Visual walkthrough — Resolving power — Rayleigh criterion

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Step 1 — What is a "point source" and why doesn't it stay a point?

WHAT. A point source is an object so far away (a star) or so tiny (a dot on a slide) that all the light we care about comes from a single location. Send that light through a hole — a lens, a pupil, a telescope opening — and you might expect a single bright point on the other side.

WHY it fails. Light is a wave. A wave passing the edge of a hole bends around it — this bending is diffraction. Only an infinitely wide hole would let the wave through untouched. Every real hole has edges, so every real image of a point is a little smeared blob, not a point.

PICTURE. Below: a flat wavefront (straight red line) arrives at a hole of width $D$. On the far side it is no longer flat — it fans out. The narrower the hole, the wider the fan. That fanning is the enemy of resolution.

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Step 2 — Build the diffraction pattern of ONE source

WHAT. Let that fanned-out light land on a screen. It does not make a uniform smear — it makes a bright centre surrounded by dark and bright rings. This is the Airy pattern.

WHY rings. Different parts of the wavefront travel slightly different distances to reach a given point on the screen. Where those distances differ by a whole wavelength in a canceling way, the waves destructively interfere and you get darkness — a dark ring. Between dark rings, light survives — bright rings.

PICTURE. The intensity (brightness) plotted against angle $\theta$ away from the centre. The tall central peak is the central maximum; the first place it hits zero is the first minimum (first dark ring), marked in red. Everything in this page hangs on the position of that red zero.

See Airy disc and Bessel functions for the full shape of that curve.

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Step 3 — Find ${\theta }_{\min }$ for a slit (the easy case first)

WHAT. Before tackling a round hole, solve the easier slit — a long thin rectangular gap of width $a$. We want the angle of its first dark fringe.

WHY the slit first. A slit is one-dimensional, so we can pair up rays by hand. Once we understand why a zero appears, the circular case is just a harder version of the same idea.

PICTURE. Split the slit into a top half and a bottom half. Pair each ray in the top half with the ray exactly $a/2$ below it (red pair). When the path difference between a pair equals half a wavelength, $\lambda /2$, that pair cancels. Every pair cancels together ⇒ total darkness.

The extra distance one ray of a pair travels is $\frac{a}{2}\sin \theta$. Set it to $\lambda /2$:

$\underset{\text{half-slit spacing}}{\underbrace{\frac{a}{2}}}\underset{\text{tilts path by this factor}}{\underbrace{\sin \theta }}=\underset{\text{half-wave}\to \text{cancel}}{\underbrace{\frac{\lambda }{2}}}$

Multiply both sides by 2:

$\underset{\text{slit width}}{\underbrace{a}}\underset{\text{direction to screen point}}{\underbrace{\sin \theta }}=\underset{\text{one wavelength}}{\underbrace{\lambda }}$

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Step 4 — Turn that into a small angle

WHAT. Solve $a\sin \theta =\lambda$ for the angle:

$\sin {\theta }_{\min }=\frac{\lambda }{a}.$

WHY approximate. For light, $\lambda$ (hundreds of nanometres) is tiny next to $a$ (millimetres), so ${\theta }_{\min }$ is a tiny angle. For tiny angles, $\sin \theta \approx \theta$ (the Small angle approximation). This lets us drop the $\sin$ and write the angle directly:

${\theta }_{\min }\approx \frac{\lambda }{a}(\text{slit}).$

PICTURE. The graph of $\sin \theta$ hugging the straight line $\theta$ near the origin — they are indistinguishable where our angles live (red band). That is our licence to swap one for the other.

Each symbol here:

• $\lambda$ — wavelength of the light (numerator: longer wave ⇒ bigger blur).
• $a$ — slit width (denominator: wider slit ⇒ smaller blur).

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Step 5 — Fix it for a ROUND hole: the 1.22

WHAT. Real lenses and pupils are circular, not slit-shaped. Redoing the ray-pairing over a disc instead of a strip changes only one thing: a numerical factor.

WHY 1.22. For a slit the first zero sat at $\lambda /a$. For a disc, the light spreads out into rings and the first dark ring is pushed slightly further out. The exact location is where a special function — the Bessel function $J_1$ — first hits zero, and that happens at $1.22$ times the slit value:

${\theta }_{\min }=1.22\frac{\lambda }{D}(\text{circular aperture}).$

Here $D$ replaces $a$ as the aperture size (diameter of the disc).

PICTURE. Side by side: the slit's dark line vs the disc's dark ring. The red ring sits at a slightly larger angle — that "slightly" is the whole $1.22$.

