hello 
everybody so today we will be starting our
first lecture on geometrical optics okay so
what is the first thing that comes to your
mind when we talk of
well geometrical optics i suppose the first
thing that comes to our mind is you know the
passage of light through lenses how curve
lenses that convex lens or a concave lens
or maybe reflection from curved mirrors so
but one has to realize that this is rather
pretty old subject is rather pretty old subject
in physics we have been we have been doing
things for 100’s of years in fact it is
the it is the use of these lenses and mirrors
which had a huge role in the development of
astronomy
for example galileo was the one who used just
two lenses he just used two lenses and then
put it in a cylindrical tube and he he he
watched the moons of jupiter and he made the
certain astronomical observations so in a
sense well this subject is it is it is it
is one of the primary things which was which
was responsible for for advancements in in
in astronomy and not only in astronomy not
only in physics for that matter
i mean if you think of the simple magnifying
glass okay to look at small things and i bet
the modern biology of owes its origin to the
observation of small things right so the study
of geometrical optics had rather far reaching
consequences not only in the development of
optics but also in other branches of physics
and also in other branches of science right
so the now that we know that it is an important
thing to study although it is an old subject
what what were rather what is the of applicability
okay so rather at what wavelength range range
should we apply geometrical optics now why
do i say that because of the back of your
mind you also have things like diffraction
interference which have been explained so
well with with electromagnetic waves okay
so what is the area of applicability here
for that let us draw a a simple diagram and
define our area of applicability
let us say we have a small slit or a small
hole okay and we are going to put some some
source in front of that okay some some source
of light in front of that right now we put
a screen a little distance away from this
hole or from the slit now what we expect to
observe well i mean we have a reasonably big
hole here so we can draw these two lines shall
come to this point that we are drawing lines
here
and then we observe the image of this of of
this slate of this hole on on this on the
screen so on this screen now which is kept
at a certain distance from this slit okay
now if you observe you have sharp boundaries
for this image okay and you you just observe
only one image of this hole then i am sure
you are going to say that we are in this regime
of geometrical optics now why i mean look
at the other look at the other extreme i mean
if you if your hole is small enough you will
have if your holy small enough
and let us say let me draw another picture
if your hole is small enough you and you can
have and then you put a source of light here
i call this some source you are going to have
you are going to have alternate regions of
light and dark regions okay now that is diffraction
so you have alternate regions of light and
dark regions here so but the field of geometrical
optics would not concern such a thing but
not with not concerned with not be too concerned
with with with this diffraction okay
but would be talking of things in which when
you have a slit okay the image of the slit
is very well defined on the on the on the
screen now when does that happen that happens
if the size of this slit that you have taken
okay let us say the size of the slit is d
now that is much much greater than the wavelength
of light that you have used okay so this is
a one important one important criterion for
the application of geometrical optics okay
and well if it is actually if it is not if
the if the slit dimension is some way it is
very near or comparable to the dimensions
of the wavelength okay you will as i said
you will have to deal with things like diffraction
okay so this is one thing about the sizes
the second thing is a bit involved but then
it is not a conceptual it is a conceptual
thing it is that the light or the energy of
the of the photon in this case okay should
be lesser than the energy sensivity of the
instruments that you are using
so it should not interact with the with the
instruments okay so basically you are not
in the quantum region quantum regime here
so we are still in this classical regime and
be using you know what what we call as these
lines or these rays of light let me draw it
once again so these are the rays of light
by which you are going to get some sharp images
on the screen okay now what are these rays
of light these rays of lights are nothing
but infinite infinitesimally thin
so these will be using the rays of light okay
so and they are the infinitesimally thin beams
of light okay now for that tree for that reason
alone for that reason geometrical optics is
also called ray optics okay so that is what
we are going to study we are going to study
ray optics or geometrical optics here okay
now let us do one more thing now now that
we know that we have we are going to deal
with rays ok so the next question to ask is
what part do these rays decay ok
so the question that we ask ourselves is what
