Resolving Power

PHYDiffraction & Resolution

Resolving power is the ability of an optical instrument to form distinct, separable images of two close objects (or to separate two nearby wavelengths). Because every aperture diffracts light, a point source images as an Airy disc rather than a point, and two such discs overlap when the objects lie too close; the Rayleigh criterion fixes the just-resolved limit.

ISC examines this as short numericals on the limit of resolution of a telescope (an angle, set by aperture $D$ ) and the resolving limit of a microscope (a distance, set by numerical aperture $n\sin\theta$ ), so you must know which formula applies and which $\lambda$ to use.

Telescope: limit of resolution (Rayleigh)

\Delta\theta = \dfrac{1.22\,\lambda}{D}

\Delta\theta

= smallest resolvable angular separation (in radian);

\lambda

= wavelength of light;

D

= diameter (aperture) of the objective. The reciprocal

1/\Delta\theta

is the resolving power.

Telescope: resolving power

R = \dfrac{1}{\Delta\theta} = \dfrac{D}{1.22\,\lambda}

R

= resolving power of the telescope; a larger

D

or smaller

\lambda

gives higher

R

(finer resolution).

Microscope: limit of resolution (smallest resolvable distance)

d_{min} = \dfrac{1.22\,\lambda}{2\,n\sin\theta} = \dfrac{0.61\,\lambda}{n\sin\theta}

d_{min}

= smallest separation of two points that can just be resolved;

n\sin\theta

= numerical aperture (NA);

n

= refractive index of the medium between object and objective;

\theta

= half-angle of the cone of light entering the objective.

Microscope: resolving power

R = \dfrac{1}{d_{min}} = \dfrac{2\,n\sin\theta}{1.22\,\lambda}

R

= resolving power of the microscope; a high NA (large

\theta

, immersion medium with large

n

) and a short

\lambda

improve it.

The factor $1.22$ comes from the first dark ring of the Airy diffraction pattern of a circular aperture; for a single slit the analogous factor is $1$ (first minimum at $\sin\theta = \lambda/a$ ), so always use $1.22$ for the circular lenses of a telescope or microscope.
Telescope resolving power depends on the objective's aperture $D$ , not on its focal length or magnification; a larger aperture both gathers more light and resolves finer detail.
$\Delta\theta = 1.22\,\lambda/D$ gives the angle directly in radian because it is already a small-angle (Rayleigh) result; convert to seconds of arc only if asked, using $1\,\text{rad} \approx 2.06\times 10^{5}$ arc-seconds.
For a microscope the key quantity is the numerical aperture $n\sin\theta$ ; using an oil-immersion medium raises $n$ (to about $1.5$ ) and so lowers $d_{min}$ , improving resolution.
Resolving power and magnification are different ideas: magnification enlarges the image, but only resolving power decides whether two close points stay separate, so empty magnification beyond the resolution limit reveals no new detail.
Shorter wavelength helps both instruments, since $\Delta\theta \propto \lambda$ and $d_{min} \propto \lambda$ ; this is why blue light resolves finer detail than red light.
By the Rayleigh criterion, two sources are 'just resolved' when the central maximum of one Airy pattern falls on the first minimum of the other.

Where the marks go

Dropping the factor $1.22$ (or using $1$ as for a single slit): circular apertures of telescopes and microscopes always carry $1.22$ , and the microscope formula carries $0.61 = 1.22/2$ .
Unit slips with the wavelength: $\lambda$ in nm must be converted to metre ( $550\,\text{nm} = 550\times 10^{-9}\,\text{m}$ ) before dividing, or the answer is wrong by a factor of $10^{9}$ .
Confusing the two formulas: a telescope's limit is an angle $\Delta\theta$ (uses aperture $D$ ), while a microscope's limit is a distance $d_{min}$ (uses numerical aperture $n\sin\theta$ ); mixing them gives nonsensical units.
Forgetting that the numerical aperture is the whole quantity $n\sin\theta$ , not just $\sin\theta$ : with an immersion medium you must include $n$ , and a stated NA of $1.0$ should be substituted directly without re-multiplying.

How the board asks it

Numericaltelescope limit of resolution $\Delta\theta = 1.22\,\lambda/D$
The objective of a telescope has an aperture of diameter $D = 0.20\,\text{m}$ . Calculate its limit of resolution for light of wavelength $\lambda = 550\,\text{nm}$ , expressing the answer in radian.
Numericalmicroscope $d_{min} = 0.61\,\lambda/(n\sin\theta)$
A microscope objective has a numerical aperture $n\sin\theta = 1.5$ and is used with light of wavelength $\lambda = 600\,\text{nm}$ . Find the smallest distance between two points that it can just resolve.
Define / statethe Rayleigh criterion
Define the resolving power of a telescope and state the Rayleigh criterion for two point objects to be just resolved.
Give reasons $\Delta\theta \propto \lambda$
Give reasons why blue light enables an optical instrument to resolve finer detail than red light.
Distinguishresolving power vs magnification
Distinguish between the magnifying power and the resolving power of a microscope, and explain why increasing magnification beyond the resolution limit reveals no new detail.
Applicationnumerical aperture and oil immersion
Account for the fact that an oil-immersion objective ( $n \approx 1.5$ ) resolves finer detail than a dry objective ( $n \approx 1$ ) used at the same semi-angle $\theta$ .

Practise this topic

PreviousSingle-Slit Diffraction NextMultiple Choice & Assertion-Reason

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.