Youngs double slit experiment - Wave Optics Video Lecture - Class 12

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with part 15 so here comes the famous
Young's double-slit experiment now why
did Thomas Young perform this experiment
what was his purpose doing this
experiment Thomas Young also wanted to
do find out that how can he prove the
wave nature of night.they because there
has to be some experiment performed to
prove the truth of a statement he was I
mean however hydrogen had given his
explanation but ITINs explenation was
all based or mostly on his observation
based on his observation he gave this
hypothesis that I think light is off
wave nature and on a pond based upon his
hypothesis he gave this concept of
wavefront and explained the phenomenon
of reflection and refraction using the
wavefront concept but what Jung did he
perform an experiment and he showed that
something the result of experiments
showed something which was not expected
if light would have been of particle
nature right so that was Young's motive
to prove the white wave nature of light
so what did Jung do
Jung performed an experiment to which
explained the phenomenon of interference
for today he made an arrangement so that
coherent setup can be made to maintain a
phase relationship between two sources
he wanted to coherent sources of light
normally as I told you whatever ordinary
light we see around us they are all
incoherent they are not coherent right
so within coherent sources you will have
light waves with different amplitudes
different phase relationships so
everything will be random if you use two
coherent sources right so Jung wanted to
coherent sources
now for example of coherent source was
laser so what he did instead of using a
they say because that was little more
humor some who use two lasers and then
perform experiment so he he found out an
alternative way what he did he took an
ordinary source of light for example a
light bulb and he made this light he
passed through a very small slit I
mentioned it before also when you made
light pass through a very small slit
which is comparable to the wavelength of
light you can observe effects you can
observe this phenomenon so what happens
when light passes through the small slit
she made another arrangement where he
had two small slits he called these two
small slits as s1 and s2 these are the
two small slits and this source was s so
these sources is so which were placed
quite close to each other they were
separated by some distance B what did he
observe and then he placed a screen in
front of them so this was his
arrangement he took a bulb made the life
of the bulb pass through a small slip
again that light which came out was made
to pass through two small slits which
were come quite close by to each other
and then a screen was placed in front of
it what did you observe he observed that
alternate dark and bright bands were
formed on the screen you have
alternatively dark and light bands on
the screen so why did this pattern come
up on the screen that means some
phenomenon would have taken place so he
thought why this kind of result was seen
what he did he did a he took another
setup he took two light bulbs two non
coherent sources of light and place the
screen in front of them what did you
observe nothing on the screen observed
light that's it
there was no fringe pattern there was no
alternate dark and light banks so what
he comes from then you concluded that
when 2 coherent sources are used some
different phenomenon take place which
does not happen within coherent sources
so what was that special phenomenon
which was
taking it that special phenomenon was
nothing but the phenomenon of
interference or the phenomenon of
overlapping of the waves from the two
sources so in this case what happened
this source s illuminated the sources or
the pinholes s 1 and s 2 therefore light
from s 1 and s 2 begin coherent why the
light became coherent because now since
X 1 and X 2 both are getting their light
from the same source s so if there is
any change in the phase that change will
reflect in both s 1 and s 2 for example
let us suppose you have two taps in your
house and the water in both the taps are
coming from the same source right so if
you mix something in that water suppose
if you mix mud in that water in the
source then when you open these tabs the
water from both the taps will be muddy
right
since so whatever changes you make in
the source you will get you will find
the same changes from all the tabs so
the same thing happened here here also
if there was any phase change or any
change in frequency at the source the
light waves coming from s 1 and s 2 will
reflect the same changes therefore s 1
and s 2 will always remain in phase
there will be no change in the phase
difference between s 1 and s 2 so that
is how in this arrangement he was able
to make two coherent sources s 1 and s 2
right now on the screen you observed
alternate - and the light brands how did
this happen let us look at it now when
this light source and this ordinary
source of light was made to pass through
the small slit these kind of wave fronts
were formed by so because normally if
you if you make this light pass Norman
what will happen light waves will come
like this from an infinite source so
