Welcome to the 10th week, we look at Waves
in Optical System or in other words the light
waves or electromagnetic radiation. Thanks
to Maxwell and many others we know thatlight
is a form of a wave. Let us see if we can
describe propagation of light and maybe more
common properties of light for example, reflection
refraction by just not worrying about the
wave form, but simply looking at it asray
that travels from point a to point b.
So, if you take this view point we need to
decide first in what limit of wave would this
be valid and how are we going to go about
working with thislimit. This sort of description
works well in what is called the limit ofray
optics. And this corresponds to taking the
limit that wavelength of light or wavelength
of the waveform is small or it tends to 0.
Let us consider a source of light here from
point P and let us say that I have a circular
aperture given by this and at the other end
I have a screen here. If P is any source of
light now question is what is it that we will
see on the screen ?
And in this case it will depend on dimension
of this aperture. Suppose this aperture is
small enough in such a situation the light
from P will of course, go like this, but we
are unlikely to see a very clear boundaries.
So, this effect that we see where you see
that there is some light even in a region
which is expected to be dark happens due to
what is called diffraction effects and this
diffraction effect is essentially an effect
that arises because of the wave nature of
light. So, we need to go to a limit where
we would not be able to see the effects of
diffraction.
In other words the conclusion is that you
will see effects of diffraction whenever lambda
is larger than 0. So, in the limit of lambda
tending to 0 even if you bring in make the
aperture sufficiently small you would not
see diffraction effects. In that case where
we can very clearly see the sharp distinction
between the region that is lit up and the
region that that falls in the shadow below
this point S prime on the screen you do see
light and above S prime on the screen its
dark.
So, in this case it goes from P to P prime
and of course, on the other side it would
also go from P to let us call this point S
double prime. So, between S and S prime and
S double prime you do see light and above
S prime below S double prime, it is very dark.
In that case we can describe our light wave
using ray diagrams basically lines like the
ones that I have drawn here. So, this is the
regime of what is called the ray optics or
geometric optics. Every time a wave of light
goes from point a to point b we can just represent
it as a line going from point a to point b.
So, the question is how do I know that light
that starting from point P takes this particular
path and this question is answered by what
is call the Fermat's principle. We need to
state the refractive index let me assume that
light travels in a medium with refractive
index mu. Let me denote by d s the distance
travelled by light in a medium with refractive
index mu if d s is the distance and v is let
us say the speed of light, then of course,
speed of light in the medium with refractive
index mu in that case I can write the time
taken to travel this distance d s as. So,
I will simply use the definition of refractive
indexto rewrite the velocity of light in the
medium that is characterized by refractive
index mu.
Now, I have assume that light travels in a
medium that is characterized by refractive
index mu and there is no change in the medium.
We can generalize this idea further by saying
that light probably travels in a in a series
of different media each one characterized
by a different refractive index. So, let me
denote by tau the time taken to reverse distance
d s. So, generalizing this formula I can write
it as. So, each one of these indices represent
a different medium and there is a change in
refractive index. So, that is characterized
by mu i.
So, d s 1 is the distance traveled in a medium
characterized by refractive index mu 1, d
s 2 is the distance traveled in a medium that
is characterized by mu 2 and so on. If there
is a continuously changing media we can replace
the summation by an integration. In which
case the distance traveled which is denoted
by tau will be given by.
So, I have this sort of formal expression
for the time taken by light to travel from
point A to point B in this picture of ray
approximation and between point A and point
B depending on where you are the refractive
index does change. So, that is taken into
account by this mu which depends on s.
Fermat's principle can be mathematically written
in the following way I have written it here
for you. So, the variation of this integral
is 0. So, the physical content of Fermat's
principle is that if you have chosen two points
and light travels from A to B you could consider
many possible paths here through which the
light might possibly travel from A to B. And
the one it actually takes is is the one for
which the variation of this quantity equals
0. Very loosely speaking its equivalent to
saying that the time taken to go from A to
B is the least, this quantity mu times d s
is called the optical path length.
So, here is Fermat's principal written out
for you . The actual path between any two
points like A and B that we had drawn in the
previous slide is the one for which the optical
path length is stationary. So, you would see
that mu d s represents the optical path length.
So, this quantity is stationary with respect
to variations in the path. So, the light actually
takes that path for which the optical path
length is an extremum.
So, Huygen's theory sort of provides this
bridge between light as a wave and light as
a ray. So, his idea is that light can be represented
as a wave front, I have this point P which
is a source of light and if that is a source
of light. So, light is going to travel in
all the directions outward from this point
P.
Now, you can ask the question what is the
locus of those points which have the same
face after all whichever direction light travels
around P its traveling with same speed. This
locus of these points with same face have
travelled a distance equal to c times t . Now,
this is a wave front for us and specifically
in this case it would be called a spherical
wave front the light coming out of point sources
of light would define a spherical wave front.
On the other hand if I had a source of light
which is linear. So, all the points with the
same face here have travelled the same distance
and the locus of such points they form a cylinder.
So, this is called a cylindrical wave front
and if you are looking at say a point source
of light which is really very far away, but
if it has traveled sufficiently long distance
and you are looking at a small portion of
that.
And that for example, something like this
would look like a plane wave. So, that would
be called a plane wave front for instance
the sunlight which is reaching us from really
longdistance. So, in such cases you could
assume that you are actually observing a plane
wave front.
So, let us say that I have this plane wave
front at some time t maybe it originally started
out as a spherical wave front, but at the
point where I am observing it what I see is
a plane wave front. Huygens idea is that each
point on this wave front acts as a source
of secondary wavelets and let us say you take
one point here its a source of secondary wavelet
which is going to be spherical.
