Light passing through any opening diffracts, and a lens or a mirror is an opening. So even a perfect telescope turns each star into a tiny blurred disk with faint rings around it (an Airy pattern). Two stars close together make two overlapping blurs. When can you still tell them apart?
Key idea
Rayleigh's criterion: two point sources can just be resolved by a circular aperture of diameter $D$ when their angular separation is $$\theta_R = 1.22\,\frac{\lambda}{D}\quad(\text{radians}).$$ Then the center of one blur falls on the first dark ring of the other. Bigger apertures and shorter wavelengths resolve finer detail.
Why it works
For a slit of width $a$, the first dark fringe is at $\sin\theta = \lambda/a$: light from the two edges of the opening then differs by one wavelength, and the slit's strips cancel in pairs. A circular opening works the same way, but its width varies across the circle; working out the sum over the circle (a calculus result that involves a Bessel function) moves the first dark ring to $\sin\theta = 1.22\lambda/D$. Rayleigh's choice of “just resolved” puts one peak on the other's first zero. There, the sum of the two blurs dips by about a quarter between the peaks, which an eye or a detector can notice.
Worked examples
Worked example: The Hubble Space Telescope
Hubble's mirror is 2.4 m across. Find its angular resolution at 550 nm, in radians and in arcseconds.
- $\theta_R = 1.22\frac{550 \times 10^{-9}\ \text{m}}{2.4\ \text{m}} = 2.80 \times 10^{-7}$ rad.
- One radian is $206{,}265''$ (arcseconds), so $\theta_R = 2.80 \times 10^{-7} \times 206{,}265 = 0.058''$.
Answer: $2.8 \times 10^{-7}$ rad $\approx 0.06''$
Worked example: Headlights at night
Your pupil is 3.0 mm across at night. Using 550 nm light, how far away can a car be with its headlights (1.4 m apart) still seen as two lights, if diffraction is the only limit?
- $\theta_R = 1.22\frac{550 \times 10^{-9}}{3.0 \times 10^{-3}} = 2.24 \times 10^{-4}$ rad.
- Small angles: separation = distance × angle, so the distance is $\frac{1.4\ \text{m}}{2.24 \times 10^{-4}} = 6.3 \times 10^3$ m.
Answer: About 6 km (in practice the eye does a little worse)
Worked example: A radio dish
A radio telescope 25 m across observes the 21 cm line of hydrogen. What is its angular resolution, in degrees?
- $\theta_R = 1.22\frac{0.21}{25} = 0.0102$ rad.
- In degrees: $0.0102 \times \frac{180}{\pi} = 0.59^\circ$, a little more than the Moon's width (about $0.5^\circ$).
Answer: About $0.6^\circ$, which is why radio astronomers link dishes across whole continents
Common mistake
Thinking enough magnification can reveal any detail. Magnifying a diffraction blur only makes a bigger blur. Once two points are closer than $1.22\lambda/D$, no eyepiece can separate them: only a wider aperture or a shorter wavelength can. Extra magnification beyond that point is called “empty magnification”.
Common mistake
Using degrees in $\theta_R = 1.22\lambda/D$. The formula gives radians; convert afterward if you need degrees or arcseconds.