in the previous lecture we saw how when a
mechanical wave coming on a string goes from
one medium to the other the other media being
another string so that there is a boundary
there is a reflection in a similar manner
we are now going to see that when electromagnetic
waves fall from on a surface from one medium
to the other there is going to be reflection
we are going to assume right in the the beginning
there is a reflected wave there is an incident
wave and there is a transmitted wave let us
see one medium on one side has permittivity
epsilon one on the other side has the permittivity
epsilon two and let's take them to be nonmagnetic
so that there permeability is same mu zero
when the wave come from the left hand side
let's take its e field to be shown by red
and arrow and because e cross b should give
me the direction of propagation i'm going
to show you the b filed by green colour is
going to be pointing this way so this is the
b field and e field is shown on that on the
other hand when it gets reflected suppose
i take e to be in the same direction as the
incoming wave then e cross b should give me
the direction of propagation and therefore
now the reflected be filled is going to be
the other way so this is b reflected let's
call the first one be incident this is e incident
and e reflected at the same time in the transmitted
wave there is is going to be e transmitted
and it is going in the same direction as the
incident wave i am going to have b incident
b transmitted as shown here
in the same direction as b incident so we
have set up what the electric field and magnetic
fields are at the boundary and now we need
to apply the conditions the boundary conditions
are e parallel is the same on both sides 
this arises simply from stokes theorem as
follows if i look at this boundary and make
a very small loop here the e field on this
the side like this on the side like this then
e dot d l is going to be equal to integration
curl of e dot d a curl of e is nothing but
d b d t it with a minus sign dot d a and this
is e on the left hand side let me write left
minus e r because we are traversing the loop
in two different directions however as we
shrink the loop we make the size the red shaded
region almost zero way right hand side becomes
zero and therefore e on the left should be
equal to e on the right side is same on both
side
and this gives me the first equation which
is e r i plus e r is equal to e transmitted
this is equation number one i will refer back
to the earlier equation that we had arrived
in the string which was y incident plus y
reflected y e is equal to y transmitted this
similar to that in a similar manner b parallel
on both sides is equal y is that that again
follows from stokes theorem because if i now
take a loop like this parallel to b then i
have b dot d l across this loop is equal to
curl of b dot d a which is equal to mu zero
displacement over d t dot d a as the area
second to zero this y goes to zero and left
hand side becomes zero which again gives like
the argument for electric field that b on
two sides must be equal
so the this on this boundary i have e e transmitted
and e reflected 
and i have b incident b reflected and b transmitted
the two equations are e incident plus e reflected
e is equal to e transmitted this my equation
one and the other equation is b incident b
reflected in the other direction b reflected
is equal to b transmitted this is equation
two but now i know from electromagnetic theory
that b incident is nothing but e incident
divided by v one in that medium minus b reflected
is nothing but e reflected over v one in that
medium this should be equal to e transmitted
over v two in that medium you may be little
un comfortable because i did not do this equation
for a wave in the medium but the relationships
are the same and this can be obtained very
easily by writing maxis equation the medium
which are let me just write the among the
side for completeness
which are divergence of d is equal to zero
and the absence of any charge curl of e remain
the same as the minus d b d t curl of b is
again there is no free current the everything
comes from displacement current is going to
be mu zero mu zero is the same d d by d t
and divergence of b is equal to zero so this
again if manipulate the relationship that
b is equal to e by v and this this immediately
to v two e i minus e r is equal to b one e
transmitted this my equation two i have got
an two equations to unknowns e transmitted
and e reflected and i can solve for them
let us multiply equation one by v one and
subtract equation two this gives me v one
minus v two e incident plus v one plus v two
e reflected is equal to zero and this gives
e reflected is equal to two minus v one over
v two plus v one e incident exactly similar
relationship in terms of velocity is in the
two medium as it was for string as similarly
e transmitted will be given by nothing but
e incident plus e reflected which will give
me two v two over v two plus v one e incident
so we have got in this medium which is epsilon
one and go inside epsilon two on the other
side and therefore there is some reflection
and transmission which are incident wave does
not transmitted completely and we have got
e reflected it was v two minus v one over
v two plus v one e incident and e transmitted
is equal to two v two over two plus v one
e incident
this two relationships are very similar to
the relationship obtain for waves on a string
notice that if v two is less then v one this
implies e r has opposite sign to e incident
that means it's reflects direction on or there's
a pie face change let us translate this equations
to equations in terms of the refractive index
now i know v is c y n and therefore e reflected
is going to be n one minus n two over one
plus n two e incident and e transmitted is
going to be two n one over one plus n two
e incident again you notice that if n two
that is a medium two have larger refractive
index then medium one the reflected wave has
a face change this you been taught in you
eleventh twelfth but now we see that this
are arries basically the boundary condition
on electric field and magnetic field
what about energy transmission so this waves
come this wave is coming gets transmitted
gets reflected i should have the pointing
vector is coming in incident should be equal
to the energy which is transmitted plus the
energy which is reflected s reflected plus
s transmitted incident one is going to be
v one times u energy density incident one-half
they should be equal to one half v one u reflected
plus one half v two u transmitted which is
indeed what the energy flow is this half cancels
this is going to be is going to be equal to
v one times epsilon one e incident square
and the right hand side is going to b one
one time epsilon one times e reflected square
plus v two epsilon two times e transmitted
square
let us see this is true let's calculate the
right hand side right hand side is v one epsilon
one e reflected square plus v two epsilon
two e transmitted square which is equal to
v one epsilon one epsilon one times v two
minus v one square over v two plus v one square
plus epsilon two v two times four v two square
over v two plus v one square times e i square
now i know that v one over square root of
epsilon mu keep in mind that mu is a same
and therefore v square epsilon one over mu
zero for both the medium and therefore i can
write this whole thing as v one epsilon one
that is v two minus v one over v two plus
v one whole square plus epsilon two v two
four v two square over v two plus v one whole
square as one over mu zero one over v one
v two minus v one over two plus v one whole
square plus epsilon two square is one mu zero
four v two over v two plus v one whole square
this is equal to one over mu zero v one v
two plus v one whole square and inside i get
v two minus v one whole square plus four v
one v two this term is nothing but v one plus
v two whole square in this cancels with this
term and therefore i get one over mu zero
v one so this whole things add up to one over
mu zero v one e incident square which is nothing
but equal to v one epsilon one e incident
square as zero we can amazing so you see that
our amplitude are such that they also satisfy
energy conservation so what are shown you
in this lecture is through the boundary condition
when an is incident electromagnetic wave comes
it gets both reflected as well as transmitted
the ratios of transmitted and ah reflected
electric field have been calculated and you
also shown through those that energy is reconserved
