Resolving Power

Resolving power and limit of resolution

An optical instrument is used not only to magnify but to see fine detail and to see nearby objects as distinct. Two fingers held up 10 m away are easily seen as two; from 50 m most eyes cannot tell whether it is one finger or two.

• Resolving power: the ability of an optical instrument (eye, microscope, telescope) to distinguish two physically separated objects as two distinct objects.
• It depends, in general, on the aperture of the instrument (usually the diameter of the lens or mirror) and on the wavelength of light used.
• Resolving ability is stated in terms of the visual angle — the angle subtended at the instrument by the two point objects to be resolved. The MINIMUM such angle that the instrument can resolve is called the limit of resolution.
• Resolving power is the reciprocal of the limit of resolution.
• Diffraction is what sets the limit: each object produces a diffraction pattern, and the patterns overlap.

7.10.1 Rayleigh's criterion

According to Lord Rayleigh, the ability of an optical instrument to distinguish between two closely spaced objects depends on the diffraction patterns of the two objects (slits, point objects, stars) produced at the screen (retina, eyepiece).

Criterion: two objects are JUST RESOLVED when the first minimum of the diffraction pattern of one source coincides with the central maximum of the diffraction pattern of the other, and vice versa.

Let $S$ be the separation between the central maxima of the two patterns and $\Delta$ the distance from the central maximum to the first minimum of either pattern.

Two linear objects

Consider two self luminous linear objects (slits) separated by some distance. By Rayleigh's criterion the first minimum of the diffraction pattern of one must coincide with the central maximum of the other. The angular position of the first principal minimum of a slit of width $a$ is

$$d\theta=\dfrac{\lambda}{a}$$

As this minimum coincides with the central maximum of the other pattern, this is the minimum angular separation between the two objects — the limit of resolution of that instrument:

$$\text{Limit of resolution}=d\theta=\dfrac{\lambda}{a}$$

The minimum separation between two linear objects that are just resolved, at distance $D$ from the instrument, is

$$y=D\cdot d\theta=\dfrac{D\lambda}{a}$$

which is simply the distance of the first minimum from the centre.

A pair of point objects

For a microscope, the objects are usually point objects. The diffraction pattern of a point object is a central bright disc surrounded by concentric rings — the Airy disc and rings (Abbe first studied this and applied it to Fraunhofer diffraction). For such objects to be just resolved, the first dark ring of the diffraction pattern of the first object must be formed at the centre of the diffraction pattern of the second, and vice versa. In other words, the minimum separation between the images on the screen is the radius of the first dark ring.

This discussion applies to any instrument — an eye, a microscope, a telescope.

7.10.2 Resolving power of a microscope

Two point objects $O$ and $O'$ are separated by a distance $a$ in front of the objective $AB$ of a microscope. The medium between the objects and the objective has refractive index $n$, and the wavelength of the light in that medium is $\lambda_{n}$. The angular separation subtended by the objects at the objective is $2\alpha$.

$I$ and $I'$ are the centres of the diffraction patterns (Airy discs and rings) due to $O$ and $O'$ on the screen, effectively at infinity. By Rayleigh's criterion the first dark ring due to $O'$ should coincide with $I$, and that of $O$ with $I'$. The illumination at a point is decided by the effective path difference there. Consider the point $I$ where the first dark ring due to $O'$ is located: the extreme rays reaching $I$ from $O'$ are $O'AI$ and $O'BI$. Since $I$ is equidistant from $A$ and $B$, the paths $AI$ and $BI$ are equal, so the actual path difference is $O'B-O'A$. From the enlarged geometry around $O$ and $O'$ it can be shown that

$$\text{path difference}=DO'+O'C=2a\sin\alpha$$

(i) A pair of non-luminous (dark) objects

In ordinary microscopy the objects are not self luminous; they are illuminated by an external source, usually obliquely, and most of them can be treated as point objects producing Airy rings. Often the space is filled with a transparent material of refractive index $n$, so the wavelength there is $\lambda_{n}=\dfrac{\lambda}{n}$, $\lambda$ being the wavelength in air.

In such a set-up the path difference at the first dark ring is $\lambda_{n}$, so

$$2a\sin\alpha=\lambda_{n}=\dfrac{\lambda}{n}$$

$$\boxed{\ a=\dfrac{\lambda}{2n\sin\alpha}=\dfrac{\lambda}{2\,\text{N.A.}}\ }$$

The factor $n\sin\alpha$ is called the numerical aperture (N.A.). The resolving power is

$$R=\dfrac{1}{a}=\dfrac{2\,\text{N.A.}}{\lambda}$$

(ii) A pair of self luminous point objects

Applying Abbe's theory of Airy discs and rings to Fraunhofer diffraction by a pair of self luminous point objects, the path difference between the extreme rays at the first dark ring is $1.22\lambda$. For just resolution,

$$2a\sin\alpha=\dfrac{1.22\lambda}{n}$$

$$\therefore\ a=\dfrac{1.22\lambda}{2n\sin\alpha}=\dfrac{0.61\lambda}{n\sin\alpha}=\dfrac{0.61\lambda}{\text{N.A.}}$$

$$R=\dfrac{1}{a}=\dfrac{\text{N.A.}}{0.61\lambda}$$

For better resolution $a$ should be as small as possible; this is achieved with an oil-immersion objective, which provides a higher value of $n$.

