc.

32. Diffraction of Light
•Light has wave nature therefore it must exhibit
diffraction phenomenon.
•Definition: The bending of the light waves round
the edge of obstacle or aperture whose size is
comparable with wavelength of light is called
diffraction of light
•This diffraction of light phenomena was first
discovered by Grimaldi in 1665.

33. Diffraction of Light
•S is a point source of monochromatic
light of wavelength λ.
•AB is a opaque obstacle with sharp
edge A.
•XY is a screen.
•Consider P is a point on screen.
•Consider light is travelling from S to P
•If light travels in straight line, then
•There should be sharp shadow of
point A at point P.
•But when we will see carefully on
screen it is seen that
•There is not sharp shadow at point P.

34. Diffraction of Light
•Alternate dark and bright bands are
observed on screen.
•Below the point P very few alternate
dark and bright bands are observed and
intensity of light start decreasing say up
point Q.
•After point Q, complete darkness will be
observed.
•Above point P, maximum number of
alternate dark and bright bands are
observed.
•The width of these bands goes decreases
if we see from point P and upward.
•These bands are called diffraction bands.

35. Diffraction of Light
•When light incident on obstacle
AB, it bends or diffracts at sharp
point A and spreads on screen.
•Light bends at sharp edge A
because the size of edge A is
considered equal to wavelength
of light λ.
•This bending of light is called
diffraction of light.
•This phenomenon was first
discovered by scientist Grimaldi
in 1665.

36. Diffraction of Light
•Newton had also the knowledge of this
phenomenon.
•Newton tried to explain this phenomena on the
basis of his corpuscular theory of light.
•Corpuscular means particles.
•Newton considered light is stream of particles.
•When these particles incident on obstacle AB,
these particles bends due to the the attractive
and repulsive forces exerted by the edge of
obstacle on flying corpuscles.
•But Newton could not explained how the bands
are formed on screen.
•So that Newton's explanation has not accepted.

37. Diffraction of Light
•Scientist Young tried to explain diffraction
phenomenon.
•He used wave theory of light.
•According to him, diffraction is due to
interference between direct light rays which are
coming from source of light S and the rays
reflected from edge A of the obstacle.
•If diffraction is due to interference then bands
formed on screen should be of equal widths and
same intensity.
•But reality was different.
•He could not gave satisfactory explanation on
the formation of bands which are observed on
the screen.
•Hence Young's explanation has also not
accepted.

38. Diffraction of Light
•Newton and Young tried to explain diffraction of light
but they failed.
•Finally the correct explanation is given by scientist
Fresnel.
•According to him, diffraction of light is due to mutual
interference of secondary wavelets originating from
various points of wavefront which is not blocked off
by the obstacle.
•Now let us see how Fresnel has explained diffraction
of light.

39. Diffraction of Light : Fresnel's explanation
•In figure, S is a point source of
monochromatic light of
wavelength λ.
•We know that, If the source is
point source it gives spherical
wavefront.
•Let XY is a spherical wavefront.
•Let this wavefront is incident on
obstacle AB.

40. Diffraction of Light : Fresnel's explanation
•According to Huygens principle, every point
on the wavefront is new source of
secondary wavelets and they emits in all
possible directions.
•In figure ,upper part of wavefront XA is not
blocked off by obstacle and lower part of
wavefront AY is blocked off by the obstacle.
•The lower part of wavefront will never
come forward.
•The upper part of wavefront XA will bend
at the sharp edge of obstacle A and
simultaneously the secondary wavelets
coming from every point on that part of
wavefront will interfere each other and
diffraction bands formed on screen.

41. Diffraction of Light : Fresnel's explanation
•While giving explanation, Fresnel used two
principles
•Huygens principle of secondary wavelets and
•Principle of interference
•He used tremendous mathematics and he
successfully calculated the positions of bands on
the screen.
•It was observed that the calculated results were
matched with observed diffraction bands.
•When Fresnel was proved this phenomenon
practically that explanation was considered to
be true and correct and was accepted also.
•That is why his explanation is called correct
explanation of diffraction of light.

42. Diffraction of Light
•The process responsible for diffraction phenomenon is going on
continuously during the propagation of every wavefront but diffraction
effects are observed only when the part of wavefront is cut by the
obstacle.
•e.g. every optical instrument like telescope ,microscope makes use of
limited portion of wavefront. So the images are not very sharp but get
diffused due to diffraction.
•These diffused images converted in to good images with help of other
lenses.

43. Types of diffraction
Fresnel Type
•In this type source of
light and screen are at
finite distances from
diffracting device.
Fraunhofer type
•In this type source of
light and screen are at
infinitedistances from
diffracting device.

