Resolving power — Rayleigh criterion: Worked examples

This page is the practice arena for the Rayleigh criterion. The parent note built the formula; here we hit every kind of question it can generate. First a map of all the cases, then worked examples — each one labelled with the map cell it fills.

The scenario matrix

Every problem on this topic is one of these cells. Skip nothing.

#	Case class	What makes it special	Example
A	Circular aperture, find angle	Straight $θ_{R} = 1.22 λ / D$	Ex 1
B	Angle → physical separation	Uses arc length $s = θ L$ (small-angle)	Ex 2
C	Slit aperture (no 1.22)	Drop the factor, use width $a$	Ex 3
D	Microscope / N.A.	Near object → $d_{m i n} = 0.61 λ / N.A.$	Ex 4
E	Limiting / proportional reasoning	Change $λ$ and $D$ , no numbers plugged	Ex 5
F	Degenerate input ( $D \to \infty$ , $λ \to 0$ )	Resolution limit vanishes	Ex 6
G	Real-world word problem	Extract $λ, D, L$ from prose	Ex 7
H	Exam twist / trap	Diameter vs radius, degrees vs radians	Ex 8

Case A — Circular aperture, find the angle

Worked example Ex 1 (Cell A) — A satellite camera

A spy-satellite camera has a circular lens of diameter $D = 0.30 m$ and images in green light $λ = 550 nm$ . What is the smallest angle between two ground points it can resolve?

Forecast: Bigger lens than your eye (2 mm) by ~150×. Guess: is $θ_{R}$ around $1 0^{- 6}$ rad? Bigger or smaller than that?

Step 1 — Pick the formula. Circular aperture, distant objects ⇒ $θ_{R} = 1.22 λ / D$ . Why this step? Cell A is the plain angular-resolution case — no slit, no N.A. The lens is round, so the $1.22$ (first zero of $J_{1}$ , see Airy disc and Bessel functions) belongs.

Step 2 — Keep units consistent. Put both lengths in metres: $λ = 550 \times 1 0^{- 9} m$ , $D = 0.30 m$ . Why this step? $λ / D$ must be dimensionless so the answer comes out in radians (an angle is a pure number). Mixing nm with m would give a nonsense factor of $1 0^{9}$ .

Step 3 — Plug in. $θ_{R} = 1.22 \times \frac{550 \times 1 0 ^{- 9}}{0.30} = 2.24 \times 1 0^{- 6} rad .$

Verify: Dimensionless ratio → radians ✓. It's smaller than your eye's $3.4 \times 1 0^{- 4}$ rad by about $150 \times$ , exactly the diameter ratio $0.30/0.002 = 150$ . Consistent, since $θ_{R} \propto 1/ D$ . ✓

Case B — From angle to a real distance on the ground

Worked example Ex 2 (Cell B) — What that camera actually sees

Same camera ( $θ_{R} = 2.24 \times 1 0^{- 6} rad$ ) orbits at height $L = 250 km$ . What is the smallest separation $s$ of two objects it can tell apart on the ground?

Forecast: Metres? Centimetres? Guess before computing.

Step 1 — Draw the geometry. Two points on the ground separated by $s$ , subtending angle $θ_{R}$ at the lens $L$ away.

Step 2 — Use the small-angle arc relation. For a tiny angle, the separation is the arc length: $s = θ_{R} L .$ Why this step? $θ_{R} \sim 1 0^{- 6}$ rad is minuscule, so $\tan\theta\approx\sin\theta\approx\theta$ holds to better than one part in $1 0^{12}$ . The "straight gap" $s$ and the "curved arc" $θ L$ are indistinguishable here.

