WEBVTT
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welcome to the lecture number twenty nine
let us continue our discussion on fluorescence
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anisotropy and till date what we have done
is to find out what will be the time dependence
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of the intensity of fluorescence when collected
in the parallel direction that means i parallel
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as a function of t that we have seen in the
in the last class let us write it once again
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so what we have seen that i parallel t is
equal to i zero by three e to the power minus
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t by tau f into one plus twice r zero e to
the power minus t by tau over similarly we
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can also calculate that the same procedure
can be used to evaluate the i perpendicular
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t and if you do so i will just write two
steps over here you will get i zero by three
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e to the power minus t by tau f and immediately
you will get one plus three i y zero minus
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i x zero minus i y zero minus i z zero divided
by i zero e to the power minus t by tau over
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so this is i zero by three e to the power
minus t by tau f from here it will be minus
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so into one minus r zero e to the power minus
t by tau over good so what i got i got two
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equation one is for i parallel t and another
is for i perpendicular t
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now i can immediately write the form of
r t right this is my anisotropy which is also
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time dependent so for this r t what i will
do i will put the this expression for this
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i parallel t minus i perpendicular t divided
by i parallel plus twice i perpendicular in
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this case i am taking the g value equal to
one but if there is any g value i can take
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but the form right will be like this only
right ok now let me write this two form so
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here will be my i zero by three e to the power
minus t by tau f one plus two r zero e to
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the power minus t by tau over minus this i
parallel and minus i perpendicular so i zero
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by three e to the power minus t by tau f one
minus r zero e to the power minus t by tau
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over divided by
this is straight forward nothing special
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just you have to substitute this expression
for i parallel and i perpendicular and you
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just need to solve it simplify it and looks
this looks like but if you simplify and which
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you can do later what you will going to see
is a very simple form of this which is nothing
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but r zero into e to the power minus t by
tau over
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so you see this simplification right i will
get like this way so ultimately what i got
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r t is r zero e to the power minus t by tau
over that means the value of r will start
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from r zero and when the t is very large eventually
the value of r will going to be equal to zero
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so the value of r which is anisotropy will
start from r zero and will decay to zero and
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that decay will be exponential in nature you
see here i have this e to the power t over
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here e to the power so if i now plot r t as
a function of t so this will be r zero and
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it will be something like this right so this
will be something like this and the corresponding
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time constant right so you feet it with e
to the power minus t by tau over so you will
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get this orientational relaxation time ok
now if if you are interested on this anisotropy
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that how this is the molecule actually reorient
then you can do this kind of prescription
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and you can get this tau over value
now as i told you that you can earlier during
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our discussion on t c s p c setup and the
fluorescence up conversion up conversion setup
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i told you that this in both the setup there
are two few two components optical elements
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was there and i told you that i will explain
those optical elements later on and now this
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is the time to discuss about that ok now consider
that you have a setup and either t c s p c
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or up conversion and you want to measure the
lifetime right so what will be your way
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to do that either you will excite vertically
and emission is collected vertically or you
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excite vertically emission is horizontal or
you excite vertically and the emission is
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unpolarised right you dont select you dont
put the analyser so let me write this
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so for lifetime measurement
i could have three different situation like
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here i want to measure tau f not tau over
right please note that so the three cases
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case one i have vertical excitation and vertical
emission this is nothing but i parallel t
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i already showed you right i can have vertical
excitation horizontal
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emission case three i can have vertical excitation
and unpolarised emission in the first case
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you already have seen what is the first case
this is nothing but i parallel t so i can
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simply write the expression of i parallel
t over here for the case one for the second
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one this is nothing but i perpendicular t
so let me write the expression and for the
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third one i have not derived but it can be
derived without any problem and let me also
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write that expression right so here this is
nothing but i parallel t which is equal to
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i zero by three e to the power minus t by
tau f to one plus two r zero e to the power
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minus t by tau over this one is nothing but
i perpendicular t so this is i zero by three
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e to the power minus t by tau f to one minus
r zero e to the power minus t by tau over
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and this is for the unpolarised so i dont
try it i parallel or perpendicular i just
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write i t this will going to be equal to some
constant will be there then i zero by three
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e to the power minus t by tau f into two plus
r zero e to the power minus t by tau over
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now you see in all these three cases right
this three cases are possible case so in all
