WEBVTT
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so welcome back today we will going to start
our lecture number thirty we are discussing
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fluorescence anisotropy and we are done till
the perrin equation so what is perrin equation
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let me write it once again so perrin equation
is written as r zero by r s s equal to one
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plus tau f divided by tau or so where this
r zero value is initial anisotropy r s s my
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steady state anisotropy and this tau f is
my fluorescence lifetime and tau over is orientational
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relaxation time
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so from here what you can see is that without
if you have some idea about these value of
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r zero right and this r s s can be measured
very easily from this fluoremeter where you
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have this polariser and analyser right normal
fluoremeter does not have this fluorophore
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polariser analyser but if you can install
a polariser analyser so that their excitation
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light is polarised light and the emission
you are going to analyse which is the perpendicular
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direction and which is in the intensity and
what is the intensity of the perpendicular
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direction and what is the intensity of the
parallel direction
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then you will be able to calculate this r
steady state easily right and if you have
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this knowledge of this fluorescence lifetime
then you will be able to calculate this tau
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over now if you recall that our stoke einstein
relationship then this tau over can be written
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as tau over equal to eta v by r t right this
is my stoke einstein relationship where this
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eta is equal to the viscosity of the medium
v is molecular volume and r is my universal
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gas constant
and t is my absolute temperature
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now if you plug in this expression for this
tau or in the perrin equation what we will
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going to see is the following one over r s
s equal to one over r zero plus tau f v by
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into r divided by r zero into v into temperature
by viscosity right it means that now if i
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plot or this is this is the measurable quantity
this is your measurable quantity these is
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the property at a particular temperature obviously
you know the viscosity so then t by eta is
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known quantity and i can measure this r steady
state for a different value of t by eta against
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the temperature obviously viscosity will change
so t by eta will change and for a different
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values of this t by eta i can measure this
steady state anisotropy and then i can plot
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it right
so let us plot that same so if i plot what
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you will going to see is one over
r steady state versus t by eta we will going
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to see a straight line and intercept will
be equal to one over r zero so from here you
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will get some idea of this r zero right and
then slope will be equal to tau f r by r zero
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v now if tau f the fluorescence lifetime is
known r obviously is a constant and r zero
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if you have fair idea of this r zero then
you can calculate v that is the volume and
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this is well used equation in application
of the fluorescence anisotropy so from with
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this you can actually determine the size of
lets lets say you are working with a protein
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and this protein is getting denatured once
the protein is getting denature the size will
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increase and if you have one fluorophore already
attached to the protein then by measuring
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the fluorescence anisotropy that means the
fluorescence this r s s value by measuring
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this r s s value what will be able to do you
will be able to see that at this particular
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condition what is the volume of the protein
or that means the size of the protein and
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so on right
now one of the important thing here is that
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what will be the value of r zero right thats
thats the biggest question right so that should
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be measured otherwise i cannot say anything
about this and we have seen that limiting
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value of r zero you remember our first equation
i have written r equal to i parallel minus
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i perpendicular divided by i parallel plus
twice i perpendicular right now i am not talking
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about i am not including this g because considering
the g equal to one g is that one then just
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a factor will be there right so considering
the g value is equal to one and i am now doing
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all those thing
now in case of r time dependence then this
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is nothing but r t is equal to i parallel
t minus i perpendicular t and so on so that
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means at time t equal to t equal to zero that
is the my initial anisotropy r zero and that
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is whatever i have written here as r zero
so in this case i have to determine what will
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be the value from that particular equation
when i said that this i parallel minus i perpendicular
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divided by i parallel plus twice i perpendicular
obviously at time t equal to zero there was
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no emission from this perpendicular component
and there is only emission from this parallel
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component makes it r zero equal to one but