your wave front will be straight but
since you have put this obstruction on
both the way both the sides so the light
waves see if the light waves were coming
coming from a distant source like this
what will
how will your wavefront look like your
wavefront will look like this but when
the light waves are coming like this but
you have put some obstruction here and
you are forcing the light waves to pass
only through this in that case what will
happen in that case these wave front
which will come out from here they will
try to spread in this way but since this
side it is locked so no light is
entering from D side so these light
which is entering from here it will
spread in this fashion as a result the
shape of your wave friend would be like
this right so that means when you put
some obstacle because of that obstacle
your light waves take this shape of
semicircles so as I result you get your
light waves in this fashion now these
light waves this is again the top you
remember the example of a stone being
thrown into water your top view looked
like this so this small slit behaves
similar to a point source so the point
source will have spherical wavefronts
now here the red bands represent your
red crests and the yellow ones represent
the trucks so if you look at the side
view it would be somewhat like this take
the side view of the wave would be
somewhere like this right so if this
this is like the top view of your water
so now again when this wave front go
passes through these two small slits
again the same phenomenon take place
lemon from these two slits the waves
will arise so these are the wave friends
of the two slits now if you look at it
very carefully you see that there are
points where red red overlap for example
these are the points where red red
overlap similarly there are points where
yellow yellow overlap right so these are
the points where red red and yellow
yellow overlaps yellow yellow red red
yellow yellow red
regnant similarly there are points where
red yellow overlaps red yellow red
yellow red yellow
so where is constructive interference
taking place and where is destructive
interference taking things the red is
your crest and yellow is your trap
so if red and yellow is combining that
means it is destructive interference
similarly if yellow and the red is
combining that means it is destructive
interference similarly if red and red is
combining it is constructive
interference if yellow and yellow is
combining it is constructive
interference again right so here if you
see this points represent constructive
interference because they are red red
yellow yellow red red yellow yellow
combines so that means these red arrows
these three red arrows represent
constructive interference so what will
happen in case of constructive
interference the resulting intensity
will be high that is why these
constructive interference gives rise to
bright bands so these regions are bright
whereas these portions where radial low
red yellow red yellow overlaps gives
rise to intensities which are lessened
so that intensity is minimum here so
these are the dark bands as a design of
these constructive and destructive
interference you get this alternate dark
and light bands pattern on the screen
right so I hope I am able to explaining
what was lungs double slit experiment
what he did he made an arrangement of
coherent sources and he found that light
waves originating from coherent sources
they overlap with each other both
constructively and destructively and
gives rise to a beautiful alternate
right and that pattern on the screen so
this phenomenon of overlapping of light
waves giving rise to regions of higher
amplitude and the regions of lower
amplitude is
known as interference so interference is
a property which is related to the wave
nature of light so interference is a
property of light obviously light waves
have the property of interference but if
but will you be able to explain this
property of interference if you consider
light wave as particle if you consider
that light waves functions of huge
number of particles will you be able to
explain this phenomenon because if if
they are particles there is no crest no
trap at all and there is no overlapping
there is no superposition of waves so
there is no interference so this
phenomenon of interference clearly told
that light is of wave nature so Young's
double-slit experiment was the one which
stamped the statement of heigen because
hyejin's principle was not very easy to
buy except for people so Young's
double-slit experiment gave a full proof
that light is of wave nature so we
understood what young then in his double
slit experiment and what he wanted him
to please now let us try to concentrate
on the fringe pattern which was always
obtained on the screen we saw that all
turning dark and light bands were formed
on this thing let us now try to analyze
the position of the dark band and the
light map where exactly the dark bands
were formed where exactly the bright
bands perform so let us analyze those
that part of it so in order to analyze
that first we we have to start from
there those bands were formed as a
result of interference correct
independence was formed as a result of
overlap right overlap was what either