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this three cases what you can see is that
you wanted to measure the tau f but you land
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up with the measurement of such kind of complex
see here here tau f is present here you have
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this tau f here you have this tau over here
you have this tau f you have the tau over
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and here also this tau f and this tau over
but you wanted to measure only the tau f end
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up with measuring a mixture of tau f and tau
over and you will never get the fluorescence
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lifetime that is defined as the tau f but
what is the prescription now for the measurement
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of fluorescence lifetime allow so that will
going to discuss now
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now let me consider that these two
this is my i perpendicular and this is my
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i parallel right let i am consider it like
that way right here and better i write the
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electric field instead of intensity which
is the square of this intensity electric field
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so i write this e parallel equal to square
root of i parallel and e perpendicular is
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equal to square root of i perpendicular so
right now what you are doing in in other cases
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either i was measuring this i parallel t or
i was measuring this i perpendicular t or
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i was measuring everything right that was
the thing what i said but is it possible that
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i measure something between i parallel and
i perpendicular why not because you remember
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that when i used my lets say this is my sample
cell and you are exciting your sample like
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this way and emission is getting collected
at the right angle and you had your polariser
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over here
now when the polariser is oriented like this
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way i assume that this is i parallel when
i rotate this polariser like this way this
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is i perpendicular right this is i parallel
because the this is like this and this is
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i perpendicular and if i rotate the polariser
like this way then it will make some angle
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theta is not parallel not perpendicular so
it is will make some angle lets say theta
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from the vertical excitation right from the
parallel component so from here if if this
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is this axes then this may be theta right
thats what i i will going to draw over here
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so now if you collect the emission then this
emission will be collected at that particular
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angle theta so thats what i am going to write
lets assume that you put your that analyser
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in such a way that the light having such kind
of polarisation is only passing through right
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this analyser so let me extend it over here
so this is the position of my analyser this
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analyser orientation
and i said that this is oriented at an angle
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theta from the parallel component
now i can simply do some trigonometry over
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here so the component of e parallel which
will projected on this angle theta i can calculate
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it i will just write draw a normal over here
right if this is i e parallel then this will
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be e parallel cos theta so this will be my
e parallel cosine theta and if this is e perpendicular
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then i will they are normal so then this part
so from here to here this part will be e perpendicular
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sin theta so let me and this is right now
my life is simple let me write it so i just
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simply write over here if it is e parallel
sin theta that means the intensity is the
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square of that when it e parallel cos theta
then its intensity is just square of that
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so now i can write it like this way you see
i is a function of t right no i is not only
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function of t it is also function of theta
now so i will so here i is a function of theta
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and t is equal to i parallel t cosine square
theta plus i perpendicular t sin square theta
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so now i will put the expression for this
i parallel t and i perpendicular t into it
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and then i will try to simplify this equation
and lets see what happen right so now i will
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put this expression first one is i zero by
three e to the power minus t by tau f to one
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plus two r zero e to the power minus t by
tau over multiplied by cosine square theta
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and i zero by three e to the power minus t
by tau f here one minus r zero e to the power
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minus t by tau over to sin square theta this
will be equal to i zero by three e to the
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power minus t by tau f i will take this common
and then i will write it once again so this
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will be cosine square theta plus twice r zero
e to the power minus t by tau over cos square
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theta plus sin square theta and then it will
come as a minus r zero e to the power minus
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t by tau over and sin square theta so i can
rewrite this as i zero by three e to the power
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minus this is important very important thing
i am doing now so into here you see cos square
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theta plus sin square theta so i have one
plus and let me take this here r zero this
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part right so then let me write this r zero
e to the power minus t by tau over and then
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this will be two cosine square theta minus
sin square theta
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so two cosine square theta minus sin square
theta this means ok i am almost done so i
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zero by e to the power minus t by tau f i
will keep this part as it is r zero e to the
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power minus t by tau over and this can be
written as three cosine square theta minus
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one i am done now you see for a certain value
of theta this three cosine square theta minus
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one this value could be zero right if so what
will happen this is zero ok let me write for
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three cosine square theta equal to one right
i can write here cos theta is equal to root
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over one upon three then when the value of
three cos square theta is equal to one then
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this will be one minus one this will be one
minus one then this quantity will going to