whether this is true and we are i also said
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that when this r value will change as time
goes because the intensity of the perpendicular
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component will increase and parallel component
will decrease because of the reorientations
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of the molecules or transition dipoles
when at at the excited state so as time goes
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there will be the the intensity of the parallel
and perpendicular components will be equal
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and in that case you will see that r will
going to be zero ultimately at a certain after
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a certain time
now let us see this things right in more detailed
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way so let us now consider that you have excited
the molecules along z axis so z axis is your
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polarization for the excitation light ok
so let me write here so polarization of the
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excitation light is along z lets i consider
it over here so if i consider that only those
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molecules thats i i said the only the molecule
whose transition moments are along the z axis
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they will be preferentially excited thats
what i said earlier again i am telling the
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same thing so those molecule will be excited
and if the direction of the absorption transition
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moment and emission transition moments are
same like they are co linear then the emission
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will be also along z axis that is my condition
now i am imposing here the direction of the
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absorption transition moment and emission
transition moment they are co linear then
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the emission will be also along the z axis
take a note on that ok
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now let me tell you that now another molecule
has been rotated an angle theta right and
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now i am looking at this molecule so now this
bold arrow is my molecule which has been rotated
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at an angle theta from the direction of the
excitation and now i am looking at this molecule
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so here you see the rotation of the molecule
makes a component of this i parallel s cosine
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theta as you know and but the molecule could
can rotate like if this is my x z axis the
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molecule can rotate from here to this side
as well as from here to this side in all the
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cases the angle is theta this side angle is
theta this side angle is theta that means
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i need to worry about all these different
angles for a fixed value of theta that means
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if i if i assign this angle as pi then the
value of pi should be zero t two pi thats
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what i have written over here so now here
you see for this perpendicular component
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this should be sin theta sin phi correct so
the intensity those are the electric field
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so those are the intensity i parallel will
be equal to cos square theta and i perpendicular
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will be sin square theta sin square phi but
here also please note that we should take
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all values of sin square phi that means i
should take the sin square phi average right
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so i should take the sin square phi average
so let me find out that
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so sin square phi average is equal to integration
zero to two pi sin square phi d phi divided
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by integration zero to two pi d phi so and
i already got this i parallel equal to like
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this i perpendicular is like this so if i
can calculate this if i can calculate this
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i can simply replace this sin square phi with
this this i can do as i said because that
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all values between all values of phi between
zero to two pi will have the equal
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probability so now i can evaluate this integration
zero to two pi sin square phi d phi right
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that part
so if you integrate this then you will
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get phi by two minus one over four sin square
phi limit i will put zero to two pi divided
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by this will going to this is simple this
is just two pi so then this will going to
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be pi by two pi this is equal to half so as
i got this sin square phi average is this
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half so i will simply write i parallel is
equal to cosine square theta and i perpendicular
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equal to half sin square theta you see that
this is now half sin square theta
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now r is equal to i parallel minus i perpendicular
divided i parallel plus twice i perpendicular
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which will going to give me cos square theta
minus half sin square theta divided by cosine
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square theta plus plus two half and two this
will cancels sin square theta so this is this
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is equal to three and then i will just simplify
this and what you will get is cosine square
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theta minus one divided by two now as you
can see here that for the value of theta if
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the theta equal to zero right the theta value
could be anything right so lets let me put
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for theta equal to zero that means it has
not oriented because you have excited along
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the z axis and just immediately after excitation
you are looking at the r value right so theta
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equal to zero so for theta equal to zero this
is nothing but three minus one divided by
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two sorry this is equal to one
for theta equal to zero the value of r equal
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to one so here i go the initial anisotropy
is one right and now as the molecule will