44. Types of diffraction
Fresnel Type
•Incident wavefront is spherical
or cylindrical.
•Lenses are not required
•Distance is important.
•Mathematical treatment is
complex.
•Less applications in designing
optical instruments.
•Diffraction at a straight edge,
diffraction by circular disc etc.
are the examples of Fresnel's
type diffraction.
Fraunhofer Type
•Incident wavefront is plane.
•Lenses are required
•Angular inclinations are
important.
•Mathematical treatment is
simple.
•Large applications in designing
optical instruments.
•Diffraction due single slit,
double slits, multiple slits
(grating),diffraction by circular
aperture etc. are the examples
of Fraunhofer type diffraction.

45. Difference between Diffraction and Interference
Diffraction
•It is the phenomenon of
interaction of light coming
from different parts of
same wavefront.
•Diffraction bands are never
be of same width.
•Dark bands are not
perfectly dark.
•Less number of bands
observed.
Interference
•It is the phenomenon of
interaction of light coming
from two different
wavefronts originating
from two coherent sources.
•Interference bands may or
may not be equal width.
•Dark bands are perfectly
dark.
•Large number of bands
observed.

46. Examples of diffraction observed in daily life
•Look at CD/DVD, rainbow pattern will
observe.

47. Examples of diffraction observed in daily life
•Ring of light around the moon or sun.

48. Examples of diffraction observed in daily life
•Rings of light around the street light lamp in
very cold season.

49. Diffraction of light
•We know that there are two types of diffraction……
•Fresnel type.
•Fraunhofer type.
•We will start with Fresnel type diffraction.
•Under this type we are going to learn……
•Fresnel's half period zones.
•Zone Plate, its action and applications.
•Diffraction at a straight edge.

50. Fresnel's half period zones
•Scientist Huygenewhose wave theory of light successfully
explained reflection, refraction, interferencebut he could
not explained rectilinear propagation of light i.e. light
travels in a straight line.
•It was left to one brilliant scientist Fresnelwho
successfully explained rectilinear propagation of light.
•He used the idea of half period zones andsuccessfully
explained rectilinear propagation of light.
•At the same time he showed that this rectilinear
propagation of light is not exactly true it is an
approximation.

51. Fresnel's half period zones
•Let s is a point source of
monochromatic light of
wavelength λ.
•Let ABCD is a plane wavefront
travelling from left to right.
•Let P is an external point where
the effect of this wavefront is to
be found.
•Let b is distance between
wavefront and point P.
•Let op=b

52. Fresnel's half period zones
•According to Huygens principle, every point
on wavefront is a new source of secondary
waves and these secondary waves emitted
in all possible directions.
•These secondary waves will reach point P
and determine the amplitude and intensity
of this wavefront.
•To calculate the amplitude and intensity
contribution of these secondary waves at
external point P is very difficult work.
•But this great scientist Fresnel found a way
and successfully calculated amplitude and
intensity of this wavefront at point P.

53. Fresnel's half period zones
•Let this wavefront is divided into the number of
zones called Fresnel's zones
•How are these zones constructed on wavefront?
•With Pas a center and radii are equal to(�+
??????
�
),
(�+
�??????
�
) , (�+
�??????
�
) ,…….. Spheres are drawn on
wavefront.
•Now this wavefront will cut into the concentric
circles with Oas a centre and radii are equal to
????????????
1, ????????????
2, ????????????
3,……..
•The area of first innermost zone is called first half
period zone.
•Area enclosed between first and second circle is
called Second half period zone.
•Here each zone differs its neighboured zone by
phase difference πOR path difference
λ
2
so that
time period is half i.e. t/2
•That is why these zones are called half period
zones.

54. Fresnel's half period zones
•Radii of half period zones:
• Let OP = b, Let �
��=�+
??????
�
, �
��=�+
�??????
�
,�
��=�+
�??????
�
,………�
��=�+
????????????
�
•Let ��
�= �
�, ��
�= �
�,��
�= �
�,…….. ��
�= �
�
•Radius of first half period zone:
• �
�= ��
�= ??????
1�
2
−��
2
•�
�=(??????+
λ
2
)
2
−??????
2
•�
�=�??????……..(1)
•Radius of Second half period zone:
• �
�=��
�= ??????
2�
2
−��
2
• �
�= (??????+
2λ
2
)
2
−??????
2
•�
�= 2??????λ……..(2)
•Thus, from above it is clear that radii of half period zones are proportional to
square root of natural numbers.

