L20 : Young's Single slit Experiment - Wave Optics, Physics, Class 12 Video Lecture

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20 so what was this phenomenon of
diffraction as I told interference and
diffraction are very closely related
phenomenon so what did young do in
Young's double-slit experiment he
replaced the two slits by a single slit
now what was observed everything else
remains the same there was one source s
which was a normal ordinary source and
then instead of two slits there was just
one slit so the light passed through
that one slit and a pattern was observed
on the screen but the unique thing was
that this time the pattern was quite
different so what was the difference in
the pattern this time the pattern which
was obtained had a central maximum which
was very wide as compared to the
interference pattern and this was
followed by alternate dark and light
banks and the intensity on both sides
keep decreasing considering that means
the central bank the central maximum was
very wide and the corresponding
secondary minima maxima and minima they
reduced in intensity to a large extent
so this broad central bright not only
dark and bright bands on both sides and
decrease in intensity this kind of
surprised people that means that there
is not only interference but there is
some other phenomena as well which also
takes place and then similar conditions
and circumstances so what was that
phenomenon that phenomenon was nothing
but diffraction so what is diffraction
of light how do we different how do we
define diffraction of light it is
nothing but the phenomenon of spreading
of light when it face
and obstacle that is how we define
diffraction so in this case the closed
region acts as the obstacle so these are
the regions which act as the obstacles
here so in this case so what was the
difference in the pattern the first
point was broad broad central bright
band was formed then what else alternate
dark and bright bands on either sides
and decreasing intensity on both sides
intensity was decreasing on either sides
right so now let us look at the the
reason why this kind of pattern was
obtained and why this experiment was
used in order to find out this new
phenomenon of diffraction well let us
look at the diffraction pattern little
more detail so this is our diffraction
takes place you see this is the incoming
wave it passes through a single slit so
this expenditure is often known as
Young's single slit experiment so
through this single slit these kind of
diffraction pattern was obtained on the
screen so if you look at the central
portion it is white white if you look at
the next bright bands it is thinner and
the next one even more thinner so that
the decrease in intensity is very
remarkable right so now let us try to
understand
I mean why these kind of Maxima take em
invite this kind of pattern was obtained
when a single slit was used well the
basic idea in this case would be I mean
how do we explain the formation of
maxima and minima in that the basic idea
would be to divide this split into
smaller parts
and then we will add the contribution of
each part I mean that is what
superposition of waves tell us right if
you have some ten waves then the
contribution at that from each of them
together add up to form the resultant
wave so similarly what we will do here
we have this slit so we are dividing
this split into smaller parts if you see
here we have marked two points and one
and M two so that we can divide the
slits into smaller parts and then we can
add up the contribution from each part
to find out the intensity at any
particular point so the incoming wave
front is incident parallel to the plane
of the slit width so it shows that the
sources are in phase so we are assuming
that the sources from I mean the source
which is giving this light they are in
phase because of which this incoming
wave fronts are parallel to each other
so let us look at let us suppose there
is some arbitrary point p on the screen
where we want to find out that intensity
so for that we have to calculate the
path difference because path difference
is the starting point to start any
discussion on the fringe pattern of
thing because path difference is
something which decides how the
interference is taking place so here
path difference will be this is my slit
Ln is the slit so the path difference is
nothing but NP minus L P so what is NP
minus L P sub C we are assuming the
rails to be parallel because we are
assuming that they are in phase with
each other
and this point P is little fine so what
would be the difference the difference
is this much that is M sure so we can
say that if this is perpendicular which
is drawn here so MD minus L P is nothing
but because this LP will be equal to Q P
so the extra distances in true and what
is this NQ if you look at M Q from this
triangle then you can say that if this
angle is Theta this entire angle is
19 so this angle will be 90 minus theta
and this is 90 so this small angle will
become theta right
so looking at this triangle you can say
that NQ is nothing but this distance is
a a is nothing but width of the slit so
if you look at the strangle sine theta
is equal to perpendicular so
perpendicular is m2 divided by
hypotenuse that is a so we can say NQ is
equal to a sine theta so here we can
write it as in sine theta like now now
this is the path difference between the