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rotate the theta will change theta will increase
and it will increase and increase and ultimately
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at some value of theta then isotropy will
be zero so when the cos square theta equal
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to one by three as you can see from this equation
that when cos square ok let me write this
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thing this is important so when for theta
equal to zero r i got equal to one and when
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theta equal to fifty four point seven five
degree it means cos square theta equal to
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one third what you get you get r equal to
zero right that means the molecule is not
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necessarily has to orient itself ninety degree
to make the anisotropy equal to zero only
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fifty four point seven degree orientations
will make the anisotropy equal to zero
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now whatever i said here right is the ideal
situation that means that your your excitation
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light is polarised along z axis only the molecules
which whose transition moment absorption transition
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moment is oriented along z axis only those
will going to be excited like that now why
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and now this molecules will reorient themself
and the theta value will change i consider
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that the direction of the transition absorption
transition and emission transitions are co
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linear so the emission will also come along
that line right as i have excited and then
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i got this equation
however it is not true that once you use a
16:14.860 --> 16:19.790
vertically polarised light or polarised light
electric filed is oscillating along the z
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axis that will only excite the molecules which
are oriented along the z direction obviously
16:26.339 --> 16:32.439
there will be some component of even the electric
field is like this and the molecules like
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this this the the component which is along
this direction the electric the component
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of the electric field which is the along this
direction that will also be used to excite
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the molecule little bit it is true that the
those molecules which are oriented just perpendicular
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to the direction of the electric field
of the excitation light those will not excited
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at all thats true but there is a possibility
that the molecules which are oriented at
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certain angle from the co linear geometry
of compared to the excitation polarization
17:09.380 --> 17:14.500
there will be some possibility of the excitation
and that is known as the excitation photo
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selection of the fluorophore let me discuss
these things over here
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so see this is the excitation photo selection
of fluorophore
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this is what i said the molecules which are
oriented like this way they will be excited
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thats sure but the molecules which are oriented
like this way will also some chance for the
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excitation right thats what i said and that
that is proportional to the cos square theta
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if the molecule is oriented at an angle theta
from the polarisation of the excitation light
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right so that will be proportional to cosine
square theta ok so let me write this over
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here so as you use ok let me write it like
this way so the excited state population will
18:39.640 --> 19:01.180
be distributed
symmetrically around z axis in this particular
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case ok
so that distribution right i can write this
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as f theta d theta that is possible to show
that the is equal to cosine square theta sin
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theta d theta right so and this anisotropy
is given by i have already seen three cosine
19:27.420 --> 19:33.050
square theta minus one by two right thats
what i already got so in this a case what
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i will do let let us continue so in this case
that cosine square theta right not here in
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this case that cosine square theta in my earlier
equation is that the value of theta is not
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a single value because now the theta value
is different and that is my distribution over
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here what i said this this this is distributed
so in a molecule so an each molecules is make
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a different angle theta and for a different
different theta value all the five values
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are possible right
so this is a like a complicated situation
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and in this case i should not write the equation
as cos square theta but i should write the
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average cos square theta so the then i have
to take i have to take the average cos square
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theta right so for this average cos square
theta what i will write sorry average cos
20:30.290 --> 20:40.130
square theta equal to cos square theta f theta
because this is my distribution f theta d
20:40.130 --> 20:50.301
theta is my distribution and that integration
from zero to pi by two divided by zero to
20:50.301 --> 21:00.820
pi by two f theta d theta and if you evaluate
this this is integration zero to pi by two
21:00.820 --> 21:16.670
cosine to the power four theta sin theta d
theta divided by zero to pi by two cos square
21:16.670 --> 21:24.250
theta sin theta d theta
so if you evaluate you will see that these
21:24.250 --> 21:33.180
value is three by five right so all this theta
what is the distribution of theta is because
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of the excitation photo selection and this
distribution is as this distribution is because
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of the excitation photo selection this is
my initial value of the cos square theta average
21:44.360 --> 21:49.500
so now if put this cos square theta average
in my original equation lets see that r equal
21:49.500 --> 21:57.260
to three cos square theta but now this is
average because i have so many things minus
21:57.260 --> 22:06.090
