Tag: resolving power of optical instruments

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.) $\propto \lambda^{-1}$