Welcome to this second part of 10th week,
we will not be studying optics in detail because
optics in itself is a vast subject. So, instead
what we are going to do is to simply confine
ourselves to few simple things which we need.
So, that we can understand how Waves and Oscillations
as phenomena work in Optical System.
To begin with let me remind you of Fermat's
principle. So, the principle simply states
that the actual path taken by the light is
the one for which the optical path is in extremum.
In the standard simple case of a homogeneous
medium and light travelling from point A to
point B its equivalent to saying in a sense
that light takes the shortest path between
those two points in this case it it would
be a straight line.
Fermat's principle can be stated as the actual
path between any two points is the one for
which the optical path is stationary with
respect to variations in the path. Today we
will apply Fermat's principle to a little
more complicated system mainly to the case
of a thin lens I am going to start with the
case of plano convex lens.
So, the lenses plane on one side and has a
convex face on the other side. So, this distance
here would define for me the thickness of
the lens all you need is a circle of which
this convex arc is a part and the radius of
that circle which have denoted by capital
R is the radius of curvature of this lens.
Whenever this d which is the thickness of
lens is much smaller than the radius of curvature
then you call it thin lens.
I am going to assume that a plane wave comes
from the left side which will be our convention
to assume this pretty much in most of the
problems that there is an incident plane wave
coming from the left side. And I am also going
to assume that the medium of this Plano convex
lens has a refractive index mu and in general
we can assume that the refractive index of
the medium outside of this lens is equal to
1.
Now, let us focus on a ray it should be at
a distance r from the optical axis. So, the
optical axis here in this case is indicated
by thisred line the origin of y axis coincides
with this red line. So, this small r would
define for me some arbitrary distance from
the optical axis. What happens to this incoming
incident plane wave after it passes through
this plano convex lens? So, if your along
the optic axis along this red line the ray
which comes along this red line would travel
and entire distance d inside the lens as you
can see.
On the other hand the ray which is hitting
the lens at a distant r from this red line
would only travel part of the distance inside
the lens there is this one part which would
be traversed inside the lens and there is
the other part which should be travelled outside
the lens.
So, let us call this time d, the time taken
for the wavefront to travel this distance
d inside the lens along the optic axis or
the central axis and that time would be given
by mu times d divided by c c is the velocity
of light d is of course, the thickness of
lens and mu is the refractive index of the
medium of lens.
Now, what happens to the wave front that hits
the lens at a distance r from this axis . So,
what we want to know is how much distance
does the wave front travel in the same time
t. You can take a look at this diagram which
is nearly the same diagram that I have here,
but I have try to focus on this particular
ray which comes and hits the lens at a distance
r from the optic axis.
Now, from the geometry of this figure you
can use Pythagoras theorem to write the following
equation R minus x whole square plus small
r square is equal to R square. So, that is
the solution for x it cannot be a value that
is larger than R, if I keep the plus sign
in the solution it would tell me that x will
have a value that is larger than R. Hence
I will take x to be R minus square root R
square minus r square.
Now, we make the binomial approximation. So,
the binomial approximation is of the following
types. So, if I havefunction of this type
1 plus x to the power half or in general I
could consider 1 plus or minus x to the power
half , this will approximately be equal to
1 plus or minus half into x. It actually an
infinite series which we are terminating at
this point and this approximation would work
only if x is much smaller than 1 this is the
result that we want.
So, the distance traversed by the wave front
at a distance R from the axis inside the lens
that is very important. So, this is the distance
travelled by the wave front, but inside the
lens its given by d minus r square by 2R.
In addition to this it also travels a certain
distance outside the lens if we indicate this
distance by, is it? So, the distance traverse
by wave front outside the lens would be given
by r square by 2R plus sum z .
So, here I have essentially equated the times
on the left hand side is the time taken to
traverse the distance d along the optic axis
and on the right inside here is the time taken
to traverse some distance in the same time
this is going to give me. So, if you remember
Z is simply some arbitrary point up till which
are ray has travelled in time t value of Z
depends on R depends on how far you are from
the optic axis and also it depends quadratically
on R.
Now, I can go back and draw figure here to
indicate this let me show that in blue, so
its going to look something like this. What
is this tell us? It tells us that all the
points which make the wave front they are
at a distant Z from this line that I have
here and the locus of all the points has this
r square dependence. So, you could say that
in three dimensions the locus of Z is going
to define for us as fear whoseradius is going
to be R by mu minus 1. This is also means
that there is going to be a point of convergence
at the centre of that sphere.
So, in other word if I had an object which
is placed at infinity and there is an incoming
wave front which you assumed to be a plane
wave its hitting this plano convex lens , what
this analysis tells us is that all the rays
would converge at this point whose radius
is given by R divided by mu minus 1. So, power
in this case would simply be mu minus 1 by
R which is equal to 1 by the focal length.
So, this is gives as an estimate of power
of a lens which is its ability to bend the
incoming beam of light.
Now, we will redo similar problem, but slightly
differently explicitly using Fermat's principle.
