WEBVTT
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in the previous lecture i described the basic
theory of interference what is difference
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between constructive and destructive interference
superposition principle methods to find the
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resultant of two or more than two sinusoidal
waves of same frequency and acting in the
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same direction at a point in a space what
is condition for constructive and interval
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destructive interference in terms of path
difference and phase difference
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how the intensity of result and wave vary
with the phase difference of the interfering
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waves what are the conditions to observe a
stationary interference pattern and how to
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get two coherent sources for interference
by using division of wave front and division
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of amplitude and i also discuss some famous
experimental setup which are used in laboratory
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to observe interference pattern based on division
of wave front such as youngs double slits
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fresnel biprism and fresnel double mirror
experiment
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mostly in all undergraduate optics labs fresnel
biprism experiment is performed with the help
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of this experiment wavelength of monochromatic
light can be determined by measuring the fringe
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width with the help of a travelling microscope
i have given the expression of fringe width
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in the previous lecture by measuring the required
parameters experimentally with the help of
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this equation wavelength can be determined
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with the help of this experiment it is also
possible to determine thickness of a transparent
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plate for this the plate is introduced in
the path of one of the interfering beams and
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resulting shift in the position of central
bright fringe is measured with the help of
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a travelling microscope the required expression
i have already obtained in the previous lecture
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here it should be noted that with monochromatic
light it is not possible to recognise the
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central bright fringe
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but if we use white light then central bright
fringe will be wide and few fringes adjacent
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to this will be coloured so shift in central
bright fringe on introducing the plate can
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be measured easily by using white light source
now i am going to describe the formation of
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interference pattern by division of amplitude
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in this case the internet beam is divided
into two persons by division of its amplitude
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by partial reflection and refraction by some
optical devices the two portions afterwards
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recombined to produce interference effect
for example if a plane wave falls on a thin
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film then the wave reflected from the upper
surface interferes with the wave reflected
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from the lower surface the colours when thrown
by thin films of oil on water by soap bubbles
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or bike racks in a piece of glass can be explained
by the phenomenon of interference of light
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by division of amplitude
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let us first consider the interference by
a plane parallel film of refractive index
03:54.180 --> 04:01.660
mu and thickness t illuminated by a plane
monochromatic light of wavelength lambda as
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shown in this figure suppose the light wave
travelling along ab is incident on the upper
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surface gh of the film part of this wave will
be reflected from the upper surface of fill
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in the direction br and remaining will be
reflected in the direction bc
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upon arrival at c part of this reflected wave
will be reflected from surface ih in the direction
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cb1 and part refracted in direction ct
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at b1 the ray cb1 will be again divided a
continuation of this process yields two sets
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of parallel rays one on each side of the film
in each of these sets the intensity decrease
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rapidly from one ray to next if the set of
parallel reflected rays: br b1 r1 b2 r2 etcetera
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is collected by lens and focus at some point
constructive and destructive interference
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will be formed at this point depending on
the phase difference between the consecutive
