Resolving power of optical instruments - class-XII

Resolving power for the instrument is found to be $\dfrac{\mu\sin\theta}{0.61\lambda}$ , UV light has short wavelength, hence higher resolving power. Oil is optically denser than air, that is, its $\mu$ is greater than that of air. Thus immersing in oil would increase the resolving power.

By definition, resolving power of an optical instrument is its ability to show two closely adjacent point (closely spaced) as distinct as possible.

the length of telescope =focal  length  of  object $(-f _0)$ +focal  length  of  eyepiece  $(f _e)$$=100+5=105cm$

In optical instruments, the lenses are used to form images by Refraction.

The image formed by the Compound microscope and Astronomical telescope is inverted,but in case of Simple microscope it form erect image.

In an Astronomical telescope, the objective lens has a greater radius than the eyepiece.

Resolving power of a telescope=$\dfrac{a}{1.22\lambda}$

Resolving power of telescope $R=\dfrac{1}{\Delta \theta}=\dfrac{a}{1.22 \lambda}$
where, $\Delta \theta$ is angular separation between two objects.
            $a$ is the diameter of the objective.
            $\lambda$ is wavelength of light.
So, clearly resolving power of a telescope depends on diameter of the objective.

Resolving power of telescope:
$R=\dfrac{1}{ \theta}=\dfrac{a}{1.22 \lambda}$

The normal pupil size of a human eye is 4 mm.which sets a minimum resolution approximately 1' to 2'.we want to pull small objects as close to our eyes as possible to be able to see them, but there is a minimum distance of comfortable viewing which is roughly at 25 cm. Hence, correct option is A.

The biggest advantage of an electron microscope over optical microscope is that they have a higher resolution and are therefore capable of a higher magnification ( up to $2 \ million$ times ). 

Resolving power of a telescope is more if the diameter of the objective lens is more because $R=\dfrac{a}{1.22 \lambda}$
where, a is diameter of the objective. objective lens of large diameter collects more light but does not increase the resolving power of the telescope because resolving power increases when angular separation increases.

Resolving power of a telescope:
$R=\dfrac{a}{1.22 \lambda}$
where, $a$ is diameter of the objective
so, $R$ increases when a is increased and $a$  increases when aperture of objective is increased

Resolving power, $R=\dfrac{a}{1.22 \lambda}$
where, $a$ is diameter of objective $\lambda$ is wavelength of light
magnifying power $m=\dfrac{-f _{0}}{f _{e}}\left ( 1+\dfrac{f _{e}}{D} \right )$
so, decreasing the wavelength of light increases the resolving power and magnifying power of telescope.

$\dfrac{1}{2}mv^{2}= eV$

$mv= \sqrt{2eVm}$

And $\lambda = \dfrac{h}{mV}$

$\dfrac{\lambda _{0}}{\lambda _{1}}= \dfrac{\sqrt{2eV _{1}m}}{\sqrt{eV _{2}m}}$

$\dfrac{\lambda _{2}}{\lambda _{1}}= \dfrac{1}{2}$

$\therefore \lambda _{2}=\dfrac{\lambda _{1}}{2}$

$R\ \propto \dfrac{1}{\lambda}$

so $R$ would change to $2R$.

     $R= \dfrac{9}{1.22\lambda }$

$\dfrac{1}{\Delta \theta }= \dfrac{5}{1.22\times 4000\times 10^{-10}}$

  $\Delta \theta = \dfrac{1}{50}sec$

$R= \dfrac{1}{\Delta \theta }= \dfrac{a}{1.22\lambda }$

$\dfrac{1}{\Delta \theta }= \dfrac{0.10}{1.22\times 5000\times 10^{-10}}$

$\Delta \theta = 6.1\times 10^{-6}\ rad$

Limit of resolution $\sin { \theta  } =\theta =\cfrac { 1.22\lambda  }{ D } $
putting the values

Resolving power(R.P.) $\propto \lambda^{-1}$

We have,
$ RP=\dfrac { D }{ 1.22  \lambda}$

$ \dfrac { 1.22\lambda  }{ 3mm } =\dfrac { 1mm }{ d } $

The mentioned instrument is compound microscope and its resolving power is $ R.P=\dfrac { 2\mu \sin { \theta  }  }{ 1.22\lambda  } =\dfrac { \mu \sin { \theta  }  }{ 0.61\lambda  } $

Resolving power is given by the distance between two objects to be distinguished per unit distance of objects from the object distinguishing them.