path does the does the let us say the ray
of light take okay now this was something
which which bothered many people and quite
quite early and apparently people had lots
of ideas well they say that oh it will follow
the least distance between two points well
if it is the same medium that is fine i mean
the least distance is fine but that may not
be the case in which you have you have in
media of different refractive indices in between
okay i will come to that again
now apparently there is a principle which
tells you what part the ray of light will
take well it is called fermat’s principle
or in its original form it is also called
the principle of least time okay so we will
be talking of the fermats principle or the
principle of least time well it simply states
that well the ray of light will follow that
path from one point to another in which the
you know i have point a and point b
well i you can take whatever refractive index
in between well it is for it will follow from
point a to point b the path in which the time
taken will be the least will be the minimum
one ok so that was fermat’s principle in
its original in its original form so that
is what fermat said that light is going to
travel from a and b ok and the path it is
going to take and it is going to take is that
for which the time is going to be minimum
so instead of this part or the other path
the one from a to b will be the one the one
the light will take and the ray of light will
take will be the one in which the time taken
is a least ok now apparently the whole of
geometrical optics can be can be derived in
principle from this one simple concept ok
so that is how important it is ok are there
the well in its this is the statement in its
original form it is the principle of least
time
but i will come a little later on on some
modifications but let me just you know tell
you two or three consequences of fermats principle
ok
so it is just do one of two consequences of
fermats principle consequences of fermats
principle well the first is that if from point
a to point b sets a light is traveling and
which has taken the path of least time so
point a to point b suppose this is the path
and this is the path of least time from point
a to point b well if i start from point b
to point a the the light will actually take
the same path back okay
so if it was the least time on one way that
will be on the other way also it will be the
least time compared to other parts okay now
this is called the principle of reciprocity
okay so it retraces its path back so this
is the principle of reciprocity okay second
thing is about well i mean what comes to my
mind is i mean i started this talking on you
with doc right by talking of some lenses okay
so let us royal lense let us draw a simple
convex lens okay
and on the axis so light let us say i have
some point on the axis on the left hand side
here okay so this is my convex lens and then
let us say the light from this point actually
comes and then he comes here and then well
i have well it comes on the other side okay
so this is the object distance and this is
the let us say the image distance which i
know denote v okay now there are well there
are infinite parts of light
let us just stick two or three of them you
know there is one path which goes entirely
through air okay so the one up at this point
and then comes and comes back here there is
one which goes partly through air which is
actually along the axis okay and then there
is one part which goes through the lens and
then comes through and then comes and travels
in air to reach the image point okay
now if you look at these two point the length
of these two points are different the path
lengths are different okay the actual length
from on the axis that is the least compared
to compared to the length when it goes to
the tip of the lengths and then comes to the
image point okay now however by the principal
release time you can you can argue that the
time taken by these two light rays okay
but this one the one in which i have drawn
the one goes to the tip and then comes to
the image point on the one in which it passes
through the you know which is along the axis
of the length okay the both of them take the
same time okay otherwise you are not going
to have an image on the other side okay so
of course i can also derive the simple laws
of reflection and refraction using fermat’s
principle okay
but before that i wish to remind you once
again about well maybe i have not have not
said this before is about the the statement
of fermat’s principle well now we do not
see that the time taken is minimum what we
see is that the light will follow that path
so from a and b so formats principle the way
we say it now formats principle is that the
light will follow that path will follow that
path for which 
the the the time taken is the time taken is
actually an extremum is actually an extremum
compared to the nearby paths okay xd mm compared
to what compared to two nearby parts when
i say an extremum i do not actually specify
whether the time is actually minimum or maximum
it is an extremum okay
so pictorially and diagrammatically how do
i write that i mean i have 0 point a and point
b okay let us say oh point a and point b now
there are actually infinite number of parts
which connect point a and point b ok now what