Now, you take the point which is very close
to this point draw a similar wave front which
came out from the second point and so on.
So, if you put all of them together what you
get is a new position of the wave front. So,
essentially you transported a wave front at
time t 0 to let us say at some time t one.
The central and core idea is that every point
on a given wave front is a source of new wavelets
and if you join the points with same face
for this secondary wavelets you are going
to produce a new wave front and you moved
your wavefront from old position to new position
and you can continue this process. It requires
a little more involved exercise including
wave phenomena to explain why it moves forward
rather than backward.
Let us say that I have some surface here which
I call S and there is an incoming wave front
like this. So, I consider secondary wavelets
and I can move it let us say to this point
which is essentially the locus of all the
secondary wavelets with the same face and
here I assume thatthe media are different
at the top and bottom. So, at the surface
a reflected component of light emerges and
that would go something like this.
So, in other words you have a light wave that
is coming let us say in this direction I have
drawn the arrows here, it gets reflected at
surface s that is a change of media there
and finally, the reflected component goes
in this direction that I have shown here.
And by the same token one can also explain
refraction, so this is for the case of refraction.
So, in the evenings the light travels longer
distance to the atmosphere hence if this is
the ray that is coming from the sun it actually
bends when it encounters the atmosphere and
reaches our eyes. And from our perspective
it would look like the sun is somewhere here
whereas, the more correct position of the
sun is somewhere way below this point and
this is because of refractive effects of the
atmosphere.
Let us say that I have a source of light S
and from here I have a diverging beams which
go out as spherical waves and after let us
say it travels over sufficient distances it
becomesplane wave. So, if you put in optical
devices you can make it finally, converge
at some other point here and of course, the
converging wave front might look like this.
The point you note is that all the points
along the wave front let us take for example,
this wave front all the points along the wave
front have traveled the same optical distance.
So, in other words they are at the same optical
distance from the source. So, the path that
light takes from let us say point S to S prime
can be accounted for by Fermat's principle
and to map this wave front we basically use
Huygens idea of secondary wavelets. So, you
can move your wave front forward using secondary
wavelets. Now, let us use all this machinery
to account for reflection and refraction.
So, what we have isthis surface S there is
an incoming incident wave which comes in a
sense from the left of this point S that I
have and gets reflected at this surface S
and you have this reflected component. And
at this point S where the reflection takes
place I have a normal drawn and with respect
to that perpendicular or normal there are
two angles defined; one is angle i which is
call the angle of incidence which is the angle
at which the incident ray hits the surface
S. And there is another angle defined which
is r this is the angle at which there is an
outgoing ray the reflected ray.
So, this entire action is taking place in
a plane that is perpendicular to S because
suppose if I take some other point let us
say S prime and I connect these points object
O to S prime and S prime to I. So, these distances
will actually be larger than O S and S I purely
from the geometry ofwhat we are looking at
and the only time this travel distance I mean
the optical distance will be smallest is when
O S I is in a plane that is perpendicular
to this surface S. So, let us apply Fermat's
principle to this problem I need to know what
is the time taken for light to travel from
O to S and from S to I.
So, to begin with I need to know first the
distances O S and S I and from the geometry
of figure I can calculateboth of them that
would be square root of x square plus y square
divided by c, where c is the speed of light.
T is equal to square root of capital X minus
small x whole square plus y square divided
by c, let us give it a different name T prime
instead of T.
Now, Fermat's theorem tells us that d by d
x of t plus T prime is equal to 0. So, substituting
expressions for T and T prime that we just
got in here I will get and this is to be say
T equal to 0 and of course, there is a minus
sign here when we differentiate and this is
going to give me the following equation that.
Now, if you look at these expressions on the
left hand side and right hand side of this
equation you will notice that this quantity
here is simply equal to sin of angle i and
similarly what would be sin of r? That is
precisely what I have here on the right hand
side of this equation.
So, Fermat's principle in this problem tells
us that sin i is equal to sin r of course,
this is a problem with multiple possible solutions,
but if we take the simplest possible solution
this tells us that i is equal to r angle of
incidence is equal to the angle of reflection.
So, this is one of the commonly stated laws
of reflection.
Now, let us do a similar kind of calculation
for refraction here again I have this surface
and presumably the media above and below the
surface are different characterized by two
different um index of refraction.
So, here I have drawn my setting the point
S is where refraction takes place is given
by capital X comma 0 and at that point we
draw this normal and defineangle of incidence
and an angle of refraction. Now, the question
is what does Fermat's principle tell us in
this scenario.
So, time to traverse O S that would begiven
by I need to calculate the time to traverse
SI let me write down that expression in the
expression I have divided it by c prime. So,
c is the velocity of or speed of light in
the first media which is characterized by
refractive index mu and c prime is the speed
of light in the second media which is characterized
by refractive index mu prime. So, now, I need
totake the sum of these two times differentiate
them with respect to x.
So, apart differentiating this expression
with respect to x this is what I am going
to get this term here and correlated with
this figure that I have it will tell you that
what I have circled here in red is simply
sin i sin of the angle of incidence this quantity
and again you correlated with this figure
you will notice that that is simply equal
to sin r sin of angle of refraction.
So, 1 by C would be mu divided by speed of
light and similarly 1 by C prime would be
mu prime divided by speed of light. Now, if
I substitute these two quantities in this
equation this quantity speed of light will
cancel out and I would get the final form
that.
So, this is the law of refraction. In the
nextlesson we will try and look at some more
examples of this, but involving optical elements,
we will see how we can map out path of array
in the presence of say lenses mirrors etcetera
.
