L19 : Young's double split: Numerical - Wave Optics, Physics, Class 12 Video Lecture

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the problem says in a Young's
double-slit experiment the slips are
separated by 0.28 millimeters and the
screen is placed 1.4 meters away so what
are the values given this tool split
between the slits is 0.2 H millimeters
distance of the screen is capital D that
is 1.4 meters distance between the
central bright fringe and the fourth
bright fringe is measured to be one
point two eight centimeters let us
suppose if the fringe pattern is
somewhat like this it says that later
suppose this is the central bright
fringe for n is equal to zero so when it
is talking about the fourth bright
fringe that means n is equal to 1 n is
equal to 2 n is equal to 3 and n is
equal to 4 so that means the distance
between the central bright fringe and
the 4 bright fringe is found to be 1
point 2 centimeters so let us try to
calculate the position of 4 bright
fringe so what would be the position of
the fourth right fringe that is X 4
which is equal to n lambda capital T
divided by small D here n is 4 what is
lambda we don't know so lambda into what
is capital D 1.4 divided by small D
which is 0.28 into 10 to the power minus
3 right now what is the position of the
fourth band fourth bright fringe it is
one point 2 centimeter so X 4 is nothing
but 1 point 2 centimeter which is 1
point 2 into 10 to the power minus 2
meters therefore we can say lambda is
equal to so we can say lambda is
when to 0.28 into 10 to the power minus
3 into 1 point 2 into 10 to the power
minus 2 divided by 4 into 1 point 4 so
this comes out to be 600 nanometer
therefore the wavelength of light used
here is 600 nanometer let us look at the
next problem it says in Young's
double-slit experiment using
monochromatic light of wavelength lambda
the intensity of light at a point on the
screen where path difference is lambda
is scale units that means the where path
difference is lambda you have an
intensity of K what is the intensity of
light atom at a point where path
difference is lambda by 3 so when you
have a path a path difference of lambda
corresponds to a phase difference of 2
pi so a path difference of lambda by 3
will correspond to a phase difference of
2 pi by 3 therefore the phase
corresponding phase differences 2 pi by
3 so from coherent and incoherent
addition of waves which we studied
before so from coherent and incoherent
addition of waves what did we find we
found that the intensity becomes equal
to 4 I not coz square Phi by 2 where Phi
is the phase difference right so from
this we can say I is equal to 4 I naught
cos square PI by 3 so this cos Phi by 3
is 1 by 2 so cos square PI by 3 will be
1 by 4 this comes out to be I naught now
what does I naught I naught is nothing
but now what is I naught I naught is
nothing but the individual intensity of
the wave right so individual intensity
of the initial
be correct so now we have to calculate
what is this I got now in the problem it
is given that when a light of wavelength
lambda was used the path the intensity
was K so that means let us calculate it
for path difference for path difference
lambda what would be the phase
difference the phase difference would be
2pi
therefore what could be the intensity in
this case intensity I dash will be equal
to 4 I not coz square Phi by 2 that is 2
pi by 2 so this comes out to be 4 I
naught therefore i dash is equal to 4/9
I naught now what is given in the
problem in the problem it says that this
I dash is equal to K units right so
corresponding to a path difference of
lambda the intensity is given as K
therefore we can say that I naught is
equal to K by 4 now using this this was
my equation 1 now put this value of I
naught in equation 1 so what do we get
we get I is equal to K by whole lunes so
this is my answer right see here I have
directly use the result which we have
obtained from coherent and incoherent
addition of waves so you can also make
use of these results let us look at the
next problem
which says that a beam of light
consisting of two wavelengths 650
nanometers and 520 nanometers is used to
obtain interference fringes in a Young's
double-slit experiment find the distance
of the third bright fringe on the screen
from the central maximum for wavelength
650 nanometers so first let us look at
the first factor C's lambda is equal to
650 nanometers which is 650 into 10 to
the power minus 9 meters let me call it
as lambda 1 we have to calculate the
distance of third bright fringe that
means n is equal to 3 so X n is equal to
n lambda capital D by small D what is n
tap 3 into 6 15 into 10 to the power
minus 9 into capital D by small D so
this comes out to be 1 9 5 0 into 10 to
the power minus 9 into capital D by
small D meters so this is the position
of the third bright fringe the next part
asks what is the least distance from the
central maximum where the bright fringes
due to both the wavelengths coincide
okay so let us suppose that for the
second part let us assume that the
innate bright fringe in it right fringe
due to lambda 1 coincides with n minus 1
net bright fringe due to lambda 2 they
