Refraction through glass slab - Physics Video Lecture - Class 11

after having the previous knowledge of a
reflection reflection and refraction now
we are going to discuss our first
specific particular topic the refraction
through a compound slab so for this
topic let's first we should know the
people should know that what is a
compound / the compounds lab it is a
rectangular bounded media in which there
are more than one medium for example
there are two medium other than air let
this is the water and this is the class
outside this compounds left there is air
now let's have the light rain which
first of all incidents on the interface
between the air and the water when it
falls on the interface this way say a B
just bands towards the honour as the
water is denser medium than the air let
the incidence angle B I and refracted
angle br1 now when this way bc goes
forward and falls on the interface
between the water and the glass of face
then we can say that there is a normal
again and this ray bc bands towards the
normal again as the glass is more denser
medium than the water so here the light
ray bans towards the not now we come to
know that this angle was arwen and by
the geometry we can see that
angle should also equal to the value our
one because the normals say it is n1 and
it is n2 these two normals are parallel
because these surfaces are also parallel
to each other thus this bc behaves like
a transversal and the two angles here
this angle n 1 dash b c & M 2 c-b both
are equal as they are alternate interior
angles now when the Raby see bands
towards the normal when entering into
the glass medium then we can say that
this angle is a when you having we can a
name it are two now this CD ray when
falls on the interface between the glass
and the air medium then it suffers
through the third refraction and as air
is a very medium than the glass this ray
bands away from the non what we observe
here is that this angle should be again
equal into our to as the reason has been
discussed earlier now we should see an
examine its geometry as we know that the
refraction is followed by Snell's law
sales law says that sine of angle of
incidence when compared with the sine of
angle of refraction then for a
particular set of medium this value
always remains the constant which is
said to be refractive index of medium 2
with respect to one now we want to have
the relationship between the refractive
index in fact a relative refractive
index of water with respect to air of
glass with respect to water and of glass
with respect to air so for getting a
common relation between the three let's
start with the ray incident ray a B when
it falls on the interface and just
reflects we can say that at the first
interface here we can apply the Snell's
law and by Snell's law we can write sine
of angle I to the sine of angle R 1
should be equal to refractive index of
water with respect to air let this is
the equation 1 a second interface we get
applying the Snell's law that sign of r1
to the silent of our to when the ratio
is taken it should be equal to the
constant which is the refractive index
of glass with respect to water now this
equation can be numbered 2 here we can
see that for the second interface are
one was the incident angle and otto was
the refracted i know are the third
interface where
garre CD falls on the interface between
the glass and the air media we can apply
the snails login and here the angle of
incidence can be taken r2 and let the
array which emerges out of the glass
medium say it is de when it emerges out
then we can say that this angle of
emergence is e-sign r2 r2 is the
incidence angle for the third interface
and sign of emergent angle e which is
the angle of refraction for the third
interface and this ratio according to
Snell's law should be equal to
refractive index of error with respect
to class now this equation may be
numbered with three let's compare the
three equations together and then get a
vertical relation between the relative
refractive index of water glass now from
the equation one equation to an equation
3 let's multiply the LHS and RHS of the
three creations together we'll get sine
of I upon sine of r1 into sine of R 1
upon sign of artoo into sign of artoo
upon sine of e this value should be
equal to refractive index of water with
respect to L into refractive index of
glass with respect to water and
refractive index of error with respect
to class now if you see the LHS of the
equation
we'll find out that this value adjusts
with this value and sign our to adjust
with the sign are 2 in the numerator and
when they compared the geometry we find
out that angle I is equivalent to angle
e that is angle of incidence is equal to
angle of emergence now as these two
angles are equal thus their ratio should
also be equal to 1 so here we get from
the above equation that refractive index
of water with respect to ear into
refractive index of glass with respect
to water should be quiz 2 1 upon
refractive index of air with respect to
class now as we know that refractive
index of medium 2 with respect to one is
always equal into wrist broken off
refractive index of medium 1 with
respect to 2 so applying this
information here you can write this
above equation as input we can rearrange
the equation as refractive index of
glass with respect to water is equal to
absolute refractive index of glass upon
absolute refractive index of water so
for the compounds lab we can have this
relationship which says that the
relative refractive index of glass with
respect to water or any medium any
medium 2 with respect to one is
equivalent to the ratio of the absolute
refractive index of medium 2 with
respect to one so weird it is the
relationship which we are getting for
the compounds lab now let's have then
all the topic for today it is the
refraction study for the glass lab here
the glass slab is again a cuboidal
bounded surface and here it is made up
of the glass we want to know that how
the light ray will behave when it goes
through a glass slab when light ray a B
forms on the interface between the glass
and the air then as you know the glass
medium is denser than the air medium
that's why this a B ray goes under the
refraction and due to the refraction
this ray-ban towards the non c'è it is
bc now for the first interface let this