points L and n now there was a scientist
named Fresnel who'd studied diffraction
in detail he was the one who was
involved in the all the rigorous
calculations related to diffraction to
find out the pattern which was obtained
on the spring so plus mu did his
calculations by dividing this slit into
smaller parts as you can see here L M 1
M 1 M 2 M 2 N and so on and then he
summed up the contributions from all the
elements in the slit so what did he find
he find that at the central point that
means at this point this is my central
maximum them in the pattern is somewhat
like this so it was found that at the
central point this theta because right
now we are assuming the point P to be
some point somewhere here right so now
when we are talking about the central
fringe we are talking about the point
which is lying linear to this center of
the fringe thank center of the slit so
that means for the central maximum theta
is equal to 0 so at central point theta
is equal to 0 now if theta is equal to 0
that implies that the path difference
will be equal to 0 right now
if the path difference is equal to zero
that means all the parts into which we
have divided it all those parts will
contribute completely therefore the
intensity will be maximum because that
is what we have seen before also
whenever the path difference is zero
that means whenever we want to find out
the intensity at such a point which is
at equal distances from all the sources
then the intensity is maximum because
all the points will contribute their
fullest so in this case also no matter
how many small parts you have divided
this Ln input but the point this centre
point is equal is at equidistant from
all of depth therefore all of them will
contribute their maximum and therefore
the intensity will be maximum so this
implies that all parts of slit will
contribute this implies that the
intensity would be maximum right
therefore we see that central maximum
occurs central maximum occurs at theta
is equal to zero so this is about the
central maximum because in the
diffraction pattern we will talk about
three things one is central maximum one
is secondary maximum secondary maximum
means the remaining Maxima which are
there because the central maxima is
little more significant here because it
is very white it has maximum intensity
right and all other Maxima will be known
as secondary maxima and be able to also
talk about the minimum correct so in
this case we talked about the central
maxima so you understood why central
maxima is occurs at theta is equal to
zero because the path difference comes
out to be in this case as a sine theta
at the central point theta is 0
therefore path difference is zero
therefore all the parts will contribute
and as a result the ink
city is maximum now let us talk about
the secondary maximum so how is it that
the secondary Maxima are little less and
are with little less intensity okay so
let us look at that part now Fresnel
with his calculations he found that the
secondary Maxima occurred at Sheeta is
equal to n plus 1 by 2 lambda by E so
this is what was found by Pleasants
calculation that all the secondary
maximum were observed for these values
of theta n plus 1 by 2 lambda by a now
let us consider the first value of n
that is n is it 1 T 1 so when n is equal
to 1 theta will come out to be 3 lambda
divided by 2 a now what is this 3 lambda
divided by 2 a so in this case Fresnel
observed that minima were found for
values of theta satisfying n lambda by a
where n was nothing but plus minus 1 2
and so on so this was the condition to
obtain minima on the diffraction pattern
suppressed and told that
stressful told that at under this
condition that is theta is equal to
n-lambda by a the minima were observed
on the diffraction pattern so let us see
how this minima was obtained at these
specific angles let us consider n is
equal to 1 so for M is equal to 1 theta
is equal to lambda by a right
now let us suppose that we have divided
this clique into small parts now if you
look at this part M 1 so for every M 1
in the portion L n there exists another
M 2 in M n that means this M 1 and these
empty these two points are such that the
contribution from M 1 and M 2 will
cancel out each other because both of
them are on the opposite side so they
will be they will have equal and
opposite contribution and as a result
contribution and F of M 1 and M 2 will
be out of phase and therefore cancel out
therefore the net contribution from all
such elements the contribution from all
elements between l and m and up will
cancel out with all the elements between
m and n as a result the net intensity
will be equal to 0 at theta is equal to
lambda by a so we can say that
contribution of all elements in L n for
example M 1 will cancel out the
contribution from all elements in MN for
example M 2 therefore net intensity will
be equal to 0 at theta is equal to
lambda by a so we can say that minima
occur at theta is equal to n
lambda by a that means minima will occur
when the path difference is equal to n
lambda by when the path difference is
equal to n lambda we get this from this
because this is a theta is equal to n
lambda so this a theta can be written as
K sine theta is equal to n lambda so a
theta is nothing but a sine theta
because theta is very small in this case
so this is equal to n lambda so the
condition for minima in diffraction