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waves it is such interference that produces
the colours of the thin films
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in order to find the phase difference between
these rays we must first evaluate the difference
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in the optical path traversed by a pair of
successive rays such br and b1r1 suppose b1d
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is perpendicular drawn from b1 on br therefore
optical path difference between the waves
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br and b1r1 is given by bc + cb1in film - bd
in air with the help of geometry we can show
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that this path difference will be = 2 mu t
cos theta therefore phase difference delta
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will be given will be = 2 pi by lambda into
path difference that is 2 pi by lambda into
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2 mu t cos theta
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if the film is optically denser than the surrounding
media then the wave br originated by reflection
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from b a point on the surface backed by a
denser medium that is film will be different
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phase by pi from the incident wave but the
wave b1 r1 which originates by reflection
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at c a point on the surface backed by a rare
medium will experience no sudden phase change
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thus the total phase difference between waves
b r and b1 r1 becomes 2 pi by lambda into
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2 mu cos theta + pi
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therefore if 2 mu t cos theta = n + half lambda
where n is integer the wave br and b1r1 interfere
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constructively and they will interfere destructively
if 2 mu t 2 mu t cos theta is = n lambda where
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n is again integer like reflected waves the
transmitted waves ct c1t1 c2t2 etcetera also
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satisfy the conditions for interference because
they originate from the same incident wave
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here ct and c1 t1 are obtained without any
sudden phase change due to reflection
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the phase difference between ct and c1t1 is
therefore only due to optical path difference
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similar to reflected waves here also we can
determine the path difference between ct and
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c1 t1 using basic geometry
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and here also it comes out to be 2 mu t cos
theta so the transmitted waves ct and c1t1
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will interfere constructively went 2 mu t
cos theta is = n lambda and destructively
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when 2 mu t cos theta is = n + half lambda
thus the condition for constructive and destructive
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interference for transmitted waves ct and
c1t1 are just reverse of that of the reflected
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waves br and b1r1
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since the geometry is same the path difference
between waves b1r1 and b2 r2 will be same
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as between waves br and b 1r1 since b1r1 and
b2r2 originate only from the internal reflection
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so there would not be phase difference of
pi due to reflection now if 2 mu t cos theta
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is = n lambda waves b1r1 and b2r2 will be
in phase and same holds for all succeeding
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pairs therefore under this condition waves
br and b1r1 will be out of phase
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but waves b1r1 and b2r2 b3r3 and so on will
be in phase with each other on the other hand
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if 2 mu t cos theta is = n + half of lambda
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waves b1r1 will be in phase with br but we
b2r2 b4r4 b6r6 and so on will be out of phase
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with b1r1 b3r3 b5r5 and so on since b1r1 is
more intense than b2r2 and b3r3 is more intense
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than b4r4 and so on these pairs cannot cancel
each other and hence there will be maximum
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intensity for the minima of intensity wave
b1r1 is out of phase with wave br but br has
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considerably greater amplitude than b1r1 so
that these two will not completely annul each
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other
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but if we add the amplitude of the waves b1r2
b2r2 b3r3 we find that the resultant amplitude
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is = the amplitude of the wave br therefore
in case of minima there will be complete darkness
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thus if the film is of such a thickness that
the condition path difference is = n lambda
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is satisfied the film appears perfectly dark
when seen by reflected light on the other
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hand if the thickness of the film is such
that path difference is = n + half lambda
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then the intensity of the film will be maximum