is the first interface and this is the
second one at this interface the angle
of incidence is I and this angle is
equivalent to the angle of refraction
and we are naming this angle as smaller
now this is the first normal for the
interface 1 n 1 n 1 dash and when the
rate bc falls on the second interface it
should get support through the second
refraction for the second diffraction we
have normal say n 2 n 2 dash and the
rain which comes at the second interface
it goes on that it bends away from the
normal so the race CD after going under
the refraction
been away from the normal when it again
enters into the air from the glass as
the air is rarer medium in the glass so
here the emergent ray which emerges out
from the glass slab we say that this
angle of emergence is e now again when
we are having the refraction at
interface one and inter phase two then
the Snell's law should be obeyed by the
two processes at interface one by
Snell's law we can write the ratio of
sine of angle of incidence to the sine
of angle of refraction this value should
be equal to the refractive index of the
ass with respect to air and that is the
equation 1 now by the geometry we can
see that this angle n 1 dash b c and n 2
c-b that is this angle should be the
same value because of being the
alternate interior angles between the
two parallel lines which is invalid
manage and in to enter dish bc behaves
like a transfer ship and that's why
these two angles are the angles
alternate interior angles and that's
where should be the same value so we are
at the interface to by the application
of snails no again we can have that the
sine of angle of refraction for the
first interface this value is the angle
of incidence for the second interface so
the ratio of sine of angle of incidence
for the second phase two days sine of
angle of emergence should be equal to
refractive index of air with respect to
glass no equation to it is and here we
can see that we have written
here refractive index of glass with
respect to hear because the second
medium for the first interface is glass
where the first one is here but for the
second medium this pair exchanges its
position and the error is a second
medium while glass is the first one so
from Equation 1 n 2 week right here that
sine of angle of incidence to the sine
of angle of refraction for the first
medium multiplied by the sine of angle
of our to the sine of angle of e should
be equal into refractive index of glass
with respect to air to the refractive
index of air with respect to glass as
this product should be equal to 1 that's
why here having these two value get
adjusted with each other so we find out
that angle I should be equal to angle e
as this ratio can be equivalent only
when the sine of angle of incidence
should be equal to sine of angle of
emergence and that's why we get the
result that angle of incidence remains
equal to the angle of incidence at the
result that angle of incidence remains
equal to the angle of emergence and
because I mathematically derived this
result we can see that the angle of
incidence when it is equivalent to angle
of emergence then the incident ray
remains parallel to the emergency we can
deal with so important concept over here
so let's see if this glass slab was not
here then this ray a B should move in
this particular direction which can be
original incident direction now because
of the same values of angle I am Inga II
we come to know that these two
directions one is the original direction
of the incident ray and the other one is
emergent raised direction these two are
parallel from each other and this
shortest distance between these two
parallel lines is said to be literal
shift yes this shortest distance between
these two parallel lines is said to be
literal shift or the literal
displacement can we have the numerical
or mathematical formula for this literal
the answer is yes and it is our
next step in the topic refraction
through the glass slab yes for getting
the mathematical formula for little
shift they are going through the diagram
this diagram once again and for getting
the value of little shift we have
extrapolated array a line from C to D
now we will deal the two triangles and
both triangles are the right angle
triangles one is be d/c and one is b/c
m1h so let's deal the first triangle
which is B D and C you can see here that
this angle of incidence is sharing a
vertically opposite angle with this
angle and when they should be d now if
this whole angle should be equal to I
and this particular angle of refraction
is our then the remaining angle which is
angle DBC should be equal to the
difference between the two that is I
minus R now we can say that in triangle
be d/c which is a right angled triangle
sine of angle i minus i should have the
value in terms of the perpendicular and
the hypotenuse it should be equal in to
c d upon b c where c is the
perpendicular and bc is the half
continues now we can see that this CD is
the little shift and we have to find out
this value so let's say this else now we
can write this value s as upon bc let
this is the equation 1 now in triangle b
c and one dash the angle R if we take
the cause of this angle I say cause our
then this value should be crystal based
upon the hypotenuse hear the bass is bn1
dash and the hypotenuse is bc then we
can write it bn1 dash upon the sea or we
can get the value of bc by rearranging
the equation and it should be equal in
to be n 1 dash of on cause of our let
this equation we named it
2 question 1 and 2 we can put the value
of BC from equation 2 in equation 1 and
we can write it as sine of i minus R is
equal in 2 s upon bc where bc is be in
one dash upon cause of our but we are
interested in the value of s so let's
rearrange this equation and should write
n since equals to be and one dash into
sine of i minus r upon cost of our now
if we see this diagram then B and one
dash this value is equivalent to the
thickness of the glass slab if we write
this thickness as T then in this
equation we can write that the lateral
shift s so this is the final equation
for the literal shift and we can see
that if this literal shift can have the
maximum value then it should be
equivalent to the thickness of the glass
slab t when the sign of i minus r is
equal to 1 and cause of our is also
equal to 1