it tells what for much principle tells is
that light will follow that path for which
for which the time taken is actually an extremum
compared to these parts compared to for example
it will be it will follow the path in which
i have shown by a solid line ok
as opposed to and that will be an extremum
path compared to let us say the the ones which
i show by the the dashed lines ok so that
is that is how we know formats principle now
now this is actually a far-reaching consequence
of the principle of least action or again
it is the hamilton’s principle mechanics
ok now let us let us as i said we can derive
the laws of reflection and refraction from
from fermat’s principle let us see how we
can do that ok
and so that will give us some more confidence
about the applicability of this of this theorem
of this principle rather ok so we want to
check what happens during reflection
draw a figure first so here we have a mirror
a plane mirror let us say and say we have
a point a and a point p so let us say we have
a point a and point b now the thing that we
want to do here is that we want a ray of light
to to come and hit the mirror reflection and
go to b now which direction should it go well
common sense and what we have learnt in school
is that well it should go in a such a way
that the angle of reflection should be equal
to the angle of riff annual of incidence now
there should be equal to the angle of reflection
ok
now why is that because you see from going
from path going from point a to b via the
mirror i can actually have an infinite number
of paths ok now why should light choose the
one in which the angle of incidence is equal
to the angle of reflection let us find out
using formats principle ok so let us draw
let us draw a normal at the point of incidence
ok and consider that this point a is a at
an height h1 from the mirror
point b is at an height h2 and the and the
distance on the mirror you know the distance
of the mirror is l let us say and let us also
think that the the distance at which the light
strikes the mirror that that is x okay so
so let me see you and have i drawn all the
all the distances yes i have taken the the
distance the the distance in the mirror the
distance where it hits the light hits the
mirror is incident on the mirror okay
so i have not drawn the angle of incidence
let me call this i ok and the other one here
is the angle of reflection let me call this
r so and this is the normal ok so the one
this this dotted line is normal i bet you
are going to say that what will be this angle
this angle is also r and this angle is i ok
these are all right angle triangles ok so
let me okay so let me i have in term labeled
these points height of this is p
let me call the point at which the light is
incident has o and let me also call this height
so b this is q okay so pq is l be 0 is x some
distance okay and the height from the mirror
is h1 here and the height from the mirror
i to the mirror of b is h2 okay now the thing
and now the question that i am trying to address
here is i mean light comes and hits the mirror
at a at a distance x here well it could have
it could have hit a little bit when if it
hits at x
what happens to the angles this is one thing
or two to put it in another way down i mean
why should it hit only at exit can hit at
the other places and come to b okay now which
is the one according to fermats principle
should be the path of the light here ok so
let us let us do this what is the time it
takes for light to go from so time to to to
to travel let us say to travel the distance
aob this distance aob let me say let me call
that time t
and the velocity of light in this medium okay
so let me say this medium itself is air so
velocity of light let us say you see how the
speed of light is c ok so what is the distance
ao well the distance ao is nothing but h1
square + x squared pi by see that is the time
it takes to travel the distance ao and the
distance ob so that will be square root of
h2 square + l - x squared divided by c okay
now what do i do now now i say that the time
will be compared to different paths compared
to its neighboring odd or nearby paths so
this is going to be going to have this is
going to be a extremum okay so that is what
i have
so i have let me let me do this dt dx so this
is equal to watch dt dx i am going to take
and 1 by c here that is the speed of light
as a constant thing i am going to differentiate
the one in the in the square root so i will
have a 2x that the top and then if i differentiate
this entire thing so i am going to have 2
times root over h1 square + x1 x squared in
the denominator what about the one on the
on the for the for the path ob
so this is 2l – x so i am going to differentiate
this point x h2 squared + l - x whole square
and the square root of that so that is equal
to twice of h2 squared + l - x whole square
okay so i have not taken the minus sign out
so because l - x ok fine i put a minus sign
here and then there is an 1 by c ok so when
i say it is an extremum i am going to put
this equal to 0 so which immediately tells
me that x by root over h1 + x square = l1
- x h2 squared + l - x root of that
so what does this tell us so it tells us that
if i want the the time to be an extremum okay
compared to all different paths light can