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be zero then this whole quantity will be zero
so what i will get i will get this equal to
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i zero by three e to the power minus t by
tau f see for a particular value of theta
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when cos theta equal to root over one by three
that means theta equal to fifty four point
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seven five degree when theta equal to fifty
four point seven five degree that means if
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the analyser is placed at an angle fifty four
point seven five degree from from the direction
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of excitation or from the i parallel like
this this is i parallel this is i perpendicular
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like this
so in that case what you will going to get
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is that these term these this part this part
vanish this part vanishes and what you will
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left over with this i zero by three e to
the power minus t by tau f that means you
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will be able to measure the lifetime otherwise
not you can do another thing if you just simply
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take i parallel plus two i perpendicular let
me write here i suppose you have collected
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the fluorescence transient or fluorescence
decay i parallel and i perpendicular and then
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you do this one i parallel plus twice i perpendicular
just put all these values i mean that expression
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of i parallel and i perpendicular and simplify
it you will also get that tau f term is there
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in this case you have to measure two times
at in the magic and if you put this angle
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during measurement then you have to measure
only single time result is same right and
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these angle is a special angle and this angle
is known as magic angle
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so now you understand the role of those things
i have drawn in t c s p c setup and up
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conversion setup those are the polariser or
analyser you have to set the analyser at the
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magic angle then only then only you will get
the correct lifetime now here one point i
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should now tell you that suppose the lifetime
but sometime you will see that people are
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measuring without an a that magic angle polarization
that means everything is wrong or under
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under some condition that the lifetime measure
without analyser without magic angle polarization
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is ok it is ok only when the lifetime is very
large compared to the rotational time right
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so in case the tau over is much smaller than
tau f much much smaller than tau f then the
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decay of this part will not going to affect
that way that tau f but if you are interested
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in the timescale which is comparable with
the tau over then your decay will be destroyed
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right
for example suppose you have fluorophores
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lifetime is one microsecond so you will going
to measure this decay from zero micro second
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to lets say lets say we are going to measure
like five microsecond like if it is one microsecond
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but the rotation of this fluorophore right
is typically is in the order of tens of picosecond
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fifty picosecond sixty picosecond hundred
picosecond so the isotropic situation will
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be achieved in the solution within lets say
two hundred picosecond but two hundred picosecond
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is nothing in front of microsecond so the
error will be very very small so for those
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kind of long lifetime sample you really dont
have to worry about this magic angle polarisation
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for the measurement of the fluorescence tangent
but if you are in the domain of few tens of
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picosecond it is it is must right otherwise
all your data will be wrong ok
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let us continue and we could see a nice equation
now what i will do now is the following i
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will integrate it what i will integrate integration
zero to infinity i parallel t d t i will do
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that i will do this integration let me write
integration zero to infinity thats the form
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of i parallel t i zero by three e to the power
minus t by tau f right see this integration
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zero to infinity i parallel t d t this is
nothing but the steady state i parallel so
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i just termed it as i parallel steady state
right so it is integration for all the time
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this i parallel t is changing with time but
this it just i have integrate it so if you
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evaluate this integral what you will get you
will get such kind of form let me write without
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evaluating which you can check later on if
you want otherwise take my word i zero by
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three tau f e to the power minus t by tau
f that limit zero infinity minus two i zero
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r zero by three into one divided by one by
this is little lengthy plus one over tau over
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into e to the power minus one over tau f plus
one over tau over and here limit is again
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zero to infinity
so if you put this value this limit so what
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will get is i zero by three into tau f plus
twice r zero divided by one over tau f plus
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one over tau over so you will get i parallel
steady state as this now similarly i can also
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calculate this i perpendicular steady state
so this i perpendicular steady state similarly
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right similarly we will do is you will get
as i zero by three to tau f minus r zero divided
27:46.520 --> 27:59.450
by one over tau f plus one over tau over and
r which is steady state you can measure i
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already told you how to measure it this r
right which is nothing but i parallel steady
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state minus i perpendicular steady state divided
by i parallel steady state plus to i perpendicular
28:15.920 --> 28:24.730
steady state and if you plug in the expression
for i parallel and i perpendicular in this
28:24.730 --> 28:34.490
equation and simplify you will be surprised
to see that after simplification this result
28:34.490 --> 28:44.669
will be very simple one plus tau f by tau
over now i can rearrange it and i can write
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it as r zero by r s s is equal to one plus
tau f by tau over very important equation
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this equation is known as perrin equation
and we will stop here today and we will continue
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our discussion on the next class
thank you