when seen by reflected light for the transmitted
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light when film thickness is such that the
condition delta of path difference is = n
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+ half lambda is satisfied
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the pair of transmitted wave ct c1t1 c2t2
c3t3 etcetera are in opposite phase and therefore
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interfere destructively in pair so the resultant
intensity will be minimum if the thickness
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of the film is such that path difference
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is n lambda is satisfied the transmitted waves
ct c1t1c2t2 etcetera are in phase with each
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other and therefore they interfere constructively
leading to maximum intensity in the transmitted
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light so when there is maximum intensity in
the reflected light there would be minimum
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intensity in the transmitted light and vice
versa thus based on the above discussions
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it is concluded that when a plane parallel
film is illuminated with parallel monochromatic
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light
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that is theta is constant over the film the
film will have maxima or minima of brightness
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all over according as the optical path difference
between the directly and the internally reflected
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waves is odd or even multiple of lambda by
2 now suppose the plane parallel thin film
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is illuminated with a parallel beam of white
light if we neglect the small variation in
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the angle of refraction with wavelength then
light waves of different colours follow approximately
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the same path within the film
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therefore the optical path difference 2 mu
t cos theta will be approximately same for
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all the colours hence the colours whose wavelength
satisfy the condition 2 mu t cos theta is
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= n lambda will be absent in the reflected
beam but they will be present in the transmitted
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beam on the other hand colours whose wavelength
satisfies the condition 2 mu t cos theta is
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= n + half lambda will be present in the reflected
beam but practically absent in the transmitted
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beam
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consequently the film will have a uniform
coloration all over but the resultant colour
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of the film by reflected light will be exactly
complementary to the resultant colour by transmitted
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light
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furthermore when film thickness t is much
smaller than lambda then 2 mu t cos theta
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is negligible in comparison to lambda in this
case phase difference delta which is = 2 pi
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by lambda into 2 mu t cos theta will be 0
for all wavelength therefore the film will
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appear perfectly dark by reflected light even
when it is illuminated with white light on
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the other hand it will appear white when seen
by transmitted light now let us consider a
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film in the shape of a thin wedge whose side
form a small angle alpha as shown in this
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figure
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now suppose this film is illuminated with
plain monochromatic light waves here also
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the directly reflected wave br and internally
reflected wave b1r1 originate from the same
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incident wave propagating along ab and hence
they are capable of producing observable interference
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effects here the two interfering waves do
not reach there along parallel path but they
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appear to diverse from a point q in the rear
of the film so destructive and constructive
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interference occurs at the point q which is
however virtual
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if the two interfering waves br and b1r1 fall
on lens they will cross each other at a real
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point q prime the focus conjugate of q as
a consequence really reinforcement are destructive
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interference would occur at 2 prime thus q
will be the position where the interference
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fringe appears to be formed from geometry
we can find the optical path difference delta
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between the waves b r and b1 r1 and this comes
out to be:
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2 mu t cos theta + alpha where t is the thickness
of film at the point b1 the path difference
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thus varies both on account of changing thickness
as well as changing angle of incidence theta
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now taking into account the abrupt phase change
of pi due to reflection at b the condition
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for constructive and destructive interference
between waves br and b1r1 becomes 2 mu t cos
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theta + alpha is = n + half lambda for maxima
and 2 mu t cos theta + alpha is = n lambda
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or minima
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if the wedge shaped film is illuminated by
a parallel beam of monochromatic light of
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wavelength lambda the angle of incidence theta
will be same at every point of the film and
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show also angle of refraction so in this case
variation in the optical path difference will
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take place only due to variation in the thickness
of the film t from point to point of the film
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at the edge of the waves film wedge films
since t is = zero the film appears perfectly
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dark because the two interfering waves are
pi out of phase
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at a distance from the edge such that the
path difference is = lambda by 2 3 lambda
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by 2 5 lambda by 2 and so on the film will
be bright while at a distance so that path
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difference is = lambda 2 lambda 3 lambda 4
lambda and so on the film will appear dark
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thus as we proceed along the film in the direction
of increasing film thickness we shall encounter
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alternatively dark and bright bands parallel
to the edge of the film deviation from perfect
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plainness of the surface will be immediately
obvious through the curvature of these bands
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because anyone band is the locus of constant
thickness of the film that is why such fringes
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are called fringes of constant thickness
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it can be shown that for a small angle of
incidence that is cos theta nearly = 1 the
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fringe width beta that is the separation between
the two consecutive dark or bright fringes
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is given by lambda upon 2 alpha mu where alpha
is the wedge angle and mu is the refractive
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index of the film now let us consider that
the wedge shaped thin film is illuminated
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with parallel beam of white light
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if the thickness at the thin edge is very
small in comparison of the wavelength of violet
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light then at the edge a truly acromatic black
fringe parallel to the edge will be seen when
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it is viewed by reflected light now if we
move along the film in the direction of increasing
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thickness we will first reach a point where
the thickness of the film is such that the
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condition of constructive interference is
satisfied for violet colour because wavelength
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of this colour is released in the visible
spectrum
20:26.960 --> 20:35.059
therefore a violet fringe is seen at this
point proceeding still further successively
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blue green yellow and red fringes will be
observed
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these fringes are called fringes of equal
chromatic order beyond certain point thickness
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of the film become so large that the condition
of constructive interference is satisfied
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for two or more colours simultaneously and
therefore a coloured banned due to overlapping
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of bright fringes of more than one colours
will be seen if we still move in the same
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direction we shall arrive at a point where
such overlapping of different colours produces
21:15.090 --> 21:18.200
uniform illumination
21:18.200 --> 21:26.119
now let us consider that a thin parallel film
is illuminated by using extended white source
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as shown in this figure
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in this case light from each point of the
source gives rise to a pair of coherent waves
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on reflection from the film and cos theta
is not same for all of them each point of
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the film is also seen by light from different
points of the extended source safe if at points
21:52.210 --> 22:02.799
a b c the condition 2 mu t cos theta 1 is
= n one + half lambda 1 2 mu t cos theta 2
22:02.799 --> 22:13.840
is = n 2 + half lambda 2 and 2 mu t cos theta
3 is = n3 + half lambda 3 are satisfied respectively
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we shall observe a bright band of lambda one
at a bright band of lambda 2 at b the bright
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band of lambda 3 at c and so on thus we shall
observe a set of equal inclination colour
22:32.730 --> 22:41.899
bands if the film is of variable thickness
we shall again observe a set of colour band
22:41.899 --> 22:49.419
each being a locus of equal thickness of the
film the colour of any particular region of
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the film changes if the i is shifted to a
new position
22:55.830 --> 23:03.679
now light from other points of extended source
is reflected from that particular region of
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the film at different angles to i in effect
2 mu t cos theta alters throughout the film
23:13.259 --> 23:21.019
leading to change in colours exhibited by
thin film this clearly explains the origin
23:21.019 --> 23:29.080
of colour exhibited by a thin film of vial
while on the surface of water or colour of
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soap bubbles by light reflected from the sky
it should be noted that bright colours will
23:37.549 --> 23:41.679
be seen only when film is extremely thin
23:41.679 --> 23:49.760
that is up to few wavelengths thick the colours
become fit for thicker films and for such
23:49.760 --> 23:57.570
cases the direction of incidence should be
kept nearly normal otherwise at other angles
23:57.570 --> 24:06.679
the directly and indirectly reflected coherent
waves may get separated to such an extent
24:06.679 --> 24:15.690
that their wave front may not interfere each
other and they may get so far apart that only
24:15.690 --> 24:24.820
one wave enters the i at a time and so all
colours effectively may vanish when viewed
24:24.820 --> 24:32.879
by naked eye now let us discuss the classification
of fringes
24:32.879 --> 24:40.399
as we have discussed when a thin film is illuminated
by light the resultant transmitted or reflected
24:40.399 --> 24:48.460
in densities will depends upon the phase difference
delta between two consecutive transmitted
24:48.460 --> 25:00.230
are reflected waves and phase difference is
given by 2 pi by lambda into 2 mu t cos theta
25:00.230 --> 25:08.879
since the value of the phase difference delta
may be varied by varying theta or t or lambda
25:08.879 --> 25:17.460
accordingly we can divide the interference
fringes of thin films into three classes
25:17.460 --> 25:25.200
if phase difference varies mainly due to variation
of theta while t remains constant then fringes
25:25.200 --> 25:32.799
are termed as fringes of equal inclination
or haidinger fringes
25:32.799 --> 25:39.440
if phase difference varies with variation
in thickness t of the film then fringes are
25:39.440 --> 25:46.480
called fringes of equal thickness or fizeau
fringes now if the phase difference varies
25:46.480 --> 25:52.720
with variation in lambda then the fringes
are known as fringes of equal chromatic order
25:52.720 --> 26:01.529
or feco fringes the term feco was used by
professor s tolansky now in this lecture i
26:01.529 --> 26:07.980
will describe some of the experimental setup
used in laboratory to observe the fringes
26:07.980 --> 26:17.499
of equal thickness and fringes of equal inclinations
one
26:17.499 --> 26:24.679
one of the most important experiments which
is based on the division of amplitude is newton
26:24.679 --> 26:27.289
ring experiment
26:27.289 --> 26:33.769
in this experiment a thin air film are progressively
increasing thickness in all direction from
26:33.769 --> 26:41.690
one point can be formed by placing a plane
convex lens of large radius of curvature on
26:41.690 --> 26:50.000
a plane glass plate such that its convex surface
faces the plate the air film does form possesses
26:50.000 --> 26:58.419
a radial symmetry about the point of contact
and when it is illuminated normally with monochromatic
26:58.419 --> 27:05.799
light the observed interference fringes are
circular ring concentrate with the point of
27:05.799 --> 27:07.139
contact
27:07.139 --> 27:13.789
these rings are known as newtons ring the
experimental arrangement to observe the newton
27:13.789 --> 27:21.841
rings in laboratory is shown in the figure
here a is a plano convex lens placed on the
27:21.841 --> 27:30.119
glass plate b the glass plate g reflect the
light down so that it is incident normally
27:30.119 --> 27:39.580
on the plates after reflection it is transmitted
through g and observed in the travelling microscope
27:39.580 --> 27:45.759
if the film in close between the lens and
the plane glass plate is extremely thin then
27:45.759 --> 27:52.690
wave single alpha can be neglected as compared
to angle of incidence theta
27:52.690 --> 28:00.340
so for normal incidence the optical path difference
between the two consecutive reflected waves
28:00.340 --> 28:08.280
become 2 mu times t thus for this case the
condition for constructive and destructive
28:08.280 --> 28:13.309
interference for two consecutive reflected
waves are given by
28:13.309 --> 28:22.879
2 mu t is = n lambda for minimum and 2 mu
t is = n + half lambda for maximum where mu
28:22.879 --> 28:30.090
is the refractive index of the film and t
the thickness of the film for air film mu
28:30.090 --> 28:39.629
will be = one as we know an interference fringe
of a given order n is the locus of the points
28:39.629 --> 28:42.049
of constant optical path difference
28:42.049 --> 28:49.909
a dark fringe of the order n is the locus
of those points where the film thickness t
28:49.909 --> 28:59.249
is n lambda upon 2 mu since the film enclosed
between the convex surface and the plane surface
28:59.249 --> 29:06.330
possesses radial symmetry about the point
of contact the dark fringe will be circular
29:06.330 --> 29:14.729
in shape so here we will observe concentric
bright and dark circular colour fringes as
29:14.729 --> 29:22.779
shown in this figure it can be shown that
the square of radius of dark ring of nth order
29:22.779 --> 29:25.679
is = n lambda
29:25.679 --> 29:36.279
r divided by mu and square of radius of bright
ring of nth order is = n + half lambda r upon
29:36.279 --> 29:43.970
mu where r is the radius of curvature of the
convex surface
29:43.970 --> 29:51.700
thus when newton rings are observed by reflected
light the diameter of the bright rings are