have two different wavelengths so we are
assuming that n it right fringe of
lambda 1 coincides with n minus 1 right
fringe due to lambda 2 or you can assume
it the other way round also you can
assume that
emeth bright fringe it into lambda 1
coincides with the n plus 1 bright
fringe due to lambda 2 so in that case
what happens the position of X
due to lambda 1 will be equal to lamp n
lambda capital T by small thing this
lambda 1 and what is X n plus 1 that
will be n plus 1 into lambda 2 capital D
by small D right so now we will equate
both therefore n lambda 1 capital D by
small D is equal to n plus 1 lambda 2
capital D by small D so this D DS will
cancel so we get n lambda 1 is equal to
n lambda 2 plus lambda 2 so from this we
can calculate the value of n so this is
n into lambda 1 minus lambda 2 is equal
to lambda 2 or n is equal to lambda 2
divided by lambda 1 minus lambda 2 what
is lambda 2 lambda 2 is given as 5 20
nanometers so this is 5 20 what is
lambda 1 that is 6 15 minus 5 20 so this
comes out to be 5 20 divided by 1 30
so this comes out to be four therefore n
is equal to 4 n plus 1 is equal to 5 so
what is my design
what did we assume we assume n it bright
fringe due to lambda 1 coincides with n
plus 1 at bright fringe due to lambda 2
that means 4th bright fringe due to
lambda 1 what is lambda 1 lambda 1 was
650 nanometers into this wavelength
coincides with n plus 1 that is when v
bright fringe due to 520 nanometers
right so what was the question the
problem asked us to find out the least
distance from the central maximum where
the bright fringes will coincide so let
us calculate the least distance least
distance is nothing but X n so what is X
n X n is X 4 that is 4 into lambda 1 so
what is lambda 1 that is 6 50 into
capital D by small D so this comes out
to be 2 6 0 0 capital D by small D
meters so at this much distance from the
central maximum the four bright fringe
due to lambda 1 will coincide with the
5th bright fringe due to lambda 2 let us
now have a look at problem 4 it says
that in a double slit experiment the
angular width of a fringe is found to
the 0.2 degrees so angular width is 0
two degrees on a screen placed one point
one meter away so capital D is one
meters wavelength of light used is 600
nanometers what will be the angular
width of the fringe if the entire
experimental apparatus is immersed in
water so we have to find out theta -
such that the refractive index is given
as four by three so what do we know we
know that angular width of a fringe or
angular fringe width is nothing but
linear fringe width divided by capital D
right now when in wattage when it is
placed inside water what happens to the
new fringe width that is beta - so beta
dash becomes equal to 1 by n into beta
so the fringe width becomes is inversely
proportional to the refractive index in
of the medium into which the the app
entire apparatus is placed so this
becomes equal to 1 by n into beta
therefore what happens to theta - theta
- becomes equal to V tau dash by D the
beta dash is beta divided by n into D
and this is equal to 1 by n into DT by D
that is heat up so this is equal to 3 by
4 into theta is 0.2 degrees so this
comes out to be 0.15 degrees
so here the catch of the problem lies in
this fact that the fringe width is
inversely proportional as the refractive
index of the material medium increases
the fringe width decreases
that's because range which depends upon
the wavelength of light used and we know
that when the medium becomes denser the
velocity decreases and velocity has a
relation to wavelength and wavelength in
terms has a relation to range with let
us look at the last problem of this
section it says that in the
experiment using light of wavelength 600
nanometers the Antonine width of a
fringe formed on a distant screen is 0.1
degrees so that means theta is equal to
0.1 degree and the wavelength of light
used is 600 nanometers so this is equal
to 600 into 10 to the power minus 9
meters so we have to calculate the
spacing between the two slits that means
small D so we know that the angular
range with theta is equal to lambda
divided by small D 8 we have already
derived this expression so from this we
can say that D is equal to lambda
divided by theta so for that we should
convert theta into radians so use for
the 0.1 into pi divided by 180 radians
so this is 0.1 into 3.14 divided by 180
radians so now we can put these values
at the 600 into 10 to the power minus 9
divided by 0.1 into 3.14 divided by 180
so this comes out to be 3 point 4 4 into
10 to the power minus 4 meters so this
is the spacing between the two sets so
now that we have finished our discussion
on the phenomenon of interference it is
time that we start with the next
phenomena that is known as diffraction
now diffraction and interference they
are very much closely related to each
other the way interference was related
to superposition similarly diffraction
is also very closely related to the
phenomenon of interference so let us see
I mean what small differences exist
between the phenomenon of interfere
and the phenomenon of diffraction thank
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