constructive overlap or destructive
overlap now constructive overlap or
destructive overlap depends on what it
depends on path difference so if your
path difference is an integral multiple
of lambda it will be constructive
interference if your path difference is
non integral multiple of lambda and
there will be destructive overlap so
path difference is your starting point
so in order to find
anything any information about the
pattern which you get on the screen
first you have to find out the path
difference so they have to start with
path difference let us start try to
calculate the path difference in this
case so in this case what where are my
sources s-one and s-two are the sources
and let us suppose I consider any point
P then I want to calculate the intensity
as I told you we have to choose one
point where I want to find out things so
let us suppose P is that point so what
will be the path difference path
difference is equal to s to P minus s 1
P this will be the path difference path
from s to minus path from s 1 so now
looking at this diagram this is a right
angle triangle similarly this is also a
right angle random so looking at this
triangle what we can say from Pythagoras
theorem we can say using pythagoras
theorem we can say that the hypotenuse
that is s 1 p square is equal to the
base square that is capital D Square
where D is the distance of the screen
from these two sources plus the
perpendicular square what is the
perpendicular distance what will be this
distance if this point T is located at
distance X from the origin so this
distance will be X minus X minus this
part what is this part this part would
be equal to this part which is D by 2
this is D by 2 so this will be X minus D
by 2 so this will be X minus D by 2
whole square similarly we can apply
pythagoras theorem in this bigger right
triangle so we can say s 2 P whole
square is equal to D square
this entire length now so what will be
this this will be nothing but X plus D
biking because this is again D Bible
right so this will be X plus D by 2
whole square now if we call this as
equation 1 we call this as equation 2 so
we can say that 1 minus 2 gives us s 1 T
whole square minus not 1 minus 2 because
we want to find s to V minus s 1 T so
let us subtract 2 minus 1 so we get s 2
T whole square minus s 1 P whole square
so this gives d squared plus X plus D by
2 whole square minus d square minus X
minus D by 2 whole square so d square d
square will get cancelled so we get X
square plus d square by 4 plus X D minus
X square minus d square by 4 plus X D so
X square X Square D square by 4 d square
by 4 gets cancelled so we get 2 X D so
what do we get s 2 P Square minus s 1 e
square is equal to 2 X D this is a
square minus B Square so it can be
written as a plus B minor into a minus B
so s 2 P minus s 1 B into s 2 P plus s 1
P is equal to 2 X D right now what is s
2 P minus s 1 T this is equal to 2 X D
divided by s 2
p+ h 5p right now if we assume that d
that is the separate small D is the
separation between these two so if we
assume that this D is very small when
compared to this distance that means if
the screen is placed quite far away from
the arrangement and if the two slits are
very nearby in that if we assume this
and we also assume that if X the point P
where it is located that X is very very
small when compared to capital D that
means basically we are assuming that if
the screen is placed quite far away from
the slit arrangement in that case we can
say that s to P plus s 1 P what is s to
D s to P and s 1 P is somewhat like this
right so then if this is very far away
we can say that H 2 P and s 1 P are
almost equal and they are almost equal
to D because if this D is very far away
this distance becomes almost equal to D
so we can write it as 2 D right it
understand it in this way it is suppose
these are the two points and this is the
screen so if the screen is here it is
like this now let us suppose if I take
this thing far away here so how will
this point look like it will look like
this right so this is your D here also
this is your D so when you increase D
here the D was next
therefore this distance and this
distance looked different but when you
have increased your D so much this
inclination has decreased and therefore
this distance appears to be same as this
deal right you understood okay so we can
write s 2 P plus s 1 P as 2 D therefore
the path difference s to P minus s 1 P
is equal to 2 X D if I'd inviting
capital B this is equal to X
my D therefore what will be conclude we
conclude that the path difference in
this case is X D divided by capital D by
small D is the distance between the two
slits capital e is the distance of the
screen from the slits and X is the
position of the point where we want to
find out the intensity right so we have
calculated the path difference now now
with the help of this path difference we
can find out where the intensity where
constructive interference will take
place and where destructive interference
will take this so what was the finding
of Young's double slit experiment it
found that light from holyland sources
reduced alternate dark and bright bands
on the screen placed some distance away
from it right thank you please visit
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