29:51.700 --> 29:59.840
proportional to the square root of the odd
natural numbers like 1357 so on whereas the
29:59.840 --> 30:08.149
diameters of dark rings are proportional to
the square root of natural number 1 2 3 4
30:08.149 --> 30:18.250
and so on it can be shown that the rings gradually
become closer as their radii increase as shown
30:18.250 --> 30:26.389
in this figure since t is = 0 at the point
of contact of the length and the glass plate
30:26.389 --> 30:34.590
at this point phase difference between the
waves reflected from the upper and lower surfaces
30:34.590 --> 30:35.759
will be pi
30:35.759 --> 30:44.279
therefore for the reflected light in the interference
pattern the central spot appears dark but
30:44.279 --> 30:51.710
if it is seen in the transmitted light the
central spot will be bright now i will discuss
30:51.710 --> 31:00.919
some important application of newtons ring
experiment one of the one example of is the
31:00.919 --> 31:06.700
determination of the wavelength of the monochromatic
light used
31:06.700 --> 31:13.080
so for this purpose we have to measure the
diameter of the newton rings with the help
31:13.080 --> 31:22.240
of the travelling microscope and by using
the relations lambda is = dn + p square - dn
31:22.240 --> 31:31.570
square upon 4 p r where dn + p square - dn
square is the difference of the squares of
31:31.570 --> 31:42.649
the diameter of the n + pth and nth dark rings
and r is the radius of curvature of the lens
31:42.649 --> 31:49.510
with the help of newtons ring experiment it
is also possible to determine the refractive
31:49.510 --> 31:52.570
index of a given liquid
31:52.570 --> 31:59.130
for this purpose also we have to measure the
diameter of the newton rings with air film
31:59.130 --> 32:11.690
as well as that as well as by introducing
the liquid in the glass plate and convex lens
32:11.690 --> 32:20.889
and by using the expression dn + p square
- dn square of air divided by dn + p square
32:20.889 --> 32:29.239
- dn square for film we can determine the
refractive index of the liquid we can show
32:29.239 --> 32:36.519
with the help of newtons ring experiment we
can determine the wavelength of the monochromatic
32:36.519 --> 32:44.619
light used and we can also measure the refractive
index of a given liquid only thing is that
32:44.619 --> 32:53.129
we have to determine the diameter of the newton
rings precisely by using the travelling microscope
32:53.129 --> 33:00.879
now i will discuss and other important experimental
setup which is based on division of amplitude
33:00.879 --> 33:06.659
is known as the michelson interferometer
33:06.659 --> 33:13.129
the main optical part of this interferometer
consists of two highly polished plane mirrors
33:13.129 --> 33:22.669
m1 and m2 two plane parallel glass plates
g1 and g2 of equal thickness as shown here
33:22.669 --> 33:30.899
in the figure the mirror m2 is fixed while
m1 can be moved such that during the motion
33:30.899 --> 33:38.020
it remains exactly parallel to its proceeding
position normally the rear side of the plate
33:38.020 --> 33:48.169
g1 is lightly silver so that the light coming
from source s is divided into reflected beam
33:48.169 --> 33:53.769
one and transmitted beam 2 of equal intensity
33:53.769 --> 34:03.979
the beams one and two are reflected back from
mirror m1 and m2 as shown in this figure respectively
34:03.979 --> 34:13.919
the reflected beam 1 is transmitted through
g1 along a and reflected beam 2 is reflected
34:13.919 --> 34:22.710
from g1 along ae these two waves travelling
along ae are derived from the same source
34:22.710 --> 34:31.610
by division of amplitude so they will satisfy
the condition of interference and hence interference
34:31.610 --> 34:41.409
fringes can be observed by looking into mirror
m1 through g1 from position e provided proper
34:41.409 --> 34:44.060
adjustments are made
34:44.060 --> 34:52.560
here the glass plate g2 is called compensating
plate its purpose is to equalise the total
34:52.560 --> 35:01.850
optical path of the being reflected from m1
and the total cost of the corresponding beam
35:01.850 --> 35:10.000
reflect it from m2 this is essential when
white light is used but not for monochromatic
35:10.000 --> 35:18.810
light in the michelson interferometer fringes
of various forms such as straight line circle
35:18.810 --> 35:27.670
parabola ellipse hyperbola can be observed
depending upon the angle between the mirrors
35:27.670 --> 35:29.850
m1 and m2
35:29.850 --> 35:37.340
but in the laboratory the interferometer is
adjusted for circular fringes for various
35:37.340 --> 35:44.210
applications therefore here i will discuss
the formation of circular fringes in the michelson
35:44.210 --> 35:46.260
interferometer
35:46.260 --> 35:52.470
to obtain the circular fringes the mirrors
m1 and m2 are made exactly perpendicular to
35:52.470 --> 36:00.480
each other with the help of screws attached
with the mirror and to see good quality fringes
36:00.480 --> 36:08.260
extended live shows is used an extended source
may be obtained by placing a ground glass
36:08.260 --> 36:15.770
screen in front of the source of monochromatic
light the origin of circular fringes can be
36:15.770 --> 36:19.320
understood with the help of this diagram
36:19.320 --> 36:28.800
here m2 prime at the virtual image of m2 formed
by reflection in g1 we can also think the
36:28.800 --> 36:39.410
extended source as being held behind the observer
and forming to virtual images l1 and l2 in
36:39.410 --> 36:48.910
m1 and m2 prime respectively these virtual
sources are coherent if t is the separation
36:48.910 --> 36:58.440
between m1 and m2 prime the virtual sources
will be separated by 2t when t is exactly
36:58.440 --> 37:06.000
and integral number of half wavelength that
is the path difference 2t is = an integral
37:06.000 --> 37:14.170
number of whole wavelength all the range of
light reflected normal to the mirrors will
37:14.170 --> 37:15.560
be in rage
37:15.560 --> 37:25.180
rays of light reflected at an angle power
