Resolving power — Rayleigh criterion — Physics notes

WHY does diffraction limit resolution at all?

So the problem is: two sources, two overlapping Airy discs. When can we separate them?

The Rayleigh Criterion

HOW: deriving the limiting angle from scratch

Step 1 — Where is the first dark ring of a single source?

For a single slit of width $a$ , the first minimum occurs when the path difference across the slit equals one wavelength: $a\sin\theta = \lambda \quad\Rightarrow\quad \theta_{\min}\approx \frac{\lambda}{a}$

Why this step? Pairing the top half of the slit against the bottom half, each ray from the top cancels its partner $a/2$ below when their path difference is $\lambda/2$ ; summed over the whole slit this gives total destructive interference at $a\sin\theta=\lambda$ .

Step 2 — Circular aperture correction.

A real lens/eye/telescope has a circular aperture (diameter $D$ ), not a slit. Integrating diffraction over a disc instead of a strip introduces a numerical factor of 1.22 (the first zero of the Bessel function $J_1$ ): $\theta_{\min}= 1.22\,\frac{\lambda}{D}$

Why this step? The slit gives $\lambda/a$ ; the circular geometry redistributes the diffracted energy and pushes the first dark ring slightly out, multiplying by $1.22$ . (For a slit aperture, drop the 1.22 and use width $a$ .)

Step 3 — Apply Rayleigh.

"Peak on first minimum" means the minimum angular separation between the two sources that we can still resolve equals exactly $\theta_{\min}$ :

Figure — Resolving power — Rayleigh criterion

Two important special cases

(a) Telescope / eye (objects far away): sources separated by angle $\theta$ . Resolvable if $\theta \ge \theta_R = 1.22\lambda/D$ .

(b) Microscope (objects close, defined by smallest distance $d$ ): Here we care about the smallest separation $d$ between two points on the slide. Deriving similarly with the objective's collecting cone gives:

Why N.A. and not $D$ ? For a near object the relevant quantity is the cone half-angle the lens collects, not just diameter; immersion oil ( $n>1$ ) widens the effective cone, shrinking $d_{\min}$ — that's why oil-immersion microscopy resolves finer detail.

Worked Examples

Worked example 1 — The human eye

Pupil $D = 2\text{ mm}$ , $\lambda = 550\text{ nm}$ . Find the smallest angle resolvable and the dot spacing readable at $25\text{ cm}$ .

Angle: $\theta_R = 1.22\dfrac{\lambda}{D}=1.22\times\dfrac{550\times10^{-9}}{2\times10^{-3}} = 3.4\times10^{-4}\text{ rad}$ . Why? Direct plug-in; keep $\lambda,D$ in metres so the ratio is dimensionless (radians).

Spacing: $s=\theta_R \cdot L = 3.4\times10^{-4}\times0.25 = 8.4\times10^{-5}\text{ m}\approx 0.084\text{ mm}$ . Why? Small angle: arc length $s=\theta L$ . So two dots closer than ~0.08 mm blur into one — matches everyday experience of print resolution.

Forecast-then-Verify

Recall Forecast before reading the answer

Q: You double the wavelength AND double the aperture diameter. What happens to $\theta_R$ ? Forecast… then check: $\theta_R \propto \lambda/D$ . Doubling both: $(2\lambda)/(2D)=\lambda/D$ → unchanged. The two effects cancel exactly.

Common Mistakes (Steel-manned)

Flashcards

State the Rayleigh criterion.

Two sources are just resolved when the central maximum of one's diffraction pattern coincides with the first minimum (first dark ring) of the other's.

Formula for limiting angular resolution of a circular aperture?

\theta_R = 1.22\,\lambda/D

, in radians.

Where does the factor 1.22 come from?

The first zero of the Bessel function

J_1

for a circular aperture (vs

\lambda/a

for a slit).

What is resolving power?

The ability to separate two close objects; equal to

1/\theta_R

— larger means finer detail resolvable.

How does resolving power depend on aperture diameter and wavelength?

Improves (θ_R smaller) with larger D and smaller λ.

Microscope resolution limit in terms of N.A.?

d_{\min}=0.61\lambda/\text{N.A.}

, where N.A.

= n\sin\beta

Why use oil immersion in microscopy?

It increases n (and thus N.A.), widening the collecting cone and shrinking

d_{\min}

for finer resolution.

Why are large telescopes good for resolution?

Larger D reduces θ_R = 1.22λ/D, so finer angular separations of stars can be distinguished.

Is magnification the same as resolving power?

No — magnification enlarges; beyond the diffraction limit it gives "empty magnification" with no extra detail.

Roughly what intensity dip exists between two just-resolved peaks?

About 26%, enough to be detected as a dip between two peaks.

Recall Feynman: explain to a 12-year-old

When light from a tiny faraway dot goes through a round hole, it doesn't stay a tiny dot — it spreads into a small fuzzy circle, like a flashlight beam getting wider. Now imagine two dots: each makes its own fuzzy circle. If the circles overlap a lot, your eyes see one fuzzy blob and you can't tell there were two dots! The Rayleigh rule says: you can just barely tell them apart when the bright middle of one fuzzy circle sits right on the dark edge-ring of the other. A bigger hole makes each circle less fuzzy, so you can tell dots apart more easily — that's why giant telescopes can see two stars that look like one star to us.

Connections

Single-slit diffraction — source of the $\lambda/a$ angle reused here.
Diffraction grating — resolving power — different formula $R=\lambda/\Delta\lambda = nN$ .
Airy disc and Bessel functions — origin of the 1.22 factor.
Numerical aperture — drives microscope resolution.
Huygens principle — why finite apertures diffract.
Small angle approximation — converting $\theta_R$ to physical spacing $s=\theta L$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab kisi door ke point source (jaise ek star) ki light kisi round hole ya lens se guzarti hai, to wo ek sharp point nahi banati — diffraction ki wajah se ek chhoti si fuzzy disc ban jaati hai jise Airy disc kehte hain. Ab agar do stars paas-paas hain, to dono ki fuzzy discs overlap kar jaati hain. Agar overlap zyada ho gaya to aankh ko sirf ek blob dikhega, do alag nahi. Toh sawaal yeh hai: kab tak hum do ko do keh sakte hain? Iska jawab Rayleigh criterion deta hai.

Rayleigh criterion bolta hai: do sources tab "just resolved" hote hain jab ek ki central maximum doosre ki first dark ring (first minimum) pe exactly baith jaaye. Is exact spacing par beech mein lagbhag 26% ka dip aata hai, jisse aankh do peaks pehchaan leti hai. Isse minimum angle nikalta hai: $\theta_R = 1.22\,\lambda/D$ (radians mein), jahan $D$ aperture ka diameter hai. Yaad rakho — 1.22 sirf round aperture ke liye, slit ke liye seedha $\lambda/a$ .

Iska practical matlