WEBVTT
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myself dr jd varma associate professor in
department of physics in this lecture i will
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discuss the basic theory of interference of
light where i will discuss the conditions
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to get interference light in the laboratory
and i will also discuss some of the famous
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experiments which are used in the laboratory
to study the interference phenomena first
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i will discuss what is interference
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so when two identical waves overlap at a point
in space the combined wave intensity at that
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point can be either greater than or less than
the intensity of either of the two waves this
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effect is called interference if the resultant
intensity is greater than the intensities
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of the individual waves the interference is
called constructive on the other hand if the
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resultant intensity is less than the intensities
of the individual wave the interference is
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called destructive whether the interference
will be constructive or destructive depends
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on the relative phase of the interfering waves
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if two identical waves arrive in phase at
the same point in space that is they line
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up crest to crest and trough to trough then
the amplitude of the resultant wave would
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be twice the amplitude of the component waves
as shown in the left panel of this figure
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this is the condition for constructive interference
now if the waves at the point of consideration
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line up crest to trough and trough to crest
the amplitude of the resultant wave will be
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zero
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as it is shown in the right panel of this
figure this is the condition for destructive
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interference in which two waves cancel one
another’s effect at a point
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although i have considered here the interference
of two waves but in principle any number of
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waves can interfere the amplitude of resultant
wave can be determined by using superposition
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principle first enunciated by thomas young
according to this principle if a medium is
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disturbed simultaneously by any number of
waves the instantaneous resultant displacement
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of the medium at every point at any instant
is that they break some of the displacement
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of the medium due to individual waves in the
absence of the other
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here it should be noted that modification
in intensity takes place only in the region
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of crossing of waves
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and out of the region of crossing they move
as if nothing had happened to them now let
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us consider the superposition of two sinusoidal
waves of same frequency at a particular point
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suppose the displacement produced by each
of the disturbance separately at the point
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are given by x1 t is =a cos omega t + theta
1 and x2 t is =a2 cos omega t + theta 2 here
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a1 and a2 are the amplitudes of the waves
and displacement x1 and x2 are in the same
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direction theta 1 and theta 2 are their initial
phases now according to the superposition
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principle the resultant displacement x would
be given by x1 + x2
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and if you substitute the value of x1 and
x2 we get a1 cos omega t + theta 1 + a2 cos
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omega t + theta 2 after simplification we
will get x is =a cos omega t + theta
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from this equation it is clear that the resultant
disturbance is also simple harmonic in character
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having the same frequency but different amplitude
and different initial phases in other words
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the sum of two simple harmonic motions of
the same frequency and along same line is
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also a simple harmonic motion of the same
frequency we can show that the amplitude a
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of the resultant wave is given by root of
a1 square + a2 square + 2 a1 a2 cos theta
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1 - theta 2
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and the initial phase of the resultant wave
theta is given by tan inverse a1 sine theta
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1 + a2 sine theta 2 divided by a1 cos theta
1 + a2 cos theta 2
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thus the resultant amplitude a depends upon
the amplitudes a1 and a2 of the component
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waves and their difference of phases theta
1 - theta 2 here we have used algebraic method
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to find the resultant of two harmonic waves
but the resultant can also be obtained quickly
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by graphical method this method is more useful
when we use we have a large number of superposing
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waves as we will see while discussing the
phenomena of diffraction
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so to obtain the resultant of two disturbance
x1 t is =a1 cos omega t + theta 1 and x2 is
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=a2 cos omega t + theta 2
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by the graphical method we draw vectors op
and oq of magnitude a1 and a2 respectively
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such that op and oq makes an angle theta 1
and theta 2 with the x axis respectively as
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shown in this figure here if we assume the
directors op and oq rotate in clockwise direction
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with angular frequency omega then the x coordinate
of op will be a1 cos omega t + theta 1 and
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that of oq will be a2 cos omega t + theta
2 at t is =0 the rotating vectors are at point
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p and q
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we use the law of hologram to find the resultant
or of the vectors op and oq the length of
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the vector or will represent
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the amplitude of the resultant wave and the
angle theta that or makes with the x axis
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is the initial phase of the resultant wave
further while the vector op and oq rotates
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on the circumference of the circles of radii
a1 and a2 the vector or rotates on the circumference
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of the circle of radius or with the same frequency
from the triangle formed by a1 a2 and a and
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using the law of cosines we can find the amplitude
of the resultant wave a is =root of a1 square
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+ a2 square + 2 a1 a2 cos theta 1 - theta
2
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from the figure we have a cos theta is =a
1 cos theta 1 + a2 cos theta 2 and a sine
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theta is = a1 sine theta 1 + a2 sine theta
2
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with the help of these expressions we can
obtain the expression for the initial phase
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of the resultant wave here we will get the
same result as we have obtained earlier using
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the algebraic method
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thus if we want to find the resultant of two
displacement then we must first draw vector
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op of length a1 making an angle theta 1 with
the x axis from the tip of this vector we
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must draw another vector pq of length a2 making
an angle theta 2 with the x axis as shown
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in this figure the length of the vector oq
will represent the resulting amplitude and
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the angle that it makes with the x axis will
represent the initial phase of the resultant
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displacement
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now if we have third displacement x3 =a 3
cos omega t + theta 3 then from the point
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q we draw a vector qr of length a3 making
an angle theta 3 with the x axis the vector
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whore will now represent the amplitude of
the wave resulting due to superposition of
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x1 x2 and x3 the angle theta made by or with
the x axis will be the initial phase of the
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resultant wave now i will discuss the conditions
for constructive and destructive interference
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as we have seen the amplitude of the resultant
wave in terms of amplitude and the initial
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phases of the interfering wave is given by
root of a1 square + a 2 square + 2 a1 a2 cos
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theta 1 - theta 2 now let us represent the
phase difference theta 1 - theta 2 by delta
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so from this equation it is clear that if
the phase difference between the waves delta
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is =0 2pi 4pi r in general 2npi where n is
=0 1 2 3 like that then a =a1 + a2
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from wave theory we know that phase difference
of 0 2pi 4pi that is 2npi means the waves
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are in phase thus if the two displacement
are in phase then the resultant amplitude
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is the sum of the amplitudes of the interfering
waves this is known as constructive interference
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and this will be the will be obtained when
delta is =2npi where n is =0 1 2 3 etcetera
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on the other hand if the phase difference
delta is equal 2pi 3pi 5pi that is in general
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delta is =2 n + 1 pi where n is =0 1 2 3 etcetera
then a is = a1 - a2 that is if two displacements
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are out of phase then the resultant amplitude
is the difference of the two amplitudes this
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is known as destructive interference and the
condition for this is delta is =2n + 1pi where
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n is =0 1 2 3 etcetera
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since phase difference delta is =2 pi by lambda
into path difference we can also write the
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above conditions for constructive and destructive
interference in terms of path difference suppose
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two waves initially in phase meet at a point
after traveling different paths at this point
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there would be constructive interference if
the path difference is = n times lambda and
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destructive interference if path difference
is = n + half times lambda where n is = 0
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1 2 3 like that so now i will discuss the
variation of intensity of resultant wave as
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a function of phase difference
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as we know the intensity of a wave is proportional
to its amplitude and proportionality constant
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is taken as 1 therefore the intensity of the
resultant wave i is = the square of the amplitude
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a of the resultant wave so we can write intensity
of the resultant wave i = i1 + i2 + 2 into
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root of i1 into i2 into cos delta here i1
is =a1 square and i2 is =a2 square are the
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intensities of the interfering waves and delta
is the phase difference theta 1 - theta 2
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between them
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since the maximum and minimum value of cos
delta are + 1 and - 1 respectively therefore
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the maximum intensity i max will be =root
of i1 + root of i2 whole square when delta
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is =2n pi
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the minimum intensity i minimum will be =root
of i1 - root of i2 whole square this occurs
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when delta is =2 n + 1 pi that is when waves
are out of phase if the amplitude of the waves
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are same that is a1 is =a2 is =a then i 1
will be =i2 will be =i naught say in this
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case the resultant intensity i will be =4a
square cos square delta by 2 or this will
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be =4 i naught cos square delta by 2 in this
case the maximum intensity i max will be =4
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i naught and the minimum intensity will be
= 0
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this figure shows i versus delta plot from
this figure it is clear that when two beams
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of light arrived at a point exactly out of
phase they interfere destructively and the
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resultant intensity is 0 on the other hand
when they meet exactly in phase that is they
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interfere constructively then the resultant
intensity is 4 times i naught here it should
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be noted that there is no violation or conservation
of energy in the phenomena of interference
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the energy which apparently disappears at
the minima is actually still present at the
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maxima where the intensity is greater than
that would be produced by the two beams acting
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separately in other words the energy is not
destroyed but merrily redistributed in the
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interference pattern
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now suppose there is no phase relationship
between the two interfering waves that is
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they are produced by incoherent light sources
then the phase difference delta will remain
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constant
for a very short duration of time which is
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of the order of 10 to the power - 10 second
and delta would also vary with time in the
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random wave in this case cos square delta
by 2 will vary randomly between 0 and 1 and
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the average of cos square delta by 2 would
be half
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so we have resultant intensity i is =i1 +
i2 =2i naught which has been marked by dashed
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line in this high versus delta plot thus for
two incoherent sources the resultant intensity
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is the sum of the intensities produced by
each one of the source independently and no
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interference pattern is observed now i will
discuss what are the various conditions that
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should be satisfied to get stable or stationary
interference patterns
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so to observe the well defined interference
pattern experimentally the experimental setup
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must fulfill the following conditions: number
1: the two interfering beams must originate
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from the same source of light this is essential
for producing stationary maxima and minima
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since for a stationary maxima and minima the
phase difference delta should not vary with
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time
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that is the initial phase difference between
the interfering waves must be zero or constant
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the sources producing this type of waves are
said to be coherent two independent sources
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can never be coherent
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this is because light from any one source
is not an infinite train of waves and there
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are sudden changes in the phase occurring
in every short interval of time of the order
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of 10 to the power - 10 second however two
coherent sources can be accomplished experimentally
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from single source by either making one source
image of the other by reflection or by dividing
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the light waves into two part by reflection
or by refraction or by partial reflection
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in these circumstances any change of phase
which the original light wave undergoes is
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served by each two part instantaneously with
the result that the phase differences between
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the waves remain constant and hence the position
of maxima and minima remain stationary the
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second condition which the sources must satisfy
is that the waves must have the same period
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and wavelength
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also the amplitude must be equal or very nearly
equal third condition is the original source
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must emit light of single wavelength that
is monochromatic source fourth the two interfering
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waves must propagate almost in the same direction
are the two interfering wave front must intersect
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at a very small angle
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in the interference experiment the conditions
1 to 3 will be satisfied if the interfering
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beams of the light are obtained by dividing
the light of a single source this can be achieved
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either by division of wave front or by division
of amplitude thus the interference experiment
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may be divided into two main categories number
1: the experiment based on division of wave
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front and number 2: the experiments based
on division of amplitude
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in the former class the wave front is divided
literally in to segments by mirrors or slits
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in the later class the amplitude of the incoming
wave of light is divided into two or more
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parts by
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partial reflection and refraction resulting
in two or more beams which are made to reunite
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to produce the interference effect i will
describe some famous experiments of both classes
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so first i will discuss some of the experiment
which is based on division of wave front
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so first experiment which i will discuss is
a famous youngs double slit experiment in
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this experimental setup monochromatic light
of wavelength lambda is allowed to pass through
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a slit s and then through two slits s1 and
s2 which are equidistance from s as shown
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in this figure the slit s s1 and s2 are perpendicular
to diagram and parallel to each other
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the two sets of cylindrical wave front from
the slits s1 and s2 interfere with each other
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to form interference pattern on the screen
ac placed at some distance in front of the
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plane containing two slits if it comes in
the light on this screen we see in any space
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bright and dark bands which are known as interference
fringes since the path s s1 = s s2 the wavelets
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arrive at slits s1 and s2 at the same instant
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therefore secondary wavelets divers to the
right from slits s1 and s2 which have precisely
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equal faces at the start furthermore their
amplitude wave length and velocity are also
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equal now let us determine the positions of
maxima and minima on this screen on this screen
23:37.739 --> 23:46.509
point o is equidistant from s1 and s2 let
us consider a point p on this screen at a
23:46.509 --> 23:59.909
distance x from o if the point p is such that
s2 p - s1 p is =n lambda where n is =0 1 2
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3 etcetera
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then the disturbance is reaching the point
