L24 : Resolving power of Microscope - Wave Optics, Physics, Class 12 Video Lecture

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with part 24 let us similarly talk about
a microscope in case of telescope video
review distant objects so resolution
plays a very important role because when
you we are seeing distant objects there
is a good possibility that you can view
two or three different objects as one
object right so resolution plays a a
very important role in case of tennis
ohms whereas in case of microscope it is
more about my magnifying however
resolution also has a role to play here
because even a microscope sometimes it
happens that few particles of the same
object are placed so close by that you
are not able to distinguish between the
two of them even though it is lying
nearby but the size of the two points
are so small that they are almost
appearing as one so in that is
resolution is an important concept what
when distance between two points in a
specimen is comparable to the wavelength
of light then diffraction occurs so that
image which is formed is nothing but a
diffraction pattern that is quite
obvious so in this case what would be
the image formed in that case that will
be nothing but a fee into schita this
was the this would have been the normal
size of the image and this theta here
into picture because of the diffraction
pattern so this would be V into 1.2 to D
right so now objects which are closer
than this distance will not be resolved
so that means this is the minimum
distance for which the object will be
resolved to beyond this the object
cannot be resolved so therefore what in
case of a microscope what should be the
minimum separation between objects so
that it can be resolved so the minimum
separation between objects to get
resolved is let us do know
the minimum which is equal to V into 1.2
to lambda R divided by capital T this
divided by magnification by what was
magnification magnification is nothing
by but signs of image by size of objects
the size of image in this case is this
and science of object is nothing but the
minimum distance between the two objects
so this comes out to be one point two
two lambda divided by D into F where F
is the focal length by F because
magnification is nothing but size of
image V divided by signs of object so in
this case the object is lying at the
focal length which is f so therefore it
is 1.2 to lambda by D into F so now let
us suppose this is my objective and this
objective makes an angle beta at its
focus and what is this this is the
diameter so from this we can say tan
beta is equal to D by 2 divided by F so
this is equal to 1.2 thing F lambda
divided by
to F into tan betta sorry so here we get
tan beta is equal to D divided by 2 F
therefore D minimum can be written as
one point two two lambda divided by tan
beta into F into F so this F and F will
get cancelled so this is one point two
two lambda divided by 2 tan beta now
when beta is very small if beta is very
small in that case we can write tan beta
is approximately equal to sine beta
which is approximately equal to beta so
this can be written as one point two two
lambda divided by two sine theta
right so this is the minimum distance in
case of a microscope for which the two
objects will be resolved now if the
objects are closer than this distance
then they cannot be resolved so this is
D minimum now let us suppose that if the
medium between the object and the
objective lens this is that this is
suppose object-- and this is the
objective lens if this is filled with
some medium of refractive index n then
what happens in that case if medium has
some refractive index say new in that
case this D minimum becomes equal to one
point two two lambda divided by 2n sine
theta so this term n sine beta is known
as numerical aperture of objective
lengths so this is termed as the
numerical aperture of a microscope so
what what about the resolving power of a
microscope this resolving power is
inversely proportional to this minimum
distance because what is minimum
distance it is the it says that
if the separation between the objects is
this much this is the minimum thing that
can be resolved by the microscope so if
this is even lesser than this that means
that microscope has a better resolving
power so if it can even distinguish
between these two so that means it has
even more
resolving power so here if you see the
resolving power is increasing as the
distance that is d minimum is decreasing
so that means the resolving power is
inversely proportional to the minimum
distance of separation for the objects
to get resolved
so therefore resolving power is equal to
2m sine beta divided by 1.2 to lambda
where beta is the angle subtended by the
object of the objective lens right so
why do we conclude from here that the
resolving power of a microscope can be
increased by choosing a medium of either
refractive index because either angle
which the object will subtend
that will depend upon the location of
the object lambda is the wavelength of
light so if you want to increase the
resolving power giving everything else
same in that case you have to increase
the refractive index of the medium so
this is how we see that the concept of
diffraction plays an important role in
determining the resolving power of
optical instruments because here also
because here also in case of the
microscope we saw that it is because of
diffraction that the image which was
formed was nothing but a diffraction
pattern and therefore we could directly
apply the the concept which we had
arrived from diffraction that is the
size of the image will be P into theta
which is equal to P into the value for
theta right because this diffraction
pattern the study of this diffraction
pattern was possible only because of our
analysis of diffraction pattern and this
becomes useful in determining the
resolving power of telescope and
microscope thank you please visit exam
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