Factors, and how to increase it

• The resolving power of a microscope depends on the numerical aperture of the objective and on the wavelength of the light used.
• It can be increased by: (a) reducing the distance of separation between the two objects; (b) filling the space at the objective with a material of high refractive index $n$, such as oil; (c) using a source of small wavelength $\lambda$.
• Since $R\propto\dfrac{1}{\lambda}$, a very short wavelength gives very high resolution. Electrons behave as waves (Chapter 14) and their wavelength can be controlled through their speed — this is the principle of the electron microscope, which can magnify about $10^{6}$ times and resolve objects less than 10 nm apart, against about $10^{-3}$ mm for an ordinary optical microscope.

7.10.3 Resolving power of a telescope

The diagram to draw

The objective $AB$ of the telescope receives two sets of parallel beams from two distant point objects (stars) whose angular separation is $\theta$. The two beams are brought to foci at $I$ and $I'$, each of which is an Airy disc with rings around it.

Derivation

• A telescope is used to see distant stars, which for us are luminous point objects; their diffraction patterns are Airy discs.
• Since the objects are far away, only the ANGULAR separation between them matters, not the linear separation.
• Resolving power of a telescope is defined as the reciprocal of the least angular separation between two objects that are just resolved.
• By Rayleigh's criterion, the minimum separation between the images $I$ and $I'$ must be equal to the radius of the first dark Airy ring.
• Applying Airy's theory to Fraunhofer diffraction of a pair of point objects, the path difference between the extreme rays at the first dark ring is $1.22\lambda$:

$$BI'-AI'=1.22\lambda$$

• If $D$ is the aperture of the telescope (the diameter $AB$ of its objective),

$$BI'-AI'\approx D\times\theta=1.22\lambda$$

$$\boxed{\ \theta=\dfrac{1.22\lambda}{D}\qquad\text{and}\qquad R=\dfrac{1}{\theta}=\dfrac{D}{1.22\lambda}\ }$$

• Factors: the resolving power of a telescope depends on the aperture of the objective and on the wavelength of the light used. For a given wavelength, the objective should be as large as possible.

Practical points

Very large lenses bring aberrations, moulding difficulties and, for space telescopes, a heavy mass that is hard to move. The preferred alternative is a front-coated curved mirror as the objective, parabolic in order to eliminate spherical aberration. A single very large mirror is also difficult, so recent telescopes use segmented mirrors made of many hexagonal segments forming one large parabolic mirror; the two largest optical telescopes under construction have mirrors of 30 m and 40 m diameter.

Radio telescope. Radio wavelengths are of the order of metres, so by $\theta=\dfrac{1.22\lambda}{D}$ an aperture of at least a few kilometres would be needed to match the limit of resolution of an optical telescope. A single dish of that size is impracticable, so arrays of antennae spread over several kilometres are used. The Giant Metre-wave Radio Telescope (GMRT) at Narayangaon near Pune, Maharashtra, is such an array: 30 dishes of 45 m diameter each, spread over 25 km, which is the largest distance between two of its antennae. Its highest angular resolution ranges from about 60 arcsec at the lowest frequency of 50 MHz to about 2 arcsec at 1.4 GHz.

Standard results

• Rayleigh's criterion: just resolved when the first minimum of one pattern falls on the central maximum of the other.
• Two linear objects: $d\theta=\dfrac{\lambda}{a}$, least separation $y=\dfrac{D\lambda}{a}$.
• Microscope, non-luminous objects: $a_{\min}=\dfrac{\lambda}{2\,\text{N.A.}}$, $R=\dfrac{2\,\text{N.A.}}{\lambda}$.
• Microscope, self-luminous point objects: $a_{\min}=\dfrac{0.61\lambda}{\text{N.A.}}$, $R=\dfrac{\text{N.A.}}{0.61\lambda}$, with $\text{N.A.}=n\sin\alpha$.
• Telescope: $\theta=\dfrac{1.22\lambda}{D}$ (radian), $R=\dfrac{D}{1.22\lambda}$.
• 1 arcsec $=\dfrac{1}{3600}$ of a degree; convert only when the question asks.