WEBVTT
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i am mk srivastav of the department of physics
iit roorkee this is the second lecture of
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the five lecture series on polarization in
the last lecture which was the first one we
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considered maxwells equations leading to electromagnetic
waves we then considered the meaning of polarized
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light polarization by reflection refraction
and scattering we study brewsterâ€™s law and
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considered working of a polaroid
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in the present lecture which is the second
one of the series we shall take up malus law
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and then shall consider superposition of two
electromagnetic waves under different conditions
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the different condition means difference in
amplitude different phase differences and
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things like that law of malus
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considered an unpolarized beam of amplitude
a and intensity i naught which is proportional
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to the square of the amplitude as we know
incident on a polarizer the incident light
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have components ax = a cos theta and ay = a
sine theta along the axis the theta varying
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randomly as the light is unpolarized
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let the pass axis of the polarizer making
an an angle alpha with the x axis the polarizer
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as we have seen is a device which allows the
light wave with electric vibrations parallel
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to the pass direction to pass through without
attenuation essentially and absorb the light
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wave electric vibrations perpendicular to
the pass direction the long arrow in the figure
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making an angle alpha with the x axis indicates
the pass direction of the polarizer
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the intensity i of the transmitted beam through
the polarizer can be obtained by taking the
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production of the components a x and a y along
the pass direction and intensity is proportional
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to the amplitude we have intensity i is proportional
to the square of a x cos alpha plus a y sine
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alpha it means the proportional to a square
times the square of cos theta cos alpha the
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sine theta sine alpha
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it means the proportional to cos is square
theta plus cos square alpha plus sine square
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theta sine square alpha plus twice sine theta
cos theta sine alpha cos alpha now and the
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angle theta varies randomly in the unpolarized
beam
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the average values are cos is square theta
the average value is half sine squared theta
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again the average value is half you see the
averages are taken over a complete cycle then
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the average value of sine theta into cos theta
again over a complete cycle is zero this leads
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to intensity is not proportional toaa square
times half of cos is square alpha plus half
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of sine square alpha which means it is proportional
to half of a not a square
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and we get the final result that i is = half
of i naught the intensity of the transmitted
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wave mean remember the incident beam is unpolarized
so the intensity of the transmitted beam is
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thus half of the intensity of the incident
unpolarized mean
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and another thing this result does not depend
on the orientation of the pass direction of
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the polarizer that is it is independent of
the angle alpha and the transmitted beam is
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plane polarized with electric vibrations along
the pass direction of the polarizer
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if the incident beam is plane polarized with
electric vibrations making an angle theta
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as before but now this angle is fixed with
the x-axis and the pass direction of the polarizer
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makes an angle alpha with the x axis as before
the intensity of the transmitted beam is again
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given by the same expression
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that is the intensity proportional to the
square of a x cos alpha plus a y sine alpha
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which means it is a proportional to a square
times cos theta cos alpha plus sine theta
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sine alpha squared so it is proportional to
a square times cos is square of theta minus
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alpha therefore the intensity i is = i naught
times cos is square of theta minus alpha you
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see no averaging need be done now because
the angle theta is fixed the incident light
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beam is polarized
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the intensity of the transmitted beam now
you find thus varies as the square of the
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cosine of the angle theta minus alpha the
angle between the plane of vibration of the
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incident beam and the pass direction of the
polarizer working here as an analyzer this
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is called law of malus
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now if a plane polarized beam polarized along
the x axis by the polarizer p1 the incident
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on another polarizer p2 here
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and if this polarizer is rotated about the
beam direction then the intensity of the emergent
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wave will vary according to the above law
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for example the polarizer p 2 is rotated in
the clockwise direction then the intensity
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will increase till the pass axis is parallel
to the x axis further rotation will result
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in a decrease in intensity till the pass axis
is perpendicular to the x axis where the intensity
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will be almost zero
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if we further rotate it will pass through
maximum and again a minimum before it reaches
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its original position
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the emergent light after passing through the
polarizer p2 is now polarized with the direction
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of electric vibrations parallel to the pass
direction of p2 working here as an analyzer
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let us now consider superposition of two waves
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consider the propagation of two plane polarized
electromagnetic waves both propagating along
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the z axis that is the same direction with
their electric vectors oscillating along the
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same direction that is the x axis for the
electric fields associated with the waves
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can be written in the form e 1 = a 1 cos of