travel now if this is the path light should
travel ok i better have x well i should have
put an h1 square + x square root over of that
that should be equal to l - x h2 squared +
l - x whole square square square root of that
what does this mean let us look at this figure
let us look at this triangle apo ok now what
is x what is x divided h1 square by x1 square
by x square ok so let us look at triangle
apo ok
let us look at the figure triangle apo now
in triangle apo what is the sine of the angle
i well it just happens to be x by x by the
distance h1 square by x square ok now compare
that in triangle boq what is the sine of the
angle r ok now these are what so sine of the
angle r is l - x divided by h2 square + l
- x whole square and take the square root
of that in the denominator so which tells
you immediately here that light will travel
that path in which i am going to have
just the you know in this same medium the
sine of this angle i is equal to the sine
of this angle r or in other words i need this
angle i to be equal to the angle r which is
actually the familiar laws of the reflection
that you have ok so well to think of this
as a mechanics problem is also quite interesting
actually if you if you if you give if you
give if you have set a problem like this and
i want to run from point a to let us say point
b by touching some wall ok which is pq
now which path should i take so that the time
taken will be the least here ok well actually
will be taking that path for which you know
the point at which you hit the wall and then
you draw in well to that then you find an
angle okay that should be equal and you still
call that an angle of incidence okay should
be equal to the the angle that you make with
the normal when you run away from the wall
to reach point b okay
this again will actually follow from the principle
of least action in mechanics but right now
we are doing geometrical optics we are dealing
with light rays of light so and we have just
now verified again the laws of reflection
using fermats principle okay now with this
a basic verification how about doing it for
the for refraction okay so you can always
say that okay fine i have done this for a
medium in which the refractive index is the
same right
so you are the reflection is in this in this
region in which you know in the in the region
aob that is in a region where the refractive
index is the same ok so let us find out what
it is going to be when you have two different
medium
let us draw figure once again so okay so we
have two media okay which is infinite on both
sides let us say so in one side the refractive
index is n1 and on the other side the refractive
index is n2 okay now you have you have you
at point a let us say and you wish to go to
point some other point ok now what is the
point what is the direction that you need
to follow okay in case you want to go to point
a to point b 
is it is it this straight line
well i should draw the point p here is it
the straight line well that is the distance
that is the least distance but then remember
that the speed of light is different in this
to media okay so the time taken is necessarily
not the same if you go which is definitely
not the same when you if you if you follow
a straight line let us say you travel this
much in in one media and then the other here
all let us say you would travel this much
in media in medium one
and then the one here well it is more like
you you all let us say n 1 so this is region
1 let us say this is land and then n to this
is some water okay you want to go from point
a to point b okay what do you do do you jump
straight you go straight into do go straight
to the you go into the show where what where
water is and then swim all the way to be okay
what do we go do you want to minimize your
swimming because you are not swimming that
that is a bit tougher than just running on
land okay because of resistance
so do you maximize your path on land okay
let us say you come here and then you swim
a little bit in sea or in water will not let
us let us not bring in sea or anything else
in water so which part should you take rather
problem is the same
so let us redraw this figure once again and
and label it so let us say you need this to
be the path and this is how you wish to go
when you are in point a and this is the point
you are incident on on the interface of both
the media so you want to go to point b here
let us say again i call this distance let
me call this distance where the distance h
this point as p this point of contact to within
these two media where light comes and hits
as o
and then b and this to be h2 and when i call
this pq so this point is q okay now i again
take this distance so this distance is fixed
this distance pq this distance pq this is
let us say let us call that as l distance
l let me call po so that is x okay so that
is variable why do i want where do i want
to hit the surface here do i want to hit it
here here here or here so that the time it
takes compared to all these paths to go from
point a to point b is an extremum let us do
that
so definitely what is point oq so this is
l - x okay so what is the time it takes to
travel this distance ao ok so the time again
to travel aob this distance aob so this is
let us say capital t so what is this distance
by the way so this distance is ao in medium
