L26 : Numerical - Wave Optics, Physics, Class 12 Video Lecture

hello friends this video on wave optics
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till part 25 before going ahead with
part 26 so let us look at the first
problem it says that a slit for
centimeter white is irradiated with
microwaves of wavelength 2 centimeter so
that means width of the slit is equal to
4 centimeters and the wavelength is also
equal to 2 centimeters find the angular
spread of central maximum assuming
incidents normal to the plate of them
normal to the plane of the sling that
means let us suppose if we have a slit
somewhat like this this is my central
maxima and then the alternate bright and
dark fringes so we have to calculate the
angular spread of the central maximum so
this would be the angular spread of the
central maximum let us call this angle
as theta this angle as theta so what
would be the angular spread so looking
at this diagram we can say that the
angular spread of central maximum will
be equal to 2 theta so we have to
calculate the value of 2 theta so what
is this point this point is secondary
minimum this point is also minimum right
so what is the condition for minima
the condition for minima is a sine theta
is equal to n-lambda and what is the
value of n for this for this point it is
n is equal to 1 this also n is equal to
1 so they put n is equal to 1 and we get
a sine theta is equal to lambda or we
get sine theta is equal to lambda by a
that is 2 by 4 which is equal to 1 by 2
so this gives the-- T is equal to 30
degrees therefore the angular spread
which is equal to 2 theta is equal to 2
into 30 degrees that is
to sixty degrees so this would be the
angular spread of the central maximum
right so what we did because when we
talk about the spread of central maximum
the end points of the central maximum
are nothing but the minimum so we can
apply the condition for minima at these
two points right okay because this angle
is the angle which is subtended by the
first minima at the center similarly
this angle is the angle which is
subtended by the first minima at the
center let us look at the next problem
which says that the light of wavelength
600 nanometers is considered - normally
on a slit of width three millimeters so
here the wavelength of light is given as
600 nanometers which can be written as
600 into 10 to the power minus 9 meters
it is incident normally on a slit of
width three millimeter so width of the
slit is three millimeter which is 3 into
10 to the power minus 3 meters spatulate
the linear width of central maxima on a
screen kept 3 meters away from the slit
so capital D is 3 meters so here we have
to dip in the previous question we had
to calculate the angular width of
central maxima now we have to calculate
the guinere will let us suppose this is
my slit this is my central maxima
this is the midpoint of the slick egg so
we have to calculate this distance right
so how do we calculate that here also we
will apply the same thing we will try to
apply the condition for the first minima
so using the condition for first minima
we can calculate the value of this theta
right and what is this this is B now let
us suppose that this distance is X so if
we are able to calculate the value of x
then we can find out the width the
linear width because the linear will be
2x right so so let us approach it in the
same way condition for minima is a sine
theta is equal to n-lambda so we are
considering the first minima because at
this point n is equal to 1 so we can
write it as a sine theta is equal to
lambda so from this we can say sine
theta is equal to lambda by a lambda is
600 into 10 to the power minus 9 divided
by a is 3 into 10 to the power minus 3
so sine theta now for very small values
of theta sine theta can be written as
theta so theta is equal to 600 into 10
to the power minus 9 divided by 3 into
10 to the power minus 3 right so we have
to calculate the linear width that means
we need to calculate the value of x now
if you look at this triangle we can say
that theta is nothing but tan theta
right this theta can be written as tan
theta because for small values of theta
sine theta is approximately going to
theta which is approximately equal to
tan theta so we can write theta as X
divided by
which is equal to 600 into 10 to the
power minus nine divided by 33 into 10
to the power minus 3 or we can say X is
equal to 600 into 10 to the power minus
9 into 3 divided by 3 into 10 to the
power minus 3 so this 3 I'm 3 will get
cancelled so we get 600 into 10 to the
power minus 6 meters therefore what
would be the width which will be 2 into
X which is equal to 1 to 0 0 into 10 to
the power minus 6 meters that is equal
to one point two millimeters so this
would be the linear width of the central
maximum now let us look at the next
problem it says a parallel beam of light
of wavelength 500 nanometer falls on a
narrow slit and the resulting
diffraction pattern is observed on a
screen 1 meters away so in this we are
given a wavelength of 500 nanometers
which is nothing but 500 into 10 to the
power 9 minus 9 meters screen is placed
1 meters away from the slit it is
observed that the first minimum is at a
distance of 2.5 millimeter from the
centre of the screen it is always better
to draw a rough diagram so that things
become simpler for you so it says that
the first minimum is at a distance of
2.5 millimeter from the centre of the
screen
that means this distances 2.5 millimeter
so we have to calculate the width of the
slit right so now again the same
condition for first minima we can say a
sine theta is equal to n lambda where n
is equal
one so we can say a sine theta is equal
to lambda or a is equal to lambda
divided by sine theta now for small
values of theta sine theta becomes equal
to theta so this will be lambda divided
by theta so what is Theta theta would be
nothing but if you compare if you try to
construct this theta theta is nothing
but this 2.5 millimeter divided by D
right so theta is nothing but 2.5
millimeters
divided by capital D which is 1 meter
therefore a will be equal to lambda that
is 500 into 10 to the power minus 9
divided by 2 point 5 into 10 to the
power minus 3 so this comes out to be
zero point 2 millimeters so this would
be the width of the slit let us look at
the last problem of this section it says
that estimate the distance for which ray
optics is a good approximation for an
aperture of hole millimeter so width of
the aperture is 4 millimeter which is 4
into 10 to the power minus 3 meters and
wavelength 400 nanometers that is 400
into 10 to the power minus 9 meters so
whenever we talk about approximation
between ray optics and wave optics I
mean the limit between the optics and
wave optics we will have to talk about
the fresnels distance because fresnels
distance tells us whether ray optics is
a good approximation or not
so entered fresnels distance is given by
n square by lambda so that is 4 into 10
to the power minus 3 whole square
divided by lambda which is 400 into 10
to the power minus 9 so this comes out
to be 40 meters so this is the distance
for which Ray of X is a good
oxidation that means for distances
lesser than crystals distance that means
for distance is lesser than 400 meters
the rain optics is a good of
approximation
whereas for distance is greater than
this ray optics is not a good
approximation so here we end our
discussion on diffraction and it's
applications and this time the next
topic which is known as hood thank you
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