k z minus omega t plus theta 1 and e 2 which
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is a 2 times cos of k z minus omega t plus
theta 2
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both of these fields are along the x-axis
a 1and a 2 reporting the amplitudes of the
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waves and x hat represent the unit vector
along the x axis theta 1 theta 2 or arbitrary
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for its constants the resultant of these two
waves would be given by
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just the vector addition so e is = e1 plus
e2 now this can always be written in a form
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like this: e is = a times cos k z minus omega
t plus theta along the x axis naturally
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where a is given by the square root of a 1
square plus a 2 square plus twice a1 a2 times
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cos of theta 1 minus theta 2 a represent the
amplitude of the resultant wave resultant
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of the superposition of the original two waves
we observe that the resultant is also a plane
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polarized wave with its electric vector oscillating
along the same direction along the x axis
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let us now consider superposition of two plane
polarized waves both propagating along the
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z axis but with their electric vectors oscillating
along two mutually perpendicular directions
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earlier they were along the same direction
now they are mutually perpendicular directions
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say along the x axis and along the y axis
remember both the waves are propagating along
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the z axis
so we can write e1 is = a 1 cos of k z minus
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omega t this is along the x-axis e 2 is = a
2 cos of k z minus omega t plus theta along
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the y direction the angle theta is the phase
difference between these two waves
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now in order to find the resultant we consider
the time variation of the resultant electric
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field at an arbitrary plane just for simplicity
at an arbitrary plane perpendicular to the
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propagation direction which is the z axis
this arbitrary plane without any loss of generality
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we can assume it to be z = 0 just for simplicity
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ex and ey represent the x and y components
of the resultant field even proceed to then
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ex will be = a 1 cos omega t and ey is = a
2 cos omega t minus theta now eliminate t
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between these two equations and for that we
rewrite the above expression for e y has e
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y is = a 2 cos omega t cos theta plus a 2
sine omega t sine theta we can write it as
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first term remaining unchanged a 2 cos omega
t cos theta plus a 2 times the square root
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of 1 minus cos square omega t times sine theta
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this can be written as = a 2 upon a 1 times
ex cos theta plus a 2 times the square root
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of 1 minus e x upon a 1 square times sine
theta or this means e y minus a 2 upon a 1
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e x cos theta whole squared is = a 2 squared
sine square theta times 1 minus e x upon a
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1 square
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we can rearrange this and when we do this
we get e x is squared upon a 1 square plus
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e y square plus a 2 square minus twice e x
e y upon a 1 a 2 times cos theta and all of
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this is = sine square theta this is the general
equation of an ellipse the general equation
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of an ellipses is an ellipse is at the center
of the origin which is the major minor axis
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making some angle with the coordinate axis
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so this represents in general the result of
superposition of two perpendicular vibrations
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of different amplitudes and the phase difference
of theta for the phase difference theta = n
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pi where n is 0 1 2 this means an integral
multiple of 1 pi 2 pi 3 pi 4 pi
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the above equation becomes e x upon a 1 minus
minus 1 raised to the power n e y upon a 2
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and the whole square = 0 which really means
e y upon e x the ratio of the y component
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to the x component is = minus 1 raised to
the power n a 2 upon a 1 in the e x e y plane
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the above expression represents a straight
line the angle phi that this line makes with
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the x axis depends naturally on the ratio
a 2 upon a 1
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in fact angle phi is tangent inverse plus
minus a 2 upon a 1 if a1 is = a2 this angle
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will be 45 degrees the condition theta = n
pi implies that the two vibrations are either
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in phase if n is even 0 to 4 which means theta
is = 0 or the phase difference of 2 pi 4 pi
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6 pi or completely out of phase which means
n is = 1 3 5 the phase difference theta is
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= 1 pi 3 pi 5 pi like this
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this means the phase difference of half a
vibration 3 half vibrations or 5 half vibrations
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that is super position of two linearly polarized
light waves with their electric fields at
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right angles to each other and oscillating
in phase are completely out of phase ok these
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just these two conditions: digital vibrations
are perpendicular to each other but they are
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in phase or completely out of phase then the
resultant is again the linearly polarized
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wave with a selective vector in general oscillating
in a direction which is different from the
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fields wider of the two making an angle of
phi
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we would say the first vibrations and this
angle phi depends on the ratio of the two
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amplitudes a2 to a1
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for theta not = n pi it is not if the phase
difference is not a integer multiple of pi
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is not = 1 pi or 2 pi or 3 pi the resultant
electric vector in this case does not oscillate
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along a straight line
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this is this figure shows the resultant for
different values of theta theta = zero pi
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by 3 pi by 2 2 pi by 3 theta = pi 4 pi by
3 different values up to a value of 2 pi is
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a complete range of theta varying from 0 to
2 pi corresponding to the simpler case when
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the amplitudes are equal a 1 is = a 2
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we shall consider these guesses in a little
more detail
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the sections a e and i of this figure correspond
