Exercises — Resolving power — Rayleigh criterion

WHAT: We are simply naming the Rayleigh angle. ${\theta }_R=1.22\frac{\lambda }{D}$ Units: radians (it is a pure ratio $\lambda /D$ times a number, so no length units survive — that ratio is dimensionless, and a dimensionless "angle" is by definition in radians).

WHY no 1.22: the factor 1.22 is the first zero of the Bessel function $J_1$, born from integrating diffraction over a disc. A slit is a strip, so we use the plain single-slit result (see Single-slit diffraction). ${\theta }_R=\frac{\lambda }{a}=\frac{600\times 10^{-9}}{0.10\times 10^{-3}}=6.0\times 10^{-3}\text{ rad}$

Direct substitution, keeping everything in metres so the ratio is dimensionless: ${\theta }_R=1.22\times \frac{550\times 10^{-9}}{2.0\times 10^{-3}}=3.4\times 10^{-4}\text{ rad}$

WHY arc length: for a tiny angle, the small-angle approximation (Small angle approximation) says the arc across distance $L$ is just $s=\theta L$ — no trig needed because $\tan \theta \approx \theta$ when $\theta$ is a fraction of a milliradian. $s={\theta }_{R}L=3.4\times 10^{-4}\times 0.25=8.5\times 10^{-5}\text{ m}\approx 0.085\text{ mm}$ Two dots closer than ~0.09 mm merge — which is why fine print at arm's length blurs.

${\theta }_R=1.22\times \frac{500\times 10^{-9}}{1.0}=6.1\times 10^{-7}\text{ rad}$ This is ~560× finer than the eye — the main reason big telescopes exist is resolution, not just light-gathering.

${\theta }_R\propto \lambda /D$.

• Halving $\lambda$: ${\theta }_R\to \frac{1}{2}{\theta }_R$ (factor 0.5).
• Doubling $D$: ${\theta }_R\to \frac{1}{2}{\theta }_R$ (factor 0.5).

They are identical. Both improve resolution by exactly a factor of 2. Neither wins — the formula is symmetric in "how you shrink $\lambda /D$."

${\theta }_R^{\prime }=1.22\frac{2\lambda }{2D}=1.22\frac{\lambda }{D}={\theta }_R$ Unchanged — the two effects cancel exactly. This is the parent note's Forecast-then-Verify result: always reason with the ratio $\lambda /D$, not the two numbers separately.

Step 1 — resolution angle: ${\theta }_R=1.22\times \frac{550\times 10^{-9}}{5.0\times 10^{-3}}=1.342\times 10^{-4}\text{ rad}$ Step 2 — invert the arc relation. We know the true angle subtended is $\theta =s/L$. Resolvable requires $\theta \ge {\theta }_R$, so the limiting distance is where $\theta ={\theta }_R$: $L_{\max }=\frac{s}{{\theta }_R}=\frac{1.5}{1.342\times 10^{-4}}\approx 1.12\times 10^4\text{ m}\approx 11\text{ km}$ Beyond ~11 km (in ideal, diffraction-limited conditions) the two lights fuse into one point. Real eyes do worse due to aberrations, but this is the hard ceiling.

WHY numerical aperture, not $D$: for a near object the lens collects a cone of light; the wider that cone, the finer the detail. The cone width is captured by $\text{N.A.}=n\sin \beta$ (see Numerical aperture), and $d_{\min }=0.61\lambda /\text{N.A.}$

First, $\sin 65^{\circ }=0.9063$.

Air: $\text{N.A.}=1.00\times 0.9063=0.9063$ $d_{\min }^{\text{air}}=\frac{0.61\times 500\times 10^{-9}}{0.9063}=3.37\times 10^{-7}\text{ m}\approx 337\text{ nm}$

Oil: $\text{N.A.}=1.515\times 0.9063=1.373$ $d_{\min }^{\text{oil}}=\frac{0.61\times 500\times 10^{-9}}{1.373}=2.22\times 10^{-7}\text{ m}\approx 222\text{ nm}$

Oil improves resolution by the factor $n=1.515$ — because immersion oil raises $n$, widening the effective cone. That is the whole reason oil-immersion lenses resolve finer.

(a) Telescope: ${\theta }_R^{\text{tel}}=1.22\times \frac{550\times 10^{-9}}{0.15}=4.47\times 10^{-6}\text{ rad}$ The stars' separation $3.0\times 10^{-6}$ rad is smaller than ${\theta }_R^{\text{tel}}$, so not resolved — even this telescope cannot split them. (b) Eye: ${\theta }_R^{\text{eye}}=1.22\times \frac{550\times 10^{-9}}{2.0\times 10^{-3}}=3.36\times 10^{-4}\text{ rad}$ Far larger still ⇒ the eye resolves nothing this fine. Both fail. Only a bigger $D$ (or shorter $\lambda$) would help; magnification alone would not — it enlarges the merged blob.

Step 1 — required angle (small-angle law): ${\theta }_{\text{req}}=\frac{s}{L}=\frac{0.30}{3.0\times 10^5}=1.0\times 10^{-6}\text{ rad}$ Step 2 — invert Rayleigh to solve for $D$. We need ${\theta }_R\le {\theta }_{\text{req}}$, i.e. the mirror must be at least big enough that its resolution angle is this small: $D_{\min }=\frac{1.22\lambda }{{\theta }_{\text{req}}}=\frac{1.22\times 500\times 10^{-9}}{1.0\times 10^{-6}}=0.61\text{ m}$ A mirror of at least ~0.61 m is required. Diffraction, not lens quality, sets this floor.

• Same idea: in both, resolution improves when the wave samples a wider extent of the aperture — more slits $N$ (grating) or larger diameter $D$ (lens). A bigger illuminated region means a narrower diffraction feature, so two nearby peaks/wavelengths separate. See Diffraction grating — resolving power.
• Different costume: the grating separates two wavelengths (spectral), the aperture separates two directions/points (spatial); but both are limited by how much of the wavefront you capture. Same physics (Huygens principle: every point of the wavefront re-emits, and a wider set of secondary sources builds a sharper combined pattern).

(a) ${\theta }_R=1.22\times \frac{550\times 10^{-9}}{3.0\times 10^{-3}}=2.237\times 10^{-4}\text{ rad}$ $s={\theta }_{R}L=2.237\times 10^{-4}\times 0.30=6.71\times 10^{-5}\text{ m}\approx 0.067\text{ mm}$ (b) Pixels per inch = one inch (25.4 mm) divided by the pitch: $\text{ppi}=\frac{25.4\text{ mm}}{0.0671\text{ mm}}\approx 378\text{ ppi}$ Displays around 300–400 ppi at typical viewing distance sit right at the eye's diffraction+cone limit — pushing far above this is invisible to the viewer, exactly the "empty magnification" idea applied to screens.

Angle from aperture ::: ${\theta }_R=1.22\lambda /D$ (circle) or $\lambda /a$ (slit). Angle ↔ separation ::: $s=\theta L$, so $L=s/{\theta }_R$ and $D=1.22\lambda L/s$. Microscope ::: $d_{\min }=0.61\lambda /\text{N.A.}$, N.A. $=n\sin \beta$; oil raises $n$. Everything improves with ::: larger aperture / cone, shorter wavelength — never with magnification.