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