Resolving Power Of Microscope And Telescope - Practice Questions & MCQ

Resolving power of optical instruments: 

Resolving power of an optical instrument is its ability to resolve or seperate the images of two nearby point objects so that they can be distinctly seen. In refrence to a microscope, the minimum distance between two lines at which they are just distinct is called resolving limit (RL) and it's reciprocal is called Resolving power (RP).

Resolving power of microscope: 

In microscope the minimum distance between two lines at which they are just distinct is called Resolving limit (RL) and its reciprocal is called Resolving power (RP)

$
\text { R.L. }=\frac{\lambda}{2 \mu \sin \theta} \text { and R.P. }=\frac{2 \mu \sin \theta}{\lambda} \Rightarrow R . P . \propto \frac{1}{\lambda}
$

$\lambda=$ Wavelength of light used to illuminate the object,
$\mu=$ Refractive index of the medium between object and objective,
$\theta=$ Half angle of the cone of light from the point object

 Rayleigh criterion for the diffraction limit to resolution states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other . We can use Rayleigh’s criterion to determine the resolving power. The angular separation between two objects must be: 

$\begin{aligned} & \Delta \theta=1.22 \frac{\lambda}{d} \\ & \text { Resolving power }=\frac{1}{\Delta \theta}=\frac{d}{1.22 \lambda}\end{aligned}$

   Thus higher the diameter d, better the resolution. The best astronomical optical telescopes have mirror diameters as large as 10m to achieve the best resolution. Also, larger wavelengths reduce the resolving power and consequently radio and microwave telescopes need larger mirrors.

Resolving power of microscope: 

The resolving power of a  microscope can be defined as the ability of the microscope to form separate images of two objects placed very close to each other. In a microscope, the minimum distance between two lines at which they are just distinct is called Resolving limit (RL) and its reciprocal is called Resolving power (RP). 

$
\text { R.L. }=\frac{\lambda}{2 \mu \sin \theta} \text { and R.P. }=\frac{2 \mu \sin \theta}{\lambda} \Rightarrow R . P . \propto \frac{1}{\lambda}
$

$\lambda=$ Wavelength of light used to illuminate the object,
$\mu=$ Refractive index of the medium between object and objective,
$\theta=$ Half angle of the cone of light from the point object

Therefore, from the above expression, we can see that, 

•  As the R.P is directly proportional to the refractive index (n), So  R.P will increases when n increases.
•   As the R.P is inversely proportional to the wavelength (λ), So  R.P will decreases when λ increases.
•   When the diameter of the objective is increased, θ increases. Hence, sinθ also increases.

           As the R.P is directly proportional to the sinθ, So  R.P will increases when the diameter of objective increases.

•  As the R.P is independent of the focal length of the lens, So R.P will remain unchanged when focal length increases.

Resolving power of telescope:

In telescopes, very close objects such as binary stars or individual stars of galaxies subtend very small angles on the telescope. To resolve them we need very large apertures. Resolving power of a telescope is defined as the reciprocal of the smallest angle subtended at the objective lens of the telescope by two point objects which can be just distinguished as separate. We can use Rayleigh’s to determine the resolving power. The angular separation between two objects must be 

$
\Delta \theta=1.22 \frac{\lambda}{d}
$

Resolving power $=\frac{1}{\Delta \theta}=\frac{d}{1.22 \lambda}$
where,
$\lambda=$ Wavelength of light used to illuminate the object,
$\mathrm{d}=$ is critical width of the rectangular slit for just resolution of two slits or objects.
$\theta=$ Half angle of the cone of light from the point object,

Thus higher the diameter d, better the resolution. The best astronomical optical telescopes have mirror diameters as large as 10m to achieve the best resolution. Also, larger wavelengths reduce the resolving power and consequently radio and microwave telescopes need larger mirrors.