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Step 6 — Now bring in the SECOND source

WHAT. We have one blurry disc. A second nearby source makes a second identical disc, shifted by the angle $\theta$ between the two sources. The question of resolution is: how far apart must the two discs be before I can tell there are two?

WHY a rule is needed. There is no single physical instant where "one blob" becomes "two". As the discs separate the dip between them deepens gradually. So we need an agreed convention — a fair line in the sand.

PICTURE. Three panels: discs too close (one hump), discs just separating (a shallow red dip appears), discs well apart (two clean peaks). The middle panel is the borderline we are about to name.

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Step 7 — State Rayleigh's line in the sand

WHAT. Rayleigh's rule: the two sources are just resolved when the central peak of one sits exactly on the first dark ring of the other.

WHY here. At this exact spacing the sum of the two patterns dips about $26\%$ below the peaks — small, but detectable by eye. Closer and the dip vanishes; farther and it is obvious. It is chosen because it is clean to compute: we already found the first-minimum angle in Steps 4–5.

So the smallest resolvable separation equals that first-minimum angle:

$\boxed{{\theta }_R=1.22\frac{\lambda }{D}}(\text{circular}),{\theta }_R=\frac{\lambda }{a}(\text{slit}).$

Term by term:

• ${\theta }_R$ — smallest angle between two sources still tellable apart (radians).
• $1.22$ — the round-hole Bessel factor.
• $\lambda$ — wavelength; bigger ⇒ worse (larger ${\theta }_R$).
• $D$ — aperture diameter; bigger ⇒ better (smaller ${\theta }_R$).

PICTURE. Peak-on-first-minimum, drawn: source A's peak (black) landing precisely on the red first zero of source B, and the resulting summed curve with its shallow dip.

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Step 8 — Edge and degenerate cases (never skip these)

WHAT. Push the formula to its extremes and check it still makes sense.

PICTURE. Four labelled limits, one panel each — the red curve shows ${\theta }_R$ vs the variable being pushed.

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Step 9 — The microscope twist: from $D$ to N.A.

WHAT. For a near object (microscope) we care about the smallest distance $d$ between two dots on the slide, not an angle in the sky. The relevant aperture measure becomes the half-angle $\beta$ of the light cone the objective collects, times the medium's refractive index $n$:

$d_{\min }=\frac{1.22\lambda }{2n\sin \beta }=\frac{0.61\lambda }{\text{N.A.}},\text{N.A.}=n\sin \beta .$

WHY the cone, not the diameter. A close object subtends the lens by a wide cone; how much of that cone the lens catches is what limits detail. Filling the gap with oil ($n>1$) widens the effective cone, so N.A. rises and $d_{\min }$ shrinks — finer detail. See Numerical aperture.

PICTURE. The objective seen from the slide: the red cone of half-angle $\beta$ entering the lens, medium of index $n$ between them.

Term by term:

• $\beta$ — half-angle of the collected cone.
• $n$ — refractive index of the medium (air $=1$, oil $\approx 1.5$).
• N.A. $=n\sin \beta$ — combines both; bigger ⇒ smaller $d_{\min }$.
• $0.61=1.22/2$ — same Bessel factor, halved by the two-sided cone geometry.

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The one-picture summary

Everything above, compressed: fan-out at the hole → Airy pattern with its first dark ring → two overlapping patterns → peak-on-first-minimum → ${\theta }_R=1.22\lambda /D$.

Recall Feynman retelling — say it back in plain words

Light is a wave, and every hole has edges, so light always spreads a little coming out — that spread is diffraction. One faraway dot therefore lands not as a point but as a bright disc ringed by dark circles (the Airy pattern). Two dots make two such discs. Slide them together and the dip between them fills in; at some point you can no longer swear there are two. Rayleigh's fair rule says "just resolved" is when one disc's bright centre sits right on the other disc's first dark ring. We already knew where that dark ring is: for a slit it is at $\lambda /a$, and for a round hole the Bessel maths nudges it to $1.22\lambda /D$. So the smallest angle you can still split is ${\theta }_R=1.22\lambda /D$ — bigger hole or shorter wave means a smaller angle means sharper eyes. For a microscope we swap the diameter for the collecting cone, and ${\theta }_R$ becomes $d_{\min }=0.61\lambda /\text{N.A.}$, which is why oil immersion, by widening the cone, lets you see finer things.

Recall Quick self-check

Peak of one pattern sits on the ...... of the other. ::: first minimum (first dark ring) Slit first-minimum angle? ::: $\theta \approx \lambda /a$ Circular-aperture limit? ::: ${\theta }_R=1.22\lambda /D$ (radians) Where does 1.22 come from? ::: first zero of the Bessel function $J_1$ Microscope limit in N.A.? ::: $d_{\min }=0.61\lambda /\text{N.A.}$