55. Fresnel's half period zones
•Area of half period zones:
•Area of first half period zone:
•Area of First half period zone = π�
�
�
• = π�??????(Bcoz �
�=�??????)
•Area of Second half period zone:
•Area of Second half period zone = π�
�
�
-π�
�
�
• = π�??????
•Thus, from above it is clear that area of each half period
zone is same.
•The number of secondary waves in each zone are same.

56. Fresnel's half period zones
•If the area is same means the number of
secondary waves in each zone are same.
•These waves on wavefront are in the
same phase
•But when they reaches the point P they
will not be in the same phase.
•It is seen that the waves from alternate
zones (1,3,5..) are in same phase and
from adjacent zones (2,4,6….) are out of
phase with those from central zone.

57. Fresnel's half period zones
•Let �
�,�
�,�
�………�
�are the amplitudes of
first, second ,third …nth zone, respectively.
•The resultant amplitude (A) at point Pdue to
entire wavefront will be,
•A= �
�−�
�+�
�−�
�+�
�−�
�…
−
+
�
�……..(1)
•Above equation can be written in following form
•A =
��
�
+
��
�
−�
�+
��
�
+
��
�
−�
�+
�
�
�
+
�
�
�
−
�
�+
��
�
+
��
�
-�
�…………
−
+
��
�
……….(2)

58. Fresnel's half period zones
•Above equation can be written in following
form
•A =
��
�
+
��
�
−�
�+
��
�
+
��
�
−�
�+
�
�
�
+
�
�
�
−�
�+
��
�
+
��
�
-�
�…………
−
+
��
�
……….(2)
•This resultant amplitude(A) is depend upon one
of the most important factor called obliquity
factor
•It is denoted by �+���??????
•It is defined as the inclination of zone with line
OP.
•If the obliquity increases amplitude is decreases.

59. Fresnel's half period zones
•If the amplitude is going on decreasing then amplitude of any
zone can be taken as mean of the amplitude of preceding and
succeeding zone
•e.g. �
�=
��+��
�
, �
�=
��+��
�
•We can rewrite Equation (2 ) in following way
•A =
��
�
+(
��
�
−�
�+
��
�
)+(
��
�
−�
�+
��
�
)+(
��
�
−�
�+
��
�
)+(
��
�
-
�
�…………
−
+
��
�
……….(3)
•A =
��
�
−
+
��
�
………..(4)
•If �is large, the term
��
�
tends to zero.
•Equation (4) becomes
•A =
��
�
…………(5)
•Thus resultant amplitude at point P due to entire wavefront is
only half the amplitude of first half period zone.

60. Fresnel's half period zones
•We know that,
•Intensity is directly proportional to
(Amplitude)
2
•Thus resultant intensity(I) at point P due
to entire wavefront is,
•I = (Amplitude)
2
=(
�
�
�
)
2
•I =
�
�
�
�
…….(6)
•Thus resultant intensity(I) at point P due
to entire wavefront is one fourth due to
the first half period zone.

61. Fresnel's half period zones
•It seen that all the results
are related to the first half
period zone only.
•It means that the first half
period is only effective from
which we get light at point
P.

62. Fresnel's half period zones
•If the obstacle whose size is
greater than first half period
zone and placed at point O.
•This obstacle will be bigger and
it cover most of the half period
zones.
•Hence no light will reach at
point P.
•This indicate that light travels in
straight line is true.

63. Fresnel's half period zones
•If the obstacle whose size is less than
first half period zone and placed at
point O.
•This obstacle will be very verysmall
and it will not cover the entire first half
period zone.
•Hence some light will definite reach at
point P.
•This indicate that light travels in
straight line is not true.
•Thus light travels in straight line is not
exactly true it is an approximation.

64. Zone Plate
•Experimental verification of Fresnel's half
period zones provided by an optical device is
called zone plate.

65. Construction of zone plate
•To construct the zone
plate……….
•Take a piece of white paper
•Draw the concentric circles
whose radii are
proportional to the square
root of natural numbers.

66. Construction of zone plate
•Paint the alternate zones
with black colour.

67. Construction of zone plate
•Then take a photograph of
this painting on thin
transparent glass plate.
•That glass plate
(photograph) is called zone
plate.
•In this photograph it is seen
that the zones which were
painted looks not painted
and vice versa.

68. Some photographs

69. Zone Plate
•There are two types of zone plates
•Positive zone plate
•Negative zone plate

70. Zone Plate
•The zone plate whose central zone is
transparent is called positive zone plate.

71. Zone Plate
•The zone plate whose central zone is opaque
is called negative zone plate.

72. Types of zone plate

73. Action of zone plate
•Consider positive zone plate.
•Let XY is section of zone plate.
•Here only 5-6 slits are shown but actually
there are number of slits.
•Let S be the point source of
monochromatic light of wavelength λ.
•Let P is position of screen where we get
bright image.
•Let ??????�=�and ��=�
•Let �??????
1= ??????
1,�??????
2= ??????
2,�??????
3= ??????
3…….
�??????
??????= ??????
??????are the radius of first, second,
third and nth half period zones
respectively.

