Resolving Power Of Microscope And Telescope

Resolving Power of Optical Instruments

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

Resolving Power of Microscope

The resolving power of a microscope is a measure of its ability to distinguish between two points that are close together. It is determined by the wavelength of light used and the numerical aperture of the microscope lens. A higher resolving power allows scientists and researchers to observe fine details and structures in biological specimens and materials, revealing insights at the cellular and even molecular levels. This capability is crucial in fields like biology, medicine, and materials science, where understanding minute details can lead to significant discoveries and advancements.

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

R.L. $=\frac{\lambda}{2 \mu \sin \theta}$ 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,

$\mu=$ The refractive index of the medium between object and objective,

$\theta=$ Half angle of the cone of light from the point object

Rayleigh's criterion for the diffraction limit to resolution states that two images are just resolvable when the centre 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.

$\Delta \theta=1.22 \frac{\lambda}{d}$
Resolving power $=\frac{1}{\Delta \theta}=\frac{d}{1.22 \lambda}$

Thus, the higher the diameter d, the 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.

Therefore, from the above expression, we can see that

• As the R.P. is directly proportional to the refractive index (n), The R.P. will increase when n increases.
• As the R.P is inversely proportional to the wavelength (λ), So R.P will decrease 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, The R.P will increase when the diameter of the objective increases.
• As the R.P. is independent of the focal length of the lens, R.P. will remain unchanged when the focal length increases.

Resolving Power of Telescope

The resolving power of a telescope is its ability to distinguish between two closely spaced objects in the sky, such as stars or planetary details. It depends on the diameter of the telescope's aperture and the wavelength of light being observed. A telescope with high resolving power can reveal fine details of distant celestial bodies, such as the surface features of planets, the structure of galaxies, and the separation of binary star systems.

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. The 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

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

where,

$\lambda=$ Wavelength of light used to illuminate the object

d = is the critical width of the rectangular slit for just the resolution of two slits or objects.

$\theta=$ Half angle of the cone of light from the point object

Thus, the higher the diameter d, the 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.

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Solved Examples Based on Resolving Power of Optical Instruments

Example 1: The value of the numerical aperture of the objective lens of a microscope is 1.25. If light of wavelength 5000 A is used, the minimum separation between two points, to be seen as distinct, will be :

1) $0.24 \mu \mathrm{m}$
2) $0.38 \mu \mathrm{m}$
3) $0.12 \mu \mathrm{m}$
4) $0.48 \mu \mathrm{m}$

Solution:

Resolving power of the microscope

$
R=\frac{2 \mu \sin \Theta}{\lambda}
$
$\mu=$ Refractive index
$\lambda=$ wavelength of light used
The minimum distance between two lines $\varepsilon=0.61 \times \frac{\lambda}{N . A}$.
where N.A is a numerical aperture
$
\begin{aligned}
& \Rightarrow \varepsilon=0.61 \times \frac{5000 \times 10^{-10}}{1.25} \\
& \Rightarrow \varepsilon=0.24 \mu \mathrm{m}
\end{aligned}
$

Hence, the answer is the option (1).

Example 2: If $R_{p=}$ Resolving power of optical instruments and $R_l=$ Resolving limit of optical instruments then $R_p$ and $R_l$ are related as

1) $R_p \propto R_l$
2) $R_p \alpha \frac{1}{R_l}$
3) $R_p \propto R_l^3$
4) $R_p \propto R_l^2$

Solution:

Resolving power of optical instruments

The resolving power of an optical instrument is its ability to resolve or separate the images of two nearby point objects so that they can be distinctly seen.

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

I.e $R_p \alpha \frac{1}{R_l}$

Hence, the answer is the option (2).

Example 3: The wavelengths of light used in an optical instrument are $\lambda_1=4000 A^0$ and $\lambda_2=5000 A^0$ then the ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$) is

1) 16:25

2) 9:1

3) 4:5

4) 5:4

Solution:

Resolving Power $\alpha 1 / \lambda$
Given,
$
\begin{aligned}
& \lambda_1=4000^{\circ} \mathrm{A} \\
& \lambda_2=5000^{\circ} \mathrm{A}
\end{aligned}
$
$
\begin{aligned}
& R P_{\lambda 1} / R P_{\lambda 2}=\lambda 2 / \lambda 1 \\
& \Rightarrow R P_{\lambda 1} / R P_{\lambda 2}=5000 / 4000=5 / 4 \\
& \Rightarrow R P_{\lambda 1}: R P_{\lambda 2}=5: 4
\end{aligned}
$

Hence, the answer is the option (4).

Example 4: Assuming the human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that the human eye can resolve at 500 nm wavelength is :

1) $1 \mu \mathrm{m}$
2) $30 \mu \mathrm{m}$
3) $100 \mu \mathrm{m}$
4) $300 \mu \mathrm{m}$

Solution:

The radius of human solution pupil=0.25cm or diameter =0.5cm=5x10^-3m

$
\begin{aligned}
& \lambda=500 \mathrm{~mm}=5 \times 10^{-7} \mathrm{~m} \\
& \text { Since } \sin \theta=\frac{1.22 \lambda}{d}=\frac{1.22 \times 5 \times 10^{-7}}{5 \times 10^{-3}}=1.22 \times 10^{-4}
\end{aligned}
$

The distance of comfortable viewing $=25 \mathrm{~cm}$
Let $x$ be the minimum separation between the two objects that the human eye can resolve them
$
\begin{aligned}
& \quad \sin \theta=\tan \theta=\frac{x}{0} \text { or } 0 \tan \theta=x \\
& x=(25 \mathrm{~cm}) \times 1.22 \times 10^{-4} \\
& =3 \times 10^{-5} \mathrm{~m}=30 \mu \mathrm{m}
\end{aligned}
$

Hence, the answer is the option (2).

Example 5: Two point white dots are 1 mm apart on black paper. They are viewed by an eye with a pupil diameter of 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light = 500 mm]

1) 6 mm
2) 3 mm
3) 5 mm
4) 1 mm

Solution:

Radius of diffraction disc

$
R=1.22 \frac{\lambda D}{b}
$
wherein
$D=$ Distance of the screen from the hole
$b=$ slit width
Resolution limit $=\frac{1.22 \lambda}{d}$
Again resolution limit
$
\begin{aligned}
& \text { Again resolution limit }=\sin \Theta=\Theta=\frac{y}{D} \\
& \therefore \quad \frac{y}{D}=\frac{1.22 \lambda}{d}
\end{aligned}
$
or $D=\frac{\lambda d}{1.22 \lambda}$
$
\xrightarrow[D]{\stackrel{1}{\longleftrightarrow}}{ }^{\uparrow}{ }^y \text { or } \quad D=\frac{\left(10^3\right) \times\left(3 \times 10^{-3}\right)}{(1.22) \times\left(5 \times 10^{-7}\right)}=\frac{30}{6.1} \approx 5 \mathrm{~mm}
$

Hence, the answer is the option (3).

Summary

The resolving power of optical instruments like microscopes and telescopes is crucial for distinguishing fine details in closely spaced objects. For microscopes, high resolving power is essential in fields like biology and materials science to observe minute structures clearly. In telescopes, resolving power allows detailed observation of distant celestial bodies, aiding in astronomical studies. Factors influencing resolving power include the wavelength of light and the aperture size of the instrument.