Resolving power and the Rayleigh limit

Why does a star look like a disc?

Look at a single star through a telescope. Even a perfect telescope shows a small bright disc with faint rings round it, not a point. The reason is diffraction. The light of the star passes through a round opening, the aperture of the telescope, and it spreads like the light from a slit.

For a circular aperture of diameter $D$, the pattern is a bright central disc, the Airy disc, with rings round it. The first dark ring is at the angle

The factor $1.22$ belongs to a circle. For a slit the factor is $1$, as in lesson one.

The Rayleigh criterion

Two stars close together make two Airy discs. When the discs overlap too much, we see only one blob. Lord Rayleigh proposed a simple rule. The two images are just resolved when the centre of one Airy disc falls on the first dark ring of the other. The smallest angle between the stars is then

At this angle, the intensity at the middle between two equally bright stars drops to about $0.735$ of the peak. The eye can see this dip. At a smaller angle there is hardly any dip.

The resolving power of the instrument is $\dfrac{1}{\theta_{\min}} = \dfrac{D}{1.22\lambda}$. It is large for a big aperture and a short wavelength.

Raise the diameter of the aperture. The limit falls, so finer detail can be seen. A telescope has a larger aperture than the eye, and this is the real reason for a large telescope, in addition to gathering more light.

The telescope, the eye and the microscope

• A telescope of diameter $D = 0.20$ m in light of $550$ nm resolves $\theta_{\min} = \dfrac{1.22 \times 550 \times 10^{-9}\ \text{m}}{0.20\ \text{m}} = 3.36\ \mu\text{rad}$, which is $0.69$ arcseconds. One arcsecond is $\dfrac{1}{3600}$ of a degree, or $4.85\ \mu\text{rad}$.
• The Hubble Space Telescope, with $D = 2.4$ m, reaches $\theta_{\min} = 0.280\ \mu\text{rad} = 0.058$ arcseconds in the same light. Above the atmosphere, nothing blurs it.
• The human eye has a pupil of about $3.0$ mm in bright light. $\theta_{\min} = \dfrac{1.22 \times 550 \times 10^{-9}\ \text{m}}{3.0 \times 10^{-3}\ \text{m}} = 2.24 \times 10^{-4}\ \text{rad}$, about $0.77$ arcminutes. Real eyes do slightly worse, near one arcminute, because of the spacing of the light cells and other faults.
• A microscope resolves points at a distance of at least $d_{\min} = \dfrac{0.61\lambda}{\text{NA}}$. The numerical aperture NA $= n\sin\beta$ describes the cone of light that the objective collects, with $n$ the refractive index of the medium in front of the lens. With $\text{NA} = 0.95$ in light of $550$ nm, $d_{\min} = 353$ nm. With oil on the lens and $\text{NA} = 1.40$, $d_{\min} = 240$ nm.

A microscope cannot beat this limit with more magnification. Making an image larger does not bring back detail that diffraction has blurred. A light microscope cannot see anything much smaller than half of a wavelength, about $200$ nm. An electron microscope uses electrons of a far shorter de Broglie wavelength, and sees atoms.

Resolving power of a grating, in brief

Can a grating separate two lines of nearly the same wavelength? Its resolving power is

where $\lambda$ is the mean wavelength, $\Delta\lambda$ is the smallest difference that can be separated, $m$ is the order and $N$ is the number of lines lit. A grating of $N$ lines gives a higher resolving power in a higher order.

A worked example. Two lines have wavelengths $500.0$ nm and $500.5$ nm. How many lines must be lit to separate them in the first order and in the second order?

1. The resolving power needed. $R = \dfrac{\lambda}{\Delta\lambda} = \dfrac{500.25\ \text{nm}}{0.5\ \text{nm}} = 1001$.
2. The first order. $N = \dfrac{R}{m} = 1001$ lines.
3. The second order. $N = 500$ lines.
4. Check. A grating with $500$ lines in each millimetre has $1001/500 = 2.0$ mm of its width lit in the first order.

Practise with new numbers

Try it in Python

The program makes the image of two stars, each an Airy pattern, and finds the intensity at the middle for four separations of the stars. The separation is in units of $\lambda/D$.

Common mistakes

• Using $\lambda/D$ without $1.22$. For a circular aperture the limit is $1.22\lambda/D$. For a slit it is $\lambda/a$.
• Thinking that more magnification brings more detail. The limit is set by the aperture and the wavelength, not by the magnification.
• Taking $\theta_{\min}$ in the wrong unit. $\lambda/D$ is in radian. Change it to arcseconds with $1\ \text{rad} = 206\,265$ arcseconds.
• Using a longer wavelength to see finer detail. A shorter wavelength resolves finer detail.

Remember these

• The image of a point through a circular aperture is an Airy disc with rings. The first dark ring is at $1.22\lambda/D$.
• Rayleigh criterion: $\theta_{\min} = 1.22\lambda/D$. Resolving power $= 1/\theta_{\min}$.
• A microscope: $d_{\min} = 0.61\lambda/\text{NA}$.
• A grating: $R = \lambda/\Delta\lambda = mN$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The aperture of a telescope is made $3$ times larger. What happens to the smallest angle that it resolves?

2. Why do astronomers like a large aperture for fine detail, in addition to the fact that it collects more light?

3. Why is a radio telescope far larger than an optical telescope?

4. Two stars are $1.0\ \mu\text{rad}$ apart. A telescope has $\theta_{\min} = 1.5\ \mu\text{rad}$. Can it tell them apart?

References

• OpenStax, University Physics Volume 3, Chapter 4 (Diffraction): circular apertures and resolution.
• NCERT Physics Class 12, Chapter 9 (Ray Optics and Optical Instruments), older editions: resolving power of the telescope and the microscope.