bill in general not be in rage the path difference
37:25.180 --> 37:31.920
between the two rays coming to the i from
corresponding points p prime and p double
37:31.920 --> 37:39.570
prime is 2t cos theta as shown here in the
figure the angle theta is necessary the same
37:39.570 --> 37:49.870
for the two rays when m1 is parallel to m2
prime so that the rage are parallel hence
37:49.870 --> 37:57.670
when the i is focused to receive parallel
rays the rage will reinforce each other to
37:57.670 --> 38:06.000
produce maximum for those angle theta satisfying
the condition 2t cos theta is = n lambda
38:06.000 --> 38:16.140
since for a given n lambda and t the angle
theta is constant the maximum will lie in
38:16.140 --> 38:26.780
the form of circle about the foot of the perpendicular
from the i to the mirrors fringes of this
38:26.780 --> 38:35.290
kind when parallel beam are brought to interfere
with the phase difference determined by angle
38:35.290 --> 38:41.310
of inclination theta are called fringes of
equal inclination
38:41.310 --> 38:48.960
in this figure effect of changing t on the
fringe pattern is shown when m1 is moved slowly
38:48.960 --> 38:58.320
toward m2 prime so that t is decreased then
the radius of a given rings given ring characterized
38:58.320 --> 39:08.220
by a given value of order n must decrease
because the product 2t cos theta must remain
39:08.220 --> 39:15.960
constant the rings therefore shrink and vanish
at the centre the ring disappearing each time
39:15.960 --> 39:18.710
t decreases by lambda by 2
39:18.710 --> 39:26.970
this follows from the fact that at the centre
cos theta is = 1 then 2t is = in lambda so
39:26.970 --> 39:40.640
the change n by unity must change by lambda
by 2 now as m1 approaches m2 prime the ring
39:40.640 --> 39:48.960
rings become more widely spaced until finally
we reach a critical position where the central
39:48.960 --> 39:58.080
fringe has spread out to cover the whole field
of view this happens when m1 and m2 prime
39:58.080 --> 40:03.160
are exactly coincidence that is = 0
40:03.160 --> 40:10.280
because under this condition in the path difference
is 0 for all angles of incidence if the mirror
40:10.280 --> 40:20.930
m1 is moved still farther it effectively passes
through m2 prime and new widely spaced fringes
40:20.930 --> 40:29.020
appear growing out from the centre these will
gradually become more closely spaced as the
40:29.020 --> 40:39.360
path difference increases now i will describe
some of the applications of michelson interferometer
40:39.360 --> 40:45.920
one important up location of this interferometer
is the measurement of wavelength of monochromatic
40:45.920 --> 40:53.600
light for the measurement of wavelength concentric
circular fringes are invariably used
40:53.600 --> 41:01.530
so first m2 is adjusted perpendicular to m1
to get concentric circular fringes now suppose
41:01.530 --> 41:10.370
the separation between m2 prime and m1 is
such that a bright fringe of the order m is
41:10.370 --> 41:18.151
formed in the centre of the field of view
we therefore have for the central bride fringe
41:18.151 --> 41:27.910
2t is = n lambda since cos theta is = one
now we moved m1 such that it increases by
41:27.910 --> 41:35.140
lambda by 2 in this case the path difference
between the normally reflected waves from
41:35.140 --> 41:41.850
the centre of field becomes to 2t + lambda
by 2
41:41.850 --> 41:52.680
and therefore now we have 2t + lambda by 2
is = m + 1 lambda that is the condition of
41:52.680 --> 42:00.160
constructive interference is again satisfied
at the centre of field but the new bright
42:00.160 --> 42:10.420
fringes now are of the order one + one instead
of m we therefore conclude that each time
42:10.420 --> 42:18.340
t is increased by lambda by 2 in the field
of view where a particular bright fringe originally
42:18.340 --> 42:24.200
appears the neighbouring bright fringe of
next higher order shall be visible
42:24.200 --> 42:31.790
that is there will be a lateral shift of one
fringe in the field of view therefore for
42:31.790 --> 42:39.570
the displacement of n bright fringes in the
centre of m1 must be moved through n lambda
42:39.570 --> 42:40.760
by 2
42:40.760 --> 42:48.420
thus to determine lambda the mirror m1 is
moved from one position corresponding to micrometre
42:48.420 --> 42:57.110
reading x1 to another position corresponding
to micrometre reading x2 and the number of
42:57.110 --> 43:04.561
the bright monochromatic changes which cross
the centre of the field is counted so on the
43:04.561 --> 43:14.220
basis of every argument we can write the relation
and lambda by 2 = x2 - x one from this relation
43:14.220 --> 43:19.910
we can compute the value of lambda by measuring
x2 - x1
43:19.910 --> 43:27.640
another important application of michelson
interferometer is the determination of difference
43:27.640 --> 43:34.500
between the wavelengths of sodium d-lines
sodium light contains two very close wavelengths
43:34.500 --> 43:46.160
which are known d1 of wavelength 5896 angstrom
and d2 lines of wavelength 5890 angstrom that
43:46.160 --> 43:53.180
difference in lambda 1 and lambda 2 can be
precisely determined with the help of michelson
43:53.180 --> 43:54.890
interferometer
43:54.890 --> 44:01.270
so now i will discuss how this is done in
this interferometer suppose for a particular
44:01.270 --> 44:10.800
position of m1 at the centre field of view
the condition 2t is = m1 lambda 1 and 2 t
44:10.800 --> 44:19.780
is = m2 lambda 2 are satisfied simultaneously
where lambda 1 and lambda 2 are the wavelengths
44:19.780 --> 44:25.360
of d1 and d2 line of sodium respectively
44:25.360 --> 44:32.990
in this case bright ring of order n1 of wavelength
lambda 1 practically coincides with the bright
44:32.990 --> 44:42.110
ring of order n2 of wavelength lambda 2 leading
to maximum visibility at the centre
44:42.110 --> 44:48.450
due to difference in lambda 1 and lambda 2
at the separation between m1 and m2 prime
44:48.450 --> 44:57.540
gradually increases the maxima at first fall