to theta = 0 pi and 2pi respectively a corresponds
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to theta = 0 that is the two vibrations are
in the same phase remember here the amplitudes
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are equal a 1 is = a 2 the e corresponds to
theta = pi they are completely out of phase
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the phase difference of half is half a vibration
and high is again when the phase difference
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is 2 pi which is essentially means phase difference
is zero
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these are the situations as i pointed out
earlier when the vibrations are in phase or
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completely out of phase we have already considered
these cases let us consider now a simple case
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corresponding to theta = pi by 2 that is very
interesting the difference is of a quarter
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vibration
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e x a 1 cos omega t e y a 1 sine omega t remember
a 1 is = a 2 e x has a cos omega t and e y
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is the sine omega t the phase difference of
pi by 2 e y upon e x comes out to be 10 omega
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t if we plot the time variation of the resultant
of these two we would find that the tip of
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the electric vector rotates from the circumference
of a circle of radius a 1 e x square plus
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e y square is = a 1 square in the anti-clockwise
direction
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the propagation of light in the pole is in
the positive direction of that axis this had
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been taken here to be coming out of the screen
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such a wave is known as right circularly polarized
wave usually denoted as rcp wave right circularly
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polarized and is shown in section c of the
figure which is a section c
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this shows a circular vibration circular because
a1 is = a2 and it is anti-clockwise direction
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remember the beam propagation is coming out
of the screen directive vector is rotating
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in the x y plane in the anti clockwise direction
and theta = 3 pi by 2
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now we have x = a 1 cos omega t as before
but now e y is = with the minus sign a 1 sine
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omega t it again represents a circularly polarized
wave as before however in this case the electric
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vector rotates in the clockwise direction
such a wave is known as left circularly polarized
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wave usually denoted as lcp wave and is shown
in section g of the figure
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23:43 figure to check if needed
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again a circular vibration the only difference
with the earlier case c is that here the electric
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vector rotates in the clockwise direction
remember the beam is propagation direction
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is coming out of the screen this is called
a left circularly polarized
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now for theta not = mpi by 2 so this means
this is a case where it is not = an integer
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multiple of pi by 2 in this case the tip of
the electric vector rotates on the circumference
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of an ellipse sections b and d of the figure
for theta = pi by 3 and theta = 2 pi by 3
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correspond to right polarized elliptical light
this is the rep light elliptically polarized
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(refer time:24:48)
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remember again the a1 is = a2 but the only
difference here is the phase difference is
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not just pi by 2 or 3 pi by 2 it is pi by
3 or in the other in the in the section d
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it is 2 pi by 3 and that is why the resultant
is is a general ellipse whose major minor
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axis makes some angle with the coordinate
axis so this is the situation when the phase
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difference is pi by 3 and 2 pi by 3
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(refer time:25:42)
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similarly now the sections f and h for theta
= 4 pi by 3 and theta equal to 5 pi by 3 they
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correspond to left elliptically polarized
lep light these are all in here section f
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again we have elliptical the only see the
sense of rotation is different compared to
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the earlier case and again remember the amplitudes
are equal so it is the wide change in the
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resultant is found depending on the phase
difference between the two vibrations
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the two vibrations are perpendicular to each
other their amplitudes are same equal a 1
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is = a 2 the only difference is the phase
difference if the phase difference is 0 or
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pi 2 pi 3 pi or 4 pi completely in phase or
out of phase the resultant is again a plane
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polarized wave linear linear vibrations making
an angle of 45 degrees or 135 degrees in this
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case a 1 is = a 2
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if the phase difference is pi by 2 or 3 pi
by 2 this means half a vibration or three
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half vibrations the result is a circular vibration
this is called a circularly polarized light
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right circularly or left circularly depending
on where the tip of the electric vector rotates
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in anti-clockwise direction or clockwise direction
and the general case when the phase difference
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is not that simple
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if it is pi by 3 in this case i showed in
the figure or 2 pi by 3 or 5 pi by 3 the result
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is elliptically polarized light right elliptically
or left elliptically depending on whether
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the electric vector rotates as can be seen
from the figure
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just let us in general degenerate into a straight
line or a circle as we have seen whenever
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theta becomes an even or an odd multiple of
pi by 2 as we have seen but in general when
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a 1 is not = a 2 we get an elliptically polarized
light waves which again degenerates into a
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straight line for theta = 0 pi 2 pi 3 pi etcetera
remember theta = 0 2 pi 4 pi means they are
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in the same phase
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and theta = pi 3 pi 5 pi means they are completely
out of phase again note one thing all these
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different states of polarization which we
have considered here depending on the value
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of theta our characteristics of transverse
waves only and electromagnetic waves are transverse
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waves so this is the end of the second lecture
i hope you enjoyed this lecture and also the
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earlier one thank you