74. Action of zone plate
•Let WW’ is a spherical wavefront
incident on zone plate.
•Let ��
�= �,��
�= �,��
�= �…..
��
�= �

75. Action of zone plate
•The position of screen is such that
there is an increasing path difference
λ
2
from zone to next zone.
•There fore we have following
relations,
•??????�+ ��= �+�
•��
�+�
��= �+�+
λ
2
…………(1)
•Similarly,
•��
�+�
��= �+�+
2λ
2
•��
�+�
��= �+�+
3λ
2
•��
�+�
��= �+�+
4λ
2
……… so on.

76. Action of zone plate
•Now consider eq.(1) and find the
values of ��
�and �
��.
•��
�+�
��= �+�+
λ
2
…………(1)
•��
�=[��
�
+��
�
�
]
�
�= [�
�
+�
�
�
]
�
�
•�
��= [��
�
+��
�
�
]
�
�= [�
�
+�
�
�
]
�
�
•By substituting these values in eq.(1) we
get,
•[�
�
+�
�
�
]
�
�+ [�
�
+�
�
�
]
�
�=�+�+
λ
2
•�[�+
�
�
�
�
�
]
1/2
+ �[�+
�
�
�
�
�
]
1/2
= �+�+
λ
2

77. Action of zone plate
•�[�+
�
�
�
�
�
]
1/2
+ �[�+
�
�
�
�
�
]
1/2
= �+�+
λ
2
…(2)
•Now by expanding the bracketed terms
with the help of binomial theorem and
neglecting higher order terms we get,
•E.g. �+??????
�
= �+�??????+�(�−�)/�!+…….
•Therefore eq.(2) becomes,
•�[�+
�
�
�
��
�
] + �[�+
�
�
�
��
�
]= �+�+
λ
2
•�+
�
�
�
��
+ �+
�
�
�
��
= �+�+
λ
2
•
�
�
�
��
+
�
�
�
��
=
λ
2

78. Action of zone plate
•
�
�
�
��
+
�
�
�
��
=
λ
2
•
�
�
�
�
[
�
�
+
�
�
]=
λ
2
•�
�
�
[
�
�
+
�
�
]=λ
•[
�
�
+
�
�
]=
λ
�
�
�
………(3)
•This eq.(3) is for radius of first half
period zone.
•For radius of n
th
we can write
•[
�
�
+
�
�
]=
�λ
�
�
�
………(4)

79. Action of zone plate
•[
�
�
+
�
�
]=
�λ
�
�
�
………(4)
•If u and v are the distances of object
and image then the formula for focal
length of thin convex lens is,
•
�
�
+
�
�
=
�
??????
…….(5)
•by comparing eq.(4) and (5) we get
•
�
??????
�
=
�λ
�
�
�
•??????
�
=
�
�
�
�λ………(6)
•This is focal length of zone plate.
•Thus zone plate is act like convex lens.

80. Multiple foci of zone plate
•We know that the focal length of zone
plate is,
•??????
�
=
�
�
�
�λ
•Where n=1,2,3….. Thus zone plate has
series of foci i.e. focal points in between
the point O and P with decreasing
intensity as we go from n=1 to onwards.
•These focal points are determined on
the condition that,
•“Each focus corresponds to the position
where each zone of zone plate contains
odd number of sub half period zones.”

81. Multiple foci of zone plate
•If P
mbe the position of image or focal
point when (2m-1) sub half period
zones present on each zone.
•And corresponding focal length f
mis
given by following equation
•??????
�=
�
�
�
��−��λ
•??????
�=
??????
�
��−�
(����????????????�??????
�
=
�
�
�
�λ)
•Where �=1,2,3…….
•If �=1, ??????
�= ??????
�
•This focal point ??????
�will be at point P and
each zone of zone plate it self half
period zone.

82. Multiple foci of zone plate
•If �=2,??????
�=
??????
�
�
•This focal point ??????
�will be at point
P
1 and here each zone of zone plate
consist of 3 sub half period zones.
•Intensity at this point P
1is less
than P
.

83. Multiple foci of zone plate
•If �=3,??????
�=
??????
�
�
•This focal point ??????
�will be at point
P
2 and here each zone of zone plate
consist of 5 sub half period zones.
•Intensity at this point P
2is less
than P
1
•If �=4, ??????
�=
??????
�
�
•If �=5, ??????
�=
??????
�
�
•If �=6, ??????
�=
??????
�
��
………..