further and further from coinciding and definitely
44:57.540 --> 45:04.620
at a particular separation the maxima of 1
practically coincides with the minima of other
45:04.620 --> 45:12.470
wavelength and vice versa thereby reducing
minimum visibility this happens when the extra
45:12.470 --> 45:20.680
path difference introduced at the centre due
to motion of m1contains 1 half wavelength
45:20.680 --> 45:23.710
more of radiation lambda 2
45:23.710 --> 45:31.390
than that of the others continuing the motion
of m1 in the same direction again gradually
45:31.390 --> 45:40.630
brings the maxima of two patterns close together
ultimately they again coincides at the centre
45:40.630 --> 45:47.580
when the extra optical path difference 2t
naught introduced at the centre contains one
45:47.580 --> 45:55.390
more wavelength of one radiation than of the
other the visibility of the fringes is again
45:55.390 --> 46:05.400
maximum and the conditions 2t + t naught is
= n1 + capital n into lambda 1 and 2t + t
46:05.400 --> 46:12.130
naught is = n1+ capital n + 1 into lambda
2 will be satisfied
46:12.130 --> 46:19.030
here it has been assumed that lambda 1 is
greater than lambda 2 and t naught if the
46:19.030 --> 46:28.040
displacement of m1 between successive visibility
of maxima and capital n represent the number
46:28.040 --> 46:36.940
of bright fringes of wavelength lambda 1 emerging
at the centre due to displacement t naught
46:36.940 --> 46:46.580
of mirror m1 therefore we can also write 2t
naught is = capital n lambda 1 and 2t naught
46:46.580 --> 46:49.970
is = capital n + one 1 into lambda2
46:49.970 --> 47:01.780
from these two equations we can write lambda
1 by lambda 2 is = n + 1 divided by n here
47:01.780 --> 47:09.260
it is very difficult to count n but n can
be determined from the relation capital n
47:09.260 --> 47:18.130
is equal to2t naught divided by lambda 1 using
this we can write lambda 1 - lambda 2 divided
47:18.130 --> 47:28.470
by lambda 2 = 1 over capital and which is
= lambda 1 divided by 2t naught finally from
47:28.470 --> 47:40.400
this we can write lambda 1 - lambda 2 is = lambda
1 into lambda 2 divided by 2t naught which
47:40.400 --> 47:46.480
will be nearly = lambda average square divided
by 2t naught
47:46.480 --> 47:53.680
where lambda average is the average of lambda
1 and lambda 2 thus by measuring the positions
47:53.680 --> 48:02.430
of m1 for two successive visibility maxima
in the central field of view lambda 1 - lambda
48:02.430 --> 48:10.200
2 can be determined since visibility minima
can be distinguished more accurately than
48:10.200 --> 48:19.490
the maxima the value of t naught is therefore
determined by the position of m1 for two successive
48:19.490 --> 48:22.680
minimum visibility positions
48:22.680 --> 48:35.650
further in order to minimize the mirrors normally
the position of mirror m1 is noted for 10
48:35.650 --> 48:45.190
successive visibility minima and from this
the separation between value t naught can
48:45.190 --> 48:54.510
be determined more accurately now i will discuss
another important interferometer which is
48:54.510 --> 48:58.790
known as the fabry perot interferometer
48:58.790 --> 49:05.540
in this interferometer the interference fringes
are produced in the transmitted light after
49:05.540 --> 49:13.710
multiple reflections in the air film between
two plane glass plates thinly silvered on
49:13.710 --> 49:20.430
the inner surfaces as shown in the figure
to observe the fringes a broad source of monochromatic
49:20.430 --> 49:29.880
light is used suppose a ray from the point
p1 on the extended source is incident at the
49:29.880 --> 49:38.770
angle theta this will produce a series of
parallel transmitter rays at the same angle
49:38.770 --> 49:47.060
which may be brought together at the point
p2 on the screen if with the help of a lens
49:47.060 --> 49:55.730
the condition for reinforcement of the transmitter
is is 2t cos theta is = m lambda where m is
49:55.730 --> 50:04.880
integer for air film of thickness t this condition
will be fulfilled by all points on a circle
50:04.880 --> 50:15.430
through p2 with their centre at o the intersection
of the axis of the lens with the screen ef
50:15.430 --> 50:22.850
when the angle theta is decreased the cosine
will increase until another maxima is reached
50:22.850 --> 50:31.810
for which m is greater by 1 2 3 and so on
so that we have for the maxima a series of
50:31.810 --> 50:35.430
concentric rings on the screen with o their
centre
50:35.430 --> 50:43.130
these rings are known as fringes of equal
inclination in this case effect of changing
50:43.130 --> 50:51.020
t on circular fringes is same as for as we
have discussed for michelson interferometer
50:51.020 --> 50:56.870
in this inter in the interferometer used in
the laboratory one plate is fixed
50:56.870 --> 51:02.710
while the other night more in both direction
keeping parallel to its preceding position
51:02.710 --> 51:09.920
to change the value of t this interferometer
is also used for the measurement of wavelength
51:09.920 --> 51:17.090
of monochromatic light and difference in wavelength
of sodium d1 and d2 lines the methods are
51:17.090 --> 51:25.660
same as described above for the michelson
interferometer so let me summarize what i
51:25.660 --> 51:28.120
have discussed in this lecture
51:28.120 --> 51:34.590
so in this lecture i have discussed how to
get interference pattern by division of amplitude
51:34.590 --> 51:42.360
by taking example of thin parallel films of
constant thickness as well as taking the film
51:42.360 --> 51:51.400
of variable thickness and i also discussed
how this colour effect is observed through
51:51.400 --> 52:01.440
thin films and i discuss three important experiments
which are usually performed in the laboratory
52:01.440 --> 52:11.730
like a newton ring experiment michelson interferometer
and fabry perot interferometer and how to
52:11.730 --> 52:20.110
determine the wavelength of monochromatic
light by using these experimental setups thank
52:20.110 --> 52:20.330
you