p from two sources will be in phase and hence
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the interference will be constructive leading
to maximum intensity
24:17.840 --> 24:29.299
on the other hand if the point p is such that
s2 p - s1 p =n + half lambda again n is =0
24:29.299 --> 24:38.080
1 2 3 etcetera then the disturbance reaching
the point p from the two sources will be out
24:38.080 --> 24:45.899
of phase and therefore the interference will
be destructive and the intensity will be minimum
24:45.899 --> 24:56.669
since the point o we have s1 o = s2 o both
waves will arrive at o in phase therefore
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a point p coincide with o the intensity will
be maximum and the point p moves up and down
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on the screen starting from point o the magnitude
of the power difference has to p - s1 p increases
25:18.320 --> 25:26.289
and hence the condition for maxima and minima
will be satisfied alternatively
25:26.289 --> 25:31.759
if we calculate the phase difference delta
at any point we can find the intensity at
25:31.759 --> 25:40.110
that point using the equation i =4a square
cos square delta by 2 where a is the amplitude
25:40.110 --> 25:47.350
of the interfering waves so now it remains
to evaluate the phase difference in terms
25:47.350 --> 25:56.529
of the distance x of the point p on the screen
from the center point o the separation d of
25:56.529 --> 26:04.950
the two slits and the distance d capital d
from the slits to the screen this can be obtained
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with the help of this figure if the point
a on s2p is such that s1 p =ap then s2 p - s1
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p will be =s2 a
26:21.460 --> 26:28.970
in the experimental setup d capital d is much
greater than small d and capital t is much
26:28.970 --> 26:37.789
greater than x hence angle theta and theta
prime are very small and practically equal
26:37.789 --> 26:48.830
under these conditions triangle s1a s2 may
be regarded as a right angled triangle and
26:48.830 --> 26:58.659
the path difference becomes d sine theta prime
=d sine theta if theta is small we have sine
26:58.659 --> 27:08.960
theta nearly =tan theta so path difference
delta will be =d sine theta will be =d xdivided
27:08.960 --> 27:11.690
by capital d
27:11.690 --> 27:19.179
and the phase difference delta will be =2
pi by lambda into path difference now for
27:19.179 --> 27:25.809
intensity to be maximum deltas should be an
integral multiple of 2pi which will occur
27:25.809 --> 27:35.250
when the path difference is an integral multiple
of lambda hence for maximum intensity x into
27:35.250 --> 27:47.129
small d divided by capital d should be = 0
lambda 2 lambda are in general = n times lambda
27:47.129 --> 27:57.470
so the position of maxima xn will be = n times
lambda into capital d divided by small d where
27:57.470 --> 28:01.669
n is =0 1 2 3 etcetera
28:01.669 --> 28:10.539
the minimum value of the intensity will occur
at those points where delta is =pi 3pi 5pi
28:10.539 --> 28:18.950
so on for this point the path difference x
into small d divided by capital d will be
28:18.950 --> 28:27.599
=lambda by 2 3 lambda by 2 like that that
is in general n + half lambda so the position
28:27.599 --> 28:38.690
of minima xn will be =n + half lambda into
capital d divided by small d the whole number
28:38.690 --> 28:46.970
n which characterizes a particular bright
fringe is called the order of interference
28:46.970 --> 28:56.610
thus the fringes with n is =0 1 2 3 are called
the 0 first second etcetera order
28:56.610 --> 29:03.919
we can calculate the distance on the screen
between two successive bright fringes and
29:03.919 --> 29:16.059
between two successive dark fringes by evaluating
xn +1- xn using the expressions of x and x
29:16.059 --> 29:23.940
corresponding to bright and dark fringes this
distance is known as fringe width w in both
29:23.940 --> 29:33.919
cases we get fringe width w is =x n + 1 – x
n =lambda into capital d divided by small
29:33.919 --> 29:34.919
d
29:34.919 --> 29:43.129
since the value of fringe width w is constant
on the screen we observe equally spaced dark
29:43.129 --> 29:51.440
and bright fringes thus fringe width varies
directly with slit screen separation capital
29:51.440 --> 29:59.669
d inversely with the separation of slits is
small d and directly with the wavelength lambda
29:59.669 --> 30:09.940
of light used now we will discuss some other
operators which are based on division of wave
30:09.940 --> 30:10.940
front
30:10.940 --> 30:18.669
now soon after the double slit experiment
was performed by young the objection was raised
30:18.669 --> 30:26.029
that the bright fringes he observed were probably
due to some complicated modification of the
30:26.029 --> 30:35.039
light by the ages of the slit and not to true
interference later fresnel brought forward
30:35.039 --> 30:42.470
several new experiments in which the interference
of two beams of light was proved in a manner
30:42.470 --> 30:45.649
not open to above of objection
30:45.649 --> 30:51.720
in the following i will discuss some experiments
in which two interfering beams are obtained
30:51.720 --> 31:00.080
by refraction and reflection of light so now
i will discuss the fresnel biprism experiment
31:00.080 --> 31:09.489
a biprism is essentially two beams each of
very small refracting angle alpha placed based
31:09.489 --> 31:18.940
to base in reality the biprism is constructed
from a single plate of glass by suitable grinding
31:18.940 --> 31:27.399
and polishing it the obtuse angle of the plates
is only slightly less than 180 degree and
31:27.399 --> 31:34.789
other angles of the order of 30 degree are
equal in the experimental arrangement the
31:34.789 --> 31:42.159
biprism is so adjusted in relation to the
source of light that the two halves of the
31:42.159 --> 31:52.369
incident wave front suffer separate simultaneous
refraction through biprism hence this single
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prism is tom reyes biprism
31:55.190 --> 32:02.909
the essential idea is to divide the incident
beam into two interfering beams by utilizing
32:02.909 --> 32:05.979
the phenomena of refraction
32:05.979 --> 32:13.159
in this experiment light from a narrow slit
s illuminated with monochromatic light of
32:13.159 --> 32:24.279
wavelength lambda is allowed to fall symmetrically
on a biprism the intersection of two inclined
32:24.279 --> 32:31.899
faces forming the obtuse angle must be adjusted
accurately parallel to the length of the slit
32:31.899 --> 32:39.070
in this figure the slit s is perpendicular
to the plane of the diagram under this condition
32:39.070 --> 32:44.590
the age b divides the incident wave front
into two parts
32:44.590 --> 32:51.700
firstly the one which in passing through the
upper half abd of the pipe of the prism is
32:51.700 --> 32:59.330
deviated through a small angle toward the
lower half of the diagram and appears to diverge
32:59.330 --> 33:08.979
from the virtual image s1 secondly the one
which is passing through the lower half cbd
33:08.979 --> 33:16.259
is deviated through a small angle towards
the upper half of the diagram and appears
33:16.259 --> 33:20.599
to diverge from the virtual image s2
33:20.599 --> 33:28.509
the two emergent wave fronts which intersect
at a small angle are derived from the same
33:28.509 --> 33:35.919
wave front and hence the fundamental condition
of the interference is satisfied the virtual
33:35.919 --> 33:46.159
images s1 and s2 being the images of the slit
as function as coherent sources in this experiment
33:46.159 --> 33:54.279
as a consequence the interference fringes
are observed on the screen in the overlapping
33:54.279 --> 33:59.629
region ef of the two emergent beams of light
33:59.629 --> 34:10.990
if s1 s2 is =d and capital d is the distance
between the plane of s1s2 and screen then
34:10.990 --> 34:18.960
like double slit experiment we can obtain
the expression for fringe width w is =capital
34:18.960 --> 34:29.560
d lambda divided by small d now i will discuss
another important experimental set up which
34:29.560 --> 34:37.380
is known as a fresnel double mirror experiment
in this experiment of fresnel to obtain the
34:37.380 --> 34:44.080
sustained interference effects the incident
wave front is divided into two coherent interfering
34:44.080 --> 34:49.159
wave fronts by utilizing the phenomena of
reflection
34:49.159 --> 34:59.190
in this experimental setup to optically plane
mirrors om1 and om2 are mounted vertically
34:59.190 --> 35:09.230
inclined to each other at a small angle theta
and touching at the point o the light from
35:09.230 --> 35:16.789
a narrow slit has illuminated with monochromatic
light of wavelength lambda is allowed to fall
35:16.789 --> 35:24.319
on the mirrors own line of intersection through
o must be adjusted parallel to the length
35:24.319 --> 35:33.230
of the vertical slit as one half of the incident
wave front is reflected from om1 and appears
35:33.230 --> 35:43.470
to diverse from the point s1 which is consequently
the virtual image of the source slit s in
35:43.470 --> 35:44.470
om1
35:44.470 --> 35:51.770
the other half of the incident wave front
is reflected from om2 giving rise to an image
35:51.770 --> 36:03.200
s2 of the source slit s obviously s1 and s2
are the virtual coherent sources in this experiment
36:03.200 --> 36:11.710
since both are the images of the same source
slit s thus the fundamental condition of interference
36:11.710 --> 36:18.609
is again satisfied here furthermore since
the angle theta between the mirror is very
36:18.609 --> 36:30.819
small the separation d between the coherent
sources s1 and s2 given by d is =2a theta
36:30.819 --> 36:36.049
where os is =a is also very small
36:36.049 --> 36:43.349
thus the condition for the observation of
widely spaced interference fringes is also
36:43.349 --> 36:50.869
satisfied the light on the screen av appears
to come from two virtual coherent sources
36:50.869 --> 37:00.539
s1 and s2 and in the region ef where the two
beam overlap interference fringes are observed
37:00.539 --> 37:11.660
parallel to the slit now let us discuss another
important experiment which is known as the
37:11.660 --> 37:14.060
lloyd’s mirror arrangement
37:14.060 --> 37:23.180
in this arrangement light from a slit s1 is
allowed to fall on a plane mirror and grazing
37:23.180 --> 37:30.960
incidence the light directly coming from this
rate s1 interferes with the light reflected
37:30.960 --> 37:39.690
from the mirror forming an interference pattern
in the region bc of this screen in this experiment
37:39.690 --> 37:50.170
the central fringes at o cannot be observed
on the screen l1 l2 unless it is moved to
37:50.170 --> 37:59.670
the position l1 prime l2 prime where it touches
the end m2 of the mirror because here light
37:59.670 --> 38:03.270
is received only from s1
38:03.270 --> 38:11.930
if the central fringe is absorbed with white
light it is found to be dark this implies
38:11.930 --> 38:20.740
that the reflected beam undergoes a certain
phase change of pi on reflection consequently
38:20.740 --> 38:31.930
when the point p on the screen is such that
s2p - s1p is =n lambda n where n is =0 1 2
38:31.930 --> 38:39.529
3 etcetera we get minimum that is destructive
interference on the other hand if s2p - s1p
38:39.529 --> 38:49.760
=n + half lambda where n is =0 1 2 3 etcetera
we get maximum
38:49.760 --> 38:56.450
thus this experiment provides an experimental
confirmation of the fact that a sudden phase
38:56.450 --> 39:03.841
change of pi occurs on reflection in a rarer
medium that is from the surface backed by
39:03.841 --> 39:12.510
a denser medium the mirror employed here should
have slurring on the front surface to avoid
39:12.510 --> 39:19.610
multiple internal reflections and its surface
should be adjusted so as to make it exactly
39:19.610 --> 39:28.640
parallel to the vertical source slit s1 so
now i will discuss the interference with white
39:28.640 --> 39:30.020
light
39:30.020 --> 39:36.240
so earlier in the youngs double-slit tunnel
biprism and double mirror experiments we have
39:36.240 --> 39:43.650
considered monochromatic light of single wavelength
to obtain large number of bright and dark
39:43.650 --> 39:51.520
fringes let us now study the changes produced
in the fringes when the monochromatic light
39:51.520 --> 39:59.260
source is replaced by white light in these
experiments white light consists of wave lengths
39:59.260 --> 40:06.390
varying from 4000 to 7000 angstrom that is
from the violet to the red end of the white
40:06.390 --> 40:07.390
light spectrum
40:07.390 --> 40:16.059
in the experimental setup at any point p naught
situated on the perpendicular bisector of
40:16.059 --> 40:24.980
the coherent sources s1 and s2 the symmetrical
path difference is 0 for the interfering light
40:24.980 --> 40:33.060
waves of all wavelengths present in the white
light here we get the light of every color
40:33.060 --> 40:42.359
in exactly the same proportions as it exists
in white light as a consequence the resultant
40:42.359 --> 40:52.039
illumination had p naught due to superposition
of 0 order or central bright fringes of all
40:52.039 --> 40:55.500
the wavelengths is white
40:55.500 --> 41:04.760
the spacing between the constructive bright
or dark fringes w is =capital d lambda divided
41:04.760 --> 41:12.950
by small d is a function of wavelength obviously
they smaller the wavelength the closure will
41:12.950 --> 41:15.490
be corresponding fringes
41:15.490 --> 41:24.289
since lambda of violet end of visible spectrum
is least therefore the condition of constructive
41:24.289 --> 41:34.769
interference that is path difference is =lambda
will be satisfied for the violet color and
41:34.769 --> 41:42.529
then for other colors in the spectral order
as you move away from the central fringe as
41:42.529 --> 41:49.109
a consequence the bright fringe nearer to
the center fringe shall have a strong violet
41:49.109 --> 41:58.240
tinge followed by other bright fringes having
a strong tinge of colors in the spectral order
41:58.240 --> 42:04.950
in reality no bright fringe is of saturated
spectral color
42:04.950 --> 42:12.900
after 8 or 10 fringes the path difference
becomes so large that the condition for constructive
42:12.900 --> 42:21.130
interference part of path difference is = n
lambda may be simultaneously satisfied for
42:21.130 --> 42:28.519
a number of wavelengths for example we have
the relation path difference is =5 lambda
42:28.519 --> 42:37.720
1 =7 lambda 2 =9 lambda 3 and also at the
same point the condition for destructive interference
42:37.720 --> 42:46.720
path difference is = n + half lambda may be
simultaneously satisfied for many colors
42:46.720 --> 42:56.029
as a consequence for large optical path difference
the dark fringes of some wavelengths are completely
42:56.029 --> 43:04.450
masked by the bright fringes of other wavelength
thus in the white light we get a white central
43:04.450 --> 43:15.319
fringe at the point of 0 path difference along
with a few color fringes on both the sides
43:15.319 --> 43:25.609
the color soon fading off to white now i will
discuss what will happen if we put a glass
43:25.609 --> 43:30.069
plate in the path of one of the interfering
beam
43:30.069 --> 43:39.950
so let us take the example of the youngs double
slit experiment so suppose t is the thickness
43:39.950 --> 43:47.769
of the transparent plate and it is introduced
in the path of one of the interfering beam
43:47.769 --> 43:54.319
has shown in this figure from the figure it
is clear that a light wave traveling from
43:54.319 --> 44:04.510
s1 to p has to traverse a distance t in the
plate while the rest s1p - t it travels in
44:04.510 --> 44:12.519
air thus the time required for light wave
to reach from s1 to the point p is given by
44:12.519 --> 44:24.920
t =s1p - t divided by c + t divided by b where
c and v are the velocities applied in air
44:24.920 --> 44:38.599
and plate respectively so capital t time taken
by the beam capital t will be =s1 p + mu - 1
44:38.599 --> 44:46.390
into t divided by c where mu is =c by v is
the refractive index of plate
44:46.390 --> 44:52.569
the physical interpretation of this equation
is that due to introduction of the plate effective
44:52.569 --> 45:03.589
path from s1 p to p becomes s1p + mu - 1 into
small t in air similarly the effective path
45:03.589 --> 45:15.010
from s1 to o the point equidistance from s1
and s2 become s1o + mu - 1 into small t in
45:15.010 --> 45:22.650
air thus when the plate is introduced the
central fringe corresponding to equal path
45:22.650 --> 45:35.760
from s1 and s2 is formed at the point o prime
where s1o prime + mu - 1 into small t =s2o
45:35.760 --> 45:46.470
prime from the symmetry we can show that s2o
prime - s1o prime =d into oo prime divided
45:46.470 --> 45:57.839
by capital d so mu - 1 into small t will be
=d into xdivided by capital d where o o prime
45:57.839 --> 46:00.690
=x
46:00.690 --> 46:11.390
thus in the fringe pattern shift by a distance
x = capital d into mu - 1 into a small t divided
46:11.390 --> 46:20.519
by small d by using white light we can measure
the displacement x in the central fringe on
46:20.519 --> 46:28.260
introduction of thin transparent plate and
with this measured value of x we can determine
46:28.260 --> 46:34.880
the thickness of the plate by using the above
expression for x
46:34.880 --> 46:42.519
now let us let me summarize what i have discussed
in this lectures so in this lecture i have
46:42.519 --> 46:50.890
discussed what is interference and what is
the basic conditions that must be fulfilled
46:50.890 --> 46:59.069
to get the stationery interference pattern
in the laboratory and we have discussed how
46:59.069 --> 47:08.370
to get the two coherent sources to get the
interference pattern that is which is that
47:08.370 --> 47:12.760
is a division of wavefront and the division
of amplitude
47:12.760 --> 47:19.600
and we have also discussed some of the famous
experimental set up which are used in the
47:19.600 --> 47:29.520
laboratory to get the stationary interference
pattern based on division of wave front in
47:29.520 --> 47:35.980
the next lecture i will discuss some of the
experimental setup which are based on division
47:35.980 --> 47:50.249
of amplitudes