Step 3 — Plug in ( $L = 250 km = 2.5 \times 1 0^{5} m$ ): $s = 2.24 \times 1 0^{- 6} \times 2.5 \times 1 0^{5} = 0.559 m \approx 56 cm .$

Verify: rad $\times$ m $=$ m ✓ (radians are dimensionless, so they "disappear" leaving metres — a units sanity check that you used radians not degrees). ~0.5 m ground resolution is realistic for such optics. ✓

Case C — A slit, not a circle (drop the 1.22)

Case D — Microscope, near object, numerical aperture

Worked example Ex 4 (Cell D) — A dry vs oil objective

A microscope objective has half-angle $β = 6 4^{\circ}$ . In air ( $n = 1$ ), $λ = 500 nm$ . (i) Find the smallest resolvable detail $d_{m i n}$ . (ii) Now add immersion oil $n = 1.52$ — recompute.

Forecast: Oil should make $d_{m i n}$ smaller (better). By how much — roughly the ratio of the $n$ values?

Step 1 — Build the numerical aperture. $N.A. = n sin β$ (see Numerical aperture). Why this step? For a near object the lens collects a cone of light; the relevant "size" is the cone half-angle $β$ , not a diameter. $sin β$ measures how wide that cone opens; $n$ scales it because oil bends more rays into the lens.

Step 2 — Air case. $sin 6 4^{\circ} = 0.8988$ , so $N.A. = 1 \times 0.8988 = 0.8988$ . $d_{m i n} = \frac{0.61 λ}{N.A.} = \frac{0.61 \times 500 \times 1 0 ^{- 9}}{0.8988} = 3.39 \times 1 0^{- 7} m \approx 339 nm .$

Step 3 — Oil case. $N.A. = 1.52 \times 0.8988 = 1.366$ . $d_{m i n} = \frac{0.61 \times 500 \times 1 0 ^{- 9}}{1.366} = 2.23 \times 1 0^{- 7} m \approx 223 nm .$

Verify: Ratio of the two answers $= 339/223 = 1.52 = n_{oil}$ ✓ — exactly the $n$ factor, since $β$ was unchanged. Oil improves resolution. ✓

Case E — Proportional reasoning (no numbers)

Case F — Degenerate / limiting inputs

Worked example Ex 6 (Cell F) — The perfect-lens myth

Two limiting thought-experiments: (i) let the aperture grow without bound, $D \to \infty$ ; (ii) let the wavelength shrink to zero, $λ \to 0$ . What does $θ_{R}$ approach in each?

Forecast: Does the blur ever fully vanish? Which limit is physically reachable?

Step 1 — Take $D \to \infty$ . $θ_{R} = 1.22 λ / D \to 0$ . Why this step? An infinitely wide aperture chops off nothing from the wavefront, so there is nothing to diffract — the parent note's "chopping a wave = diffraction" logic runs in reverse. Perfect point, but $D = \infty$ is unbuildable.

Step 2 — Take $λ \to 0$ . $θ_{R} = 1.22 λ / D \to 0$ . Why this step? Zero wavelength is the geometric-optics (ray) limit: waves that don't wave don't diffract. This is why electron microscopes ( $λ$ thousands of times smaller than light) beat optical ones.

Step 3 — Guard against nonsense. $D = 0$ (no aperture) or $λ = \infty$ gives $θ_{R} \to \infty$ : light spreads over all angles, zero resolving power — you image nothing.

Verify: Both "good" limits push $θ_{R} \to 0$ (infinite resolving power $1/ θ_{R} \to \infty$ ), both "bad" limits push it to $\infty$ . The formula behaves monotonically and sensibly at every extreme. ✓

Case G — Real-world word problem

Worked example Ex 7 (Cell G) — Reading a car number plate from a drone

A camera-drone hovers at $L = 120 m$ . Its lens aperture is $D = 8.0 mm$ , $λ = 550 nm$ . Number-plate characters have strokes about $4 mm$ apart. Can it read the plate?

Forecast: Gut call — readable or not?

Step 1 — Find the angular limit. $θ_{R} = 1.22 \times \frac{550 \times 1 0 ^{- 9}}{8.0 \times 1 0 ^{- 3}} = 8.39 \times 1 0^{- 5} rad .$ Why this step? "Can it tell two features apart?" is exactly the resolution question — Cell A gives the finest angle.

