Real and Apparent Depth - Physics, Class 12 Video Lecture

we are dealing with the topic real and
apparent herbs and it is the result of
the refraction process which happens
from the surface in this we have
observed in fact all the time when we
observe anything which is inside a
liquid and we are observing it from a
different medium say it is air then we
look to the object and we observe it add
an apparent position now this apparent
position is the result of the refraction
which is taken placed at the interface
of the two medium now you can see here
that object o is lie down at the base of
the container now this container is
filled up of water say it is water now
this object oh when it is observed are
they just above point D then we just do
not observe any shift of the position
but when we observe it from a different
point than the D then we can observe its
apparent position at ah now where the
refracted ray a B is met another
refracted ray CD when it is extrapolated
back and this point is I and we observe
the another position or the apparent
position of object o at I now if we
observe this diagram then simply we can
say that whenever the incident ray or a
leaf falls on the interface of the water
and air then it reflects away from the
normal and go through the park AV now
this angle is the angle of incidence I
and this angle is instead
is the angle of refraction now if we go
through the geometry then these join the
shells the alternate interior angles
that's why these two angles should be
same well this angle R is sharing the
corresponding angle with this angle
c.i.a and that's why these two angles
should have the same value defi now if
we consider the triangle see a 0 then in
triangles CA 0 sine of angle I is equal
to this is the angle I this range c 0
behaves like the base and see a is the
perpendicular while a 0 is the
hypotenuse as the sine of angle I is the
ratio of perpendicular to the hypotenuse
that's why I sign I is equals to see a
upon ok now if we consider the another
triangle which is the triangle see a I
then it is also a right angle triangle
and from this triangle we can get the
value of sine of angle of refraction and
this sign r is equal to it is CA which
is the perpendicular again and here in
this triangle the hypotenuse is a I now
you have kept the two equations equation
1 which is dealing with the value of
sine of angle of incidence in equation 2
which is dealing with the sine of angle
of refraction if we apply the Snell's
law at the interface then we can write
my snails long sine of angle of I chose
a sine of angle of refraction should be
equivalent to refractive index of second
medium which is here
with respect to what or we can say that
sign I value is ca upon OA in to see a
bond I and it is equivalent to one by
refract your neck soft water with
respect to hear so this value reduces to
retract of an ex of water with respect
to air is equal in 20 a coupon I a now
if we observe this diagram then here
this point oh and this point a if it is
compared with the OC then we can say
that C and a are the very cloves points
OA is equivalent to OC and AI is equal
to AC so applying the assumption of C is
very close to a then we can say that OA
is equivalent to OC and oh I is equal in
two I see so this equation radio system
refractive index of water is equal in 20
see upon I see
where oh say is the real depth and the
ICS apparent app so this equation again
we can write it as 0 refractive index of
water is equal in to readapt upon a
parent so students let's take the
lemarchal based on the reflection and
refraction of light as we have studied
the refraction so let's start with a
marker on refraction let's have an
American and in this numerical we have
given the medium and the light ray is
moving from air to glass its frequency
is given that is a 6 into 10 / 14 Hertz
and we have to calculate its speed and
wavelength first per glass medium and
second for air meeting as we know the
frequency remains the same when the
light rate reflects or refracts so let's
calculate the velocity the speed and
leave a link for the light the
refractive index of glass with respect
to air it is given and it is 1.5 as
refractive index is a scalar quantity
and as well as it is unitless and
dimensionless now let's solve
this Christian as we know that the
velocity in medium 1 with respect to
velocity in medium 2 is given as the
refractive index of the medium so in
this question we have the velocity of
air with respect to velocity in the
glass as you know the relation between
the speed of light in first medium to
the speed of light in second video is
given by this formula where mu is the
refraction index of medium 2 with
respect to one so we can have for Erin
glass which is the combination of the
media and it is given in this question
so it is the ratio of the velocity of
the ratio of the speed of light in air
to the speed of light India a glass is
equal to 1.5 which is the refractive
index of glass with respect to air and
it is given in the question so the
velocity the speed of claw arm the speed
of light in glass is to be calculated in
the first part and the first part deals
with glass as a medium so VG that is a
speed of light in mass is equal to VA
upon would point 5 and as we know that
this fear of light in air or vacuum is a
constant and it is equal into 3 into 10
to power 8 meter per second so let's put
the value and find out the speed of
light in glass it is equivalent to 2
into 10 power 8 meter per second so the
one part for the part first of the
numerical has been solved and the speed
is calculated which is equal to 2 into
transfer 8 meter
second for the wavelength of light in
glass medium we can use the formula
which deals with the speed and the
frequency which is denoted venue and the
wavelength of the wave so they had the
speed we have the speed border right in
glass that is 2 into 10 power 8 meter
per second and we have to find out the
wavelength and if we shift the frequency
here it will divide with the velocity it
will be divided with the velocity and
velocity so it will be divided with the
speed of light and it is given 6 into
transfer 14 Hertz that is per second
then it is so lambda is equal into when
my 3 of 10 is to power minus 6 10 to
power minus 6 is the raising of 10 and
the unit is in meter so the wavelength
in glass is equivalent to point 3 3 into
10 to power minus 6 meter so this is the
value for the wavelength of light wave
in the medium glass so the first part of
the question has been discussed and we
have got
the wavelength value as 0 point 3 3 into
10 power minus 6 meter well the speed
magnitude as tinto into transfer 8min
Aversa be so let's now calculate the
speed and the wavelength for the error
now for air we know the velocity is we
know the speed is VA is equal to 3 into
tons per 8 meter per second so the
wavelength can be calculated by the
formula v EI upon frequency of light as
it is not changing for the media so VA
value is 3 into 10 to power 8 and
frequency is given as 6 into 10 spot 14
Hertz so when we rearrange the numerical
data and calculate it then we'll get it
as point 5 into 10 to power minus 6
meters so it is the value of wavelength
for your and finally we know that the
speed of a speed of light in air a
strange trance / 8 meter per second
let's have another new Merkel which
actually is based on the real and
apparent depth so in this question we
are dealing with the real and aprender
comes it with
Chile takes place because of the
refraction of light now the question
gives us that we have about a time and
in this water time the depth of water is
equivalent to four meter also it is
given that this water is having the
refractive index equivalent to four by
three now we have to tell that how much
distance does the bottom of ten appear
to be raised so to get the solution we
can just apply the formula which we have
studied in the lobster refractive index
is equivalent to three left upon the
actual depth here the refractive index
is of water with respect to air now we
have the relation between the relative
refractive index and they drill that and
the abundant of the object in a liquid
that are I supposed to reflect to index
is supposed to real depth upon apparent
depth if you have to calculate the
apparent depth then we can rearrange the
formula and can get a print weight
apparent depth as real that upon the
refractive index in the question we are
given that the refractive index of glass
with respect to water it is equal to 4
by 3 and the real depth is also given
and it is equivalent to four meter so
when you put the venues in the formula
you get the value of a parent depth
equivalent to
for me turn phone 4 by 3 and it is
equals to 3 meet so this numerical is
also discussed and it gives us the
parent depth of the water bottom of the
bottom of the tank but as the question
asks us that how much distance the water
tank is seem to be raised so we do not
have to tell that what is the apparent
depth we have to tell that how much the
level is raised so let's make a simple
calculation that if the actual tap is
equals to but if the rain that is equals
20 meter and the up green death is
equals to 3 meter then the level raised
is equal to 4 minus 3 that is 1 meter so
these were the simple blue iris based on
the concept of refraction