Numerical: Refraction at Spherical Surface - Ray, Optics & Optical Instruments Video Lecture - Class 12

hello friends this video on ray optics
part 27 is brought to you by exam for
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make sure that you have watched all the
videos to the part 26 before going ahead
with part 27 so on few problems on
refraction by spherical surfaces so here
comes problem 8 in this problem
determine the image formed by
diffraction in this problem determine
the image formed by refraction so here
again we have a spherical surface so
this is the object okay so we have to
find out the image formed by diffraction
so in this case V is what we have to
calculate so yet u that is the distance
of the object is given as 25 centimeters
this will be negative because it is
opposite to the direction of the
incident ray the radius of curvature R
is given as 20 centimeter because R is
nothing but the distance of the centre
of curvature from the pool so this would
be 20 centimeter and this would be
positive mu 1 is equal to 1 and mu 2 is
equal to 1.5 so now we will apply the
expression which we derived in the
previous slide that is the refraction
through a spherical surface so we found
that N 2 by V minus N 1 by U is equal to
n 2 minus N 1 divided by I so now if we
put these values which are given above 1
point 5 divided by V minus 1 divided by
minus 25 is equal to n 2 minus N 1
divided by R so we get 1 point 5 divided
by 3 plus 1 by 25 is equal
to 0.5 divided by 20 so we get 1 point 5
divided by V is equal to 0.5 divided by
20 minus 1 by 25 so this value comes out
to be minus 1 point 5 divided by hundred
or we can say that V comes out to be
minus 100 centimeter so what we see that
the image is formed at a distance of 100
centimeter from the pole and since the
sine is negative that means this
negative sign shows that image has
formed on left hand side of the pole
right so let us look at the neck let us
look at the next problem it says that a
small pin fixed on a tabletop is viewed
from above from a distance of 50
centimeters by what distance would the
pin appear to be raised if it is viewed
from the same point through a 15
centimeter thick glass slab held
parallel to the table refractive index
of glass is equal to 1.5 does the answer
depend on the location of the slab so it
is the if this problem is related to the
real depth apparent depth concept there
we saw that the fisherman or the eye was
located at the same position just that
due to the presence of the water that is
a different medium the depth of the fish
changed
similarly here due to the presence of
the glass slab the distance or the
position of the small pin will change so
let us suppose that the distance by
which the pin appear to be raised by
which pin appear to be raised is equal
to a
so that means what is X X is nothing but
real depth that means the actual depth
of the pin - apparent depth that is the
depth which we are viewing so X is
nothing but real depth - apparent depth
that is ad minus a now we already saw
what do you fact it would be oh god did
we observe there we saw that the ratio
of real depth to apparent depth is equal
to the refractive index of the medium so
here the medium is glass so this is
equal to Meucci so we can say that
apparent depth is equal to real depth
divided by refractive index of glass
therefore we can say X is equal to R
minus R by mu G so what is this real
depth this real depth into 1 minus 1 by
mu G so what is the real depth the real
depth is nothing but the thickness of
the glass slab which is given here as 15
centimeter because the location from
where we are viewing the small pin that
really doesn't matter because for
example
oh please compare it with the scenario
which I showed you in real depth
apparent depth that is the fish example
there the eye was located somewhere here
so the location of the eye I mean how
far the ID is located from the object
that really doesn't matter what matters
is the amount of water or the thickness
of the water which is present there so
here ad is nothing but the thickness of
the slab which is introduced so what is
the thickness of the slab here it is 15
centimeters and this is 1 minus 1 by mu
G that is 1 going to 5 so this comes out
to be 5 centimeters so that means this
object appears to be raised by 5
centimeter as the answer depends on the
location of the slab no right
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