1 okay where i have the refractive index n1
now how is the velocity or the speed of light
related with the refractive index well if
the speed of light in vacuum air in vacuum
let us say is c and then i divide that in
a medium where the speed of light is v
then the refractive index is n right so what
do i get so the velocity of light in a certain
medium which has a refractive index n which
has a refractive index n is nothing but c
by n where c is the velocity of light and
all the speed of light in vacuum and then
n is the 1 in n is the refractive index here
ok so the time it takes to go to to do the
distance ao what is that well let us um find
the distance or the distance ao so this is
x ok
so this is nothing but again h1 square by
x square now we need to divide this by the
velocity in medium 1 or the speed in medium
1 which is c by n1 because you are a medium
one now we need to add again so this is the
time it takes for light to cover travel assistance
ao ok now what about this distance ob well
that is h2 squared + l - x whole square divided
by c by n to remember the speed of light in
the in medium to why call that velocity v2
and that is again dependent on the refractive
index the speed and the c is the speed of
light in free space or the speed of light
in vacuum is not right
now we need to take the next team of this
so which means we simply take dt dx = 0 so
which tells us that well c and n and all these
things are constants here so we need to differentiate
this entire expression with with respect to
with respect to x so what is that so that
is n1 by c i am going to take it to x upstairs
and so and then a differentiate that that
will be so that is h1 square + x square so
differentiate 2x squares i get a 2x here and
then again h1 square by x square so this half
by this entire thing to the bottom 1 –half
so that is ok
so square root comes down and then i have
again in 2 by c the other one i get ok so
l - l - x so i will have a minus sign here
and so i get a 2 of l - x ok and again h2
squared + l - x whole squad ok ok so so what
is the point well what is the time light will
take once it goes from point a to b and via
this via this from one medium to another so
we will be calculating their time and the
time it takes to travel from a to aob we have
calculated that
now we are extra miser we are doing dt dx
= 0 we are trying to find out what is the
extremum of that we found a condition ok now
what is that condition
once again you just found the condition that
n1 over c remember n1 is the refractive index
in in medium 1 so time x by root over h1 squared
+ x squared and that should be equal to n2
over c root over in to over c and then l - x
root over h2 squared + x square ok now what
was the thing that we were doing we were we
were looking at refraction case from medium1
to medium2 ok
now this angle was i let us say that is the
angle of incidence and then this was height
h1 so this was x so this was a it said b and
then the angle of refraction was r ok so this
distance was h2 and so obq that is also r
now you have got our relation regarding some
the refractive index times something times
x by something some distance ok so let us
find what the angles is by the way so this
if this angle is the angle of incidence the
angle pao ok that angle is also i right
so in in triangle apo what do i have sign
i to be i have sign i to be x by the distance
ao which is nothing but root over h1 squared
+ x squared and in triangle boq what is the
the sine of angle r well sine of angle r nothing
but l - x divided by root over of h2 squared
+ x square therefore when we come back to
our old equation which we have got by equating
the time to be by by making the time to be
an extremum okay is if we substitute it back
into our old equation whenever this equation
here this one
what we get is n1 sine of i is equal to the
is equal to this value n2 sine of r okay where
i is the angle of incidence that is the angle
at which the light makes or the ray of light
mix with normal when it strikes the boundary
of this to medium and then r is the angle
that is the angle in which light emerges in
medium2 okay now this you know everybody knows
as this as snells law in optics
now see that we have simply obtained the snells
law in optics using fermats principle okay
and remember this mechanics problem is telling
you so if you wish to go from point a to point
b let us say and then the region one was land
and region two was water okay so what is the
what is the root you need to take when you
better take the route a or b okay that is
also the thing you can work out using the
principle of least action of the in mechanics
right
so to summarize what we have done today so
we have talked of the beginning of geometrical
optics we have talked of the the fundamental
guiding principle in geometrical optics which
is fermat’s principle which actually tells
you how the path of a ray of light okay so
we have done that for reflection from a plane
mirror and we have also seen the path of light
the ray of light will take when it goes from
one one medium to another okay
and in doing so what we have also found is
that we are right because we have reproduced
the familiar laws of reflection and the and
the familiar laws of refraction here find
that snell’s law so thank you very much