84. Multiple foci of zone plate
•We know that area of each zone is π�??????.
•It is depend upon distance between zone plate and
screen.
•If screen is moved towards the zone plate ,area
decreases and more and more sub half period zones
creates on each zone.
•When area of sub half period zones decreases
intensity also decreases.
•That is f
1 > f
2 > f
3 > f
4 >…..

85. Multiple foci of zone plate
•The images which are obtained on right side of zone
plate is called real images.
•These images are obtained when n is odd.

86. Multiple foci of zone plate
•If n be of odd negativevalues, then virtual images
with focal lengths –f
n, -f
n/3, -f
n/5… are also formed
on left side of zone plate.
•Thus we can say that zone plate act like both convex
and concave lens.

87. Difference between zone plate and Convex lens
Convex lens
•Image is obtained by refraction of
light
•
�
�
+
�
�
=
�
??????
is used for finding the
focal length.
•For given wavelength of light it has
singlefocal length
•It is used in only visible region of
light
Zone plate
•Image is obtained by Diffraction of
light
•??????�=
��
�
�λ
is used for finding the
focal length
•For given wavelength of light it has
multiplefocal length i.e. foci
•It is used beyond the visible region
of light i.e. X-ray , gamma rays etc

88. Applications of Zone plate
•There are many wavelengths outside the visible light region like
ultraviolet rays, X-rays, Gamma rays etc. where glass lens become not
transparent (because of very small wavelength the photons of ultraviolet
rays, X-rays, Gamma rays are absorbed by glass material).
•This difficulty is solved by zone plate.
•Zone plates have much more applications in photography, microscopy
and gamma ray holography etc. Particularly X-ray zone plates are
prepared at nanometre scale to achieve desired focussing effect.
•Typically X-ray zone plates have diameter of about 4 mm and zone
thickness is about 50 to 300 nanometre. Such type of zone plates gives
resolution up to 10 nanometre.
•They are used to study biological materials (e.gBlood, bacteria, viruses
etc.) and crystal structures (to study the arrangement ofatoms,ionsor
moleculesin amaterial) etc.

89. Applications of Zone plate
•Zone plates are also used in the field of medical imaging holography.
•Holography is one type of photography where image is captured in
interference pattern.
•This technique is used to take images of regions around the isotopes in the
body.
•In this technique firstly the radioactive source illuminate zone plate then
zone plate produces the image on photographic film.
•Then image reflect the interference pattern created by zone plate in 3-D.
Lastly photographed image illuminated by ordinary light to reconstruct the
image which gives in detail information of structure around the isotopes.
•Zone plates are also used in gun sight, because in gun sight the gun that
enables it to aimed accurately by using expensive optical sights. As
compare to expensive optical sights zone plates are very cheap and
reliable and convenient.
•Zone plates have been suggested as an alternative to bifocal lenses or
progressive lensesfor correction of Presbyopia.

90. Information about Presbyopia
•Presbyopia is an age-related eye condition that makes it more
difficult to see very close.
•When you are young, the lens in your eye is soft and flexible. The
lens of the eye changes its shape easily, allowing you to focus on
objects both close and far away.
•After the age of 40, the lens becomes more rigid. Because the lens
can’t change shape as easily as it once did, it is more difficult to
read at close range. This normal condition is called presbyopia.
•Since nearly everyone develops presbyopia, if a person also has
myopia (near-sightedness), hyperopia (far-sightedness) or
astigmatism, the conditions will combine.
•People with myopia may have fewer problems with presbyopia.

91. Problem
•If the diameter of central zone is 2.3 mm and a
point source of light of wavelength 5000 A
0
is
placed 5 meter away from zone plate . Find the
position of primary image and first secondary
image.
•Given : λ= 5000 A
0
=5 x 10
-7
m
•a = 5 m
•Diameter of zone = d = 2.3 mm= r = 1.15 x 10
-3
m

92. Answer
•First find focal length,
•�
�=
�
�
�
λ
•??????
�= 2.64 m
•Now, for position of primary
image,
•
�
�
+
�
�
=
�
??????
�
•
�
�
+
�
�
=
�
�.��
•b = 5.614 m

93. Answer
•Now, for position of first
secondary image ,
•�
�=
??????
�
3
•�
�=
�.��
3
•�
�= 0.8816
•
�
�
+
�
��
=
�
??????
�
•
�
�
+
�
��
=
�
�.����
•b1 = 1.0703 m

94. Thank you……
Have a nice day