Step 2 — Convert to a ground distance at $L = 120 m$ : $s_{m i n} = θ_{R} L = 8.39 \times 1 0^{- 5} \times 120 = 1.01 \times 1 0^{- 2} m \approx 10 mm .$ Why this step? Compare like with like: the smallest resolvable gap $s_{m i n}$ versus the actual $4 mm$ stroke spacing.

Step 3 — Decide. $s_{m i n} \approx 10 mm > 4 mm$ ⇒ the strokes are closer than the resolution limit ⇒ they blur together ⇒ cannot read it.

Verify: rad $\times$ m $=$ m ✓. To read a 4 mm feature you'd need $s_{m i n} \leq 4$ mm, i.e. $D \geq 8.0 \times (10/4) = 20 mm$ — a bigger lens. Physically sensible. ✓

Case H — Exam twist / classic trap

Worked example Ex 8 (Cell H) — Radius given, degrees demanded

A telescope mirror has radius $r = 1.25 m$ ; observe $λ = 500 nm$ . Express the resolution limit in arcseconds. (An arcsecond is $1/3600$ of a degree.)

Forecast: Two traps are hidden here — spot them before computing.

Step 1 — Trap 1: diameter, not radius. The formula uses diameter $D = 2 r = 2.50 m$ . Why this step? $θ_{R} = 1.22 λ / D$ — the aperture's full width diffracts the wave; halving it (using $r$ ) would double $θ_{R}$ and halve resolution.

Step 2 — Compute in radians first. $θ_{R} = 1.22 \times \frac{500 \times 1 0 ^{- 9}}{2.50} = 2.44 \times 1 0^{- 7} rad .$ Why radians first? The formula always outputs radians (it's a length ratio). Convert only at the end.

Step 3 — Trap 2: radians → arcseconds. Multiply by $\frac{180}{π}$ (rad→deg) then by $3600$ (deg→arcsec): $θ_{R} = 2.44 \times 1 0^{- 7} \times \frac{180}{π} \times 3600 = 0.0503 arcsec .$

Verify: $1 rad = 206265 arcsec$ , so $2.44 \times 1 0^{- 7} \times 206265 = 0.0503$ arcsec ✓ — matches. About $0.0 5^{''}$ is right for a 2.5 m mirror. ✓

Recall Which cell is which? (self-test)

Camera resolution in radians ::: Cell A — plain $1.22 λ / D$ Ground separation from an angle ::: Cell B — $s = θ_{R} L$ , small angle Slit-shaped aperture ::: Cell C — $λ / a$ , no 1.22 Microscope detail from N.A. ::: Cell D — $0.61 λ / N.A.$ "What if both change?" ::: Cell E — track the ratio, don't plug " $D \to \infty$ " ::: Cell F — $θ_{R} \to 0$ , perfect but unbuildable Radius given / degrees asked ::: Cell H — use $D = 2 r$ , convert last

Flashcards

Satellite lens D = 0.30 m, λ = 550 nm — angular resolution?

1.22 \times 550 \times 1 0^{- 9} /0.30 = 2.24 \times 1 0^{- 6}

rad.

At 250 km, what ground separation does θ_R = 2.24×10⁻⁶ rad give?

s = θ_{R} L = 2.24 \times 1 0^{- 6} \times 2.5 \times 1 0^{5} \approx 0.56

Slit aperture a = 0.80 m, λ = 600 nm — resolution?

λ / a = 7.5 \times 1 0^{- 7}

rad (no 1.22).

Oil objective N.A. = 1.37, λ = 500 nm — smallest detail?

0.61 λ / N.A. \approx 223

nm.

Halve λ and halve D together — effect on θ_R?

Unchanged; the ratio λ/D is preserved.

2.5 m mirror, λ = 500 nm — resolution in arcsec?

≈ 0.050 arcsec (use D = diameter, convert radians last).