So, its not a lens what I have is a surface
a refracting surface. So, the surface itself
is given by this S P M and we have an object
let us say at positionO here and Q is any
point such that O P Q together form a straight
line. Of course, given that O is the some
object or some position where would this B
missed after refraction on the other side
of this curved surface. Capital R is the radius
of curvature of this curved surface.
Now, let us start by looking at this triangle
SOC using the cosine law of triangles . I
can write the following equation for OS you
can use cosine law of triangles and write
this equation. Before I do anything with these
expression I should point out that we will
be working in the regime where thetas small.
So, this is an approximation in which we will
work with and this sort of approximation where
the rays or close to the optic axis is called
paraxial approximation in optics. So, now
I can write O S as follows now this expression
can be simplified, I am once again going to
use a binomial approximation here, we have
an expression for O S now.
And you can do a very similar calculation
to find out S Q. So, I will not do that and
directly write the result, but its fairly
straight forward just follow the recipe that
we just adopted. So, I have OS and SQ the
total optical path length would be. So, the
path length OS as you can see happens in a
medium where the refractive index is mu 1
and the path length SQ takes place in a medium
where refractive index is mu 2.
Now, I have collected all the terms together
added them and this is my expression for optical
path length L. So, I want to find out now
d L by d theta and set it equal to 0, R is
the radius of curvature capital R. So, its
not equal to 0 . So, what could be 0 or only
these two things; one is its possible that
theta could be 0 or this entire expression
within the square brackets that could be 0.
So, theta equal to 0 gives me a straight line
path and there is only one possiblestraight
line path. Now, the other way by which d L
by d theta can be 0 is when this quantity
within the square brackets is 0 this is what
we haveobtained from the condition that d
L by d theta should be equal to 0.
If I designate this P I to be some y 0 or
the distance of the image from the point P
in that case all that this equation tells
me is that many possible paths are allowed.
So, the point here is that any path of type
O S I is allowed provided of course, I identify
y with y 0 the distance of the image.
So, typically these kind of equations follow
some convention for instance if you go back
to this figure that we have here the distances
to the left of P generally take a negative
sign and the distances on the right of P point
P would take a positive sign. So, if I actually
apply this convention here this equation could
be rewritten as.
Once again you might recollect that this is
simply equal to the powerof the lens. So,
now I have read on the figure slightly differently.
So, that we can obtain the same result using
Snell's law of reflection. So, there is a
beam that goes from point O to S and gets
refracted and possibly converges at point
I there are various angles it defined alpha
1, alpha 2 beta and there is also this perpendicular
which is of height h 1. So, what is of interest
for us are these three distances x which is
the distance from point P to O and the distance
from P to I which is called y and then there
is radius of curvature which is P C in this
diagram.
At the refracting point I should be able to
write an equation of this type this is simply
a statement of law of refraction, I am going
to assume that phi 1 and phi 2 are small in
which case. So, this is what is called the
paraxial approximation. So, this line which
goes from C to S is the normal at normal to
the curved surface at the point S hence this
would be our phi 1 and similarly this should
be my phi 2.
So, if I look at this triangle OSC the sum
of the angles of this triangle should be 180degrees
this equation would simplified to the following
phi 1 is equal to alpha 1 plus beta. Now,
let us also look at the triangle SCI sum of
all the three angles is 180 degrees this would
give me an expression for phi 2 which would
be beta minus alpha 2.
Remind yourself that we are looking at and
approximation where all these angles are very
small in which case the sin phi 1 can be approximately
written as tan alpha 1 plus tan beta. So,
I just expand sin a plus b and consistently
apply that all the angles are small enough
and then every time you have a sign of theta
you can replace it by tan of theta. Again
remember that we are working in small angle
approximation in which case this distance
P D is so small that we can ignore the distance
and take this distance O and D to be distance
O D to be S self.
So, taking P D small would correspond to again
thin lens approximation P and D are so close
that we can ignore that small difference and
take the distance between P and C to be R
itself. Now, by the same token I can write
expressions for sin phi 2 as well that would
be tan beta minus tan alpha 1 .
Now I have expressions for sin phi 1 and sin
phi 2 I simply need to substitute in this
equation. This is the expression that I get
when I substituted back in this equation which
is the law of refraction . Now, by simply
rearrange this equation I can write it in
a moresuggestive form. So, this is the final
result I get, but again to put it in a form
that is useful for us let us call this distance
u to be x, but remember that any distance
which is to the left of P is going to get
a negative sign. So, u will be minus x and
v will be y .
So, this is exactly the result that we just
got earlier on. So, we got it by two different
methods; one is by actually applying the Fermat's
principle and other by using the law of refraction
it shouldnt be surprising simply because the
law of refraction itself was obtained by using
Fermat's principle.
So, it looks like everything is consistent
for us and the quantity here on the right
hand side mu 2 minus mu 1 divided by R is
again simply equal to the power or P which
is the power of lens. You have light beam
going from one medium to another medium characterized
by refractive indices mu 1 and mu 2. So, we
have mu 2 minus mu 1 in the denominator in
instead of mu minus 1 that we had earlier
on. Now, in the next lecture we will try and
put together all these results to obtainequation
for thin lens .