one divided by two so then this is equal to
for the initial value three into five minus
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one by two this is nothing but zero point
four so r zero what i will get is a zero point
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four and this is true for when the absorption
and emission transition moments are co linear
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with each other if they are not then this
value will not like that if not if the absorption
22:42.890 --> 23:00.160
and emission
transition moments will make an angle beta
23:00.160 --> 23:15.480
in between them right
then this r zero equation right now i am writing
23:15.480 --> 23:24.390
r zero i am not writing r right now so r zero
will can be written as two by five to three
23:24.390 --> 23:34.890
cosine square beta divided by say minus one
by two right so when this beta equal to zero
23:34.890 --> 23:39.800
that means they are co linear right immediately
what you will going to see is that r zero
23:39.800 --> 23:47.130
equal to zero point five if beta equal to
fifty four point seven five then what will
23:47.130 --> 23:53.340
immediately get that r zero equal to zero
if beta equal to ninety degree that means
23:53.340 --> 24:00.030
if this absorption and emissions are perpendicular
to each other right then for example for example
24:00.030 --> 24:05.030
let me give you example that will then it
will be much clear to you
24:05.030 --> 24:15.960
lets say i have this molecule our famous example
many times i use this molecule enthroning
24:15.960 --> 24:25.880
right and i told you this absorption for the
s zero s one is along this direction and s
24:25.880 --> 24:35.750
zero s two is along this direction i consider
that emission is also the same but if you
24:35.750 --> 24:41.540
now excite from s zero s two and but you will
get the emission from the s zero s one from
24:41.540 --> 24:48.330
s one to s zero because of the so you are
exciting along this direction so here this
24:48.330 --> 24:59.160
direction is your excitation because you are
exciting from s zero to s two but emission
24:59.160 --> 25:06.600
is along this direction is different right
so they are perpendicular to each other
25:06.600 --> 25:13.130
now if you excite emission is from s one to
s zero now if you excite from s zero to s
25:13.130 --> 25:17.290
one and then s one to s zero then they are
co linear to each other right then they are
25:17.290 --> 25:22.710
collinear so in that case that is the value
beta value is equal to zero and in this case
25:22.710 --> 25:28.000
the beta value is equal to ninety degree so
for beta value equal to ninety degree what
25:28.000 --> 25:35.560
you will get is this r zero value equal to
minus zero point two
25:35.560 --> 25:42.530
so what you can see is that the the value
of r zero or initial anisotropy or fundamental
25:42.530 --> 25:53.290
anisotropy right it depends it depends on
many factors right many factors means basically
25:53.290 --> 26:00.360
this is what is the angle between the absorption
transition moment and the emission transition
26:00.360 --> 26:05.720
moment right so for a different different
molecules this angle will be different so
26:05.720 --> 26:13.850
the value will be different so this has to
be measured right and we can measure it
26:13.850 --> 26:20.060
by time resolve anisotropy decay as i told
you if you have a very very high time resolution
26:20.060 --> 26:25.290
like a ten to second time resolution using
the fluorescence up conversion system then
26:25.290 --> 26:31.760
by putting this polariser in the ten to second
up conversion system either at the parallel
26:31.760 --> 26:37.820
or in the perpendicular position compared
to the direction of the excitation or the
26:37.820 --> 26:42.340
polarisation of the excitation is lesser been
you will be able to find out the value of
26:42.340 --> 26:48.310
r at very very early time with a very a high
time resolution and that will be going to
26:48.310 --> 26:56.650
be your r zero value right
ok so this all about this fluorescence anisotropy
26:56.650 --> 27:04.950
and as we have little more time then i think
that we should continue our discussion on
27:04.950 --> 27:09.850
little more atmospheric of this fluorescence
anisotropy which is the associated anisotropy
27:09.850 --> 27:29.460
decay associated anisotropy decay so right
now what i said is that the decay our anisotropy
27:29.460 --> 27:38.360
decay of a fluorophore is present in a particular
media right in a homogeneous media now i am
27:38.360 --> 27:47.830
thinking of that if i have two def two fluorophores
right but present together in a solution but
27:47.830 --> 27:53.390
their environment are different their the
environment of two different types of fluorophores
27:53.390 --> 27:58.750
are different but they will going to give
me give the contribution to the anisotropy
27:58.750 --> 28:09.630
decay then what will be the effect
so lets take this as that i have i of the
28:09.630 --> 28:21.830
i th component is defined as alpha i so this
my lifetime part right t by tau f i and my
28:21.830 --> 28:37.210
r t part r i t is equal to r zero i into
e to the power minus t by tau or i right so
28:37.210 --> 28:54.360
i can write the fractional intensity of the
i the component if it is only two then first
28:54.360 --> 29:11.460
and second component at any time t just i
write this as f i t equal to alpha i e to
29:11.460 --> 29:26.400
the power minus t by tau f i divided by sum
over i alpha i e to the power minus t by tau
29:26.400 --> 29:34.450
f i that means the time dependent anisotropy
that r t r t should be given by sum over i
29:34.450 --> 29:44.070
because if it has more weightage the early
time then the weightage of the r r r i
29:44.070 --> 29:48.210
t will be there because it is now more weightage
and in the later time
29:48.210 --> 29:57.320
so i can simply write r t equal to sum over
i f i t r i t so you see this is a complicated
29:57.320 --> 30:04.370
thing right and as expected the anisotropy
will also be complicated usually anisotropy
30:04.370 --> 30:08.510
looks like that is started from some value
r zero and exponentially decay but in this
30:08.510 --> 30:15.020
case if you see this anisotropy decay it will
look something like this but using such kind
30:15.020 --> 30:23.820
of formula you can analyse it without any
problem so if i now plot r t as a function
30:23.820 --> 30:37.000
of t over here then it could looks like this
like this or it could look like this so it
30:37.000 --> 30:43.340
doesnt look is anisotropy but this is because
of this two different fluorophore two or more
30:43.340 --> 30:48.850
different fluorophores present whose environments
are quite different one quick example is that
30:48.850 --> 30:54.010
the tryptophan right
so if you have tryptophan in a in a protein
30:54.010 --> 31:00.950
right when tryptophan is present in the protein
the r t value right or this tau over value
31:00.950 --> 31:06.800
is very large because the protein is very
large but if it is now free right then the
31:06.800 --> 31:11.970
tau over value is really small and the lifetime
is also quite different so for such kind of
31:11.970 --> 31:17.400
complicated cases what you will going to see
such kind of associated anisotropy decay and
31:17.400 --> 31:22.950
you need to really careful you have to be
really careful for the analysis of such kind
31:22.950 --> 31:26.370
of anisotropy decay ok so we will finish here
and
31:26.370 --> 31:28.230
thank you for your kind attention