$\sqrt{\dfrac{16}{4}}=\dfrac{R}{x}$

Resolving power of an optical instrument $\displaystyle \propto \dfrac {1}{\lambda}$

Resolving power $= \displaystyle \dfrac{\lambda}{d \lambda}$ plane transmission granting 

Larger aperture help in high (increased) resolution of telescope and large amount of light collection.

The resolution of the human eye is the smallest object our  eye can see. This is limited by the diffraction limit, which is approximated by the angular size ratio of the object's size versus the distance to the object.
The normal pupil size of a human eye is 4 mm, which sets a minimum angular resolution of the eye  and to able to see the small objects we bring them as close to our eyes as possible, but there is a minimum distance for comfortable viewing which is roughly at 25 cm.But quoted figure for the smallest resolvable size is 0.1 mm, showing that the diffraction limit is a crucial factor in visual resolving power.

If the angular limit of resolution of human eye is R then
R = $\displaystyle\frac{1.22\lambda}{a}$ = $\displaystyle\frac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}}$ rad
         
      = $\displaystyle\frac{1.22 \times 5 \times 10^{-7}}{2 \times 10^{-3}} \times \frac{180}{\pi} \times 60$ minute = 1 minute
  

magnifying power of telescope  (M P)$=8$
The magnifying power of an astronomical telescope is $=\cfrac{focal  \ length \ of \ object (-f _0)  }{focal \ length \ of \ eyepiece(f _e)}=8$
hence, the ratio of the focal length of the objective to the focal length of the eyepiece is $8$.

We have, $\dfrac{y}{D}\ge 1.22 \dfrac{\lambda}{d}$

The resolving power of an instrument is given by the formula, $R.P=1.22\times \frac{\lambda D}{d}$
Here, d is aperture of the instruments, D is distance of satellite from the earth. Here eye is the optical instruments.
$\displaystyle R.P=\frac{1.22\times 500\times 10^{-9}}{5\times 10^{-3}}\times 400\times 1000$
          $\displaystyle =1.22\times \frac{10^{-2}}{10^{-3}}\times 4 = 1.22\times 40=50 m$

The resolving power of a telescope increases as diameter of objective lens increases.
Resolving Power =D1.22λD1.22λ
where D is diameter of objective and λλ is wavelength of light used.

High-quality optical systems use combinations of concave and convex lenses (achromatic doublets) to correct for chromatic and spherical aberrations.

For a microscope, the limit of resolution is given by,
 $X =  \dfrac{\lambda}{2A} $
where $ \lambda $ is the wavelength of light used, and A is the numerical aperture.
Hence, substituting the values,  $X =  \dfrac{600}{2 \times 0.12} $,
which gives, $X= 2.5  \mu m $

Resolving power of eye = $ 1' = \dfrac {\pi} {60 \times 180} radians$ (approximately)

Resolution power $= \dfrac {d\lambda}{D} = \dfrac {1000 \times 5000 \times 10^{-10}}{10 \times 10^{-2}} = 5mm$

The aperture of an astronomical telescope is defined as the diameter of the objective lens.
In telescopes since, the stars are very far away from away us and emit light internsities, we need a large aperture to increase the amount of light entering the telescope thereby increasing the resolution

Limit of resolution of an optical instrument is the minimum angle that two points on an object have to subtend at the eye so that they are just resolved. (C)

we know,mirror formula
$\cfrac {1}{f}=\cfrac{1}{v}+\cfrac{1}{u}, magnification(m)=\cfrac{-v}{u}$
$\cfrac{1}{u}=\cfrac{1}{f}-\cfrac{1}{v}$
$\cfrac{1}{u}=\cfrac{v-f}{fv}$
${u}=\cfrac{fv}{v-f}$
$(m)=\cfrac{-v}{u}$,substituting $u$.
$m=\cfrac{-v}{1}\times\cfrac{v-f}{fv}$
$m=\cfrac{f-v}{f}$

Limit of resolution of telescope = $\dfrac{1.22 \lambda}{D}$
$\theta = \dfrac{1.22 \times 500 \times 10^{-9}}{200 \times 10^{-2}} = 305 \times 10^{-9}$ radian

Electron microscope is a microscope that can magnify very small details with high resolving power due to the use of electrons as the source of illumination. Since the wavelength of electrons are 100,000 times shorter than visible light the electron microscopes have greater resolving power
We have the Abbe's formula as resolution limit $d=\dfrac{0.61\lambda}{NA}$(NA is the numerical aperture)

Here, $f _o+f _e=36 cm$
$M=-8$ (magnifying power is negative)
Now, $M=\cfrac {f _o}{f _e}$
$\therefore -8=-8=-\cfrac {f _o}{f _e}$
$\Rightarrow f _o=8f _e$