pattern is a sine theta is equal to n
lambda where a is the width of the slit
lambda is the wavelength of light which
is used now let us look at the
diffraction patterns secondary maximum
so till now we talked about the central
maxima which occurs have kita is equal
to 0 then we talked about the minimum
for which the condition is a sine theta
is equal to n lambda now let us consider
the secondary maximum for secondary
maxima the condition was found to be
theta is equal to n plus 1 by 2 lambda
by ay so here also let us consider n is
equal to 1 so what do we get theta will
be equal to 3 lambda divided by 2 a now
what does this 3 lambda by 2 a 3 lambda
by 2 a is nothing but it is a mean it
lies midway between two dark fringes if
you look at the pattern the pattern is
somewhat like this right so this
corresponds to n is equal to 0 so this
is my first minima this is my second
minima so the first minima corresponds
to n lambda which is 1 lambda so this
first minima corresponds to 1 lambda
second minima corresponds to 2 lambda
third minimum corresponds to 3 lambda so
what is this
this corresponds to 1.5 lambda by a
right so this basically corresponds to
1.5 lambda that means this lies between
two dark fringes so it is talking about
this region right so this is midway
between two dark fringes you since it is
midway between two dark fringes we can
assume that it must be bright but any
less we will prove it now let us suppose
that we divide this slick into three
parts
this entire slit we have divided into
three parts so what happens now if I
consider this is my slip I divided it
into three parts so let us suppose I
consider first to third as one portion
so what would be the path difference of
this part the first to third of the slit
this is 2/3 and this is remaining 1/3
right this is my slip mnscu thank so now
first we consider consider first 2/3 of
the slit so what would be the path
difference so the path difference would
be equal to 2 by 3 ei into theta 2/3 of
the slip so 2/3 of a that is the width
of the slit for now multiplied by theta
and what is Theta theta is 3 lambda by 2
a so this will be 2 by 3 ei into 3
lambda by 2 a so this comes out to be
lambda so now this lambda so that means
this is
the path difference for the first two
third portion of them sick now this
lambda can be divided into two halves
with lambda by two path difference this
wavelength lambda can be divided into
two halves so now if you divide this
part into two halves with lambda by two
wavelength the contribution from each
half will cancel out so that means the
contribution from this upper 2/3 is
cancelling out amongst itself so this
2/3 portion also internally we divided
it into lambda by two this lambda was
divided into lambda by 2 and lambda by
two so this cancels out each other as a
result only 1/3 is remaining so that
means only this 1/3 contributes to the
intensity right so that means only this
portion contributes to the intensity and
as a result of this the intensity in
this case is not zero that means there
is some intensity but the intensity is
greatly reduced because out of this
whole slit only one-third of the slip is
contributing for the intensity in case
of central maxima all portions of the
slit were contributing to the intensity
so the intensity was maximum but in this
case only 1/3 is contributing to the
intensity therefore even though the
intensity is not zero at this portion
right now we are talking about this
point so it says that here the intensity
is not zero so we have the right two
things intensity is not zero but
intensity is reduced correct okay so now
here we had considered n is equal to 1
similarly if you consider n is equal to
2 then what happens to theta in that is
theta becomes 5 by 2 lambda by 8
right so when theta becomes 5 by 2
lambda by a if you follow similar
calculation in that case you divide the
slit into five equal parts and again you
will see that only 1/5
of the part with Lenin and contribute in
intensity similarly when you put n is
equal to three Peter becomes equal to
seven lambda by 2 a so in this case 1/7
of the portion remains because here you
divide in the first case you divide it
into five different parts you consider
the first four fifth of the part you
will see that it comes out to the
integral multiple which will again
cancel out amongst each other right
because when you consider so this is how
we see that as the value of n increases
that means as we go away from the
central fringe the contribution of the
slit to the intensity decreases and as a
result the intensity keeps decreasing
right so there could with increase in
value of n the intensity weekends so
this is how we were able to explain the
diffraction pattern that is the
coordination of central maximum
formation of secondary Maxima whose
intensity keeps decreasing as we move
away from the center and the formation
of minimum however the exact
calculations which derive to these
values by then by Fresnel by dividing
this slit into smaller parts and then
taking into consideration the
contribution of each part and then
summing them up so we will not discuss
all those things in that details now for
now you should know why these kind of
pattern or observer right so we have
already learned that so I hope now
diffraction is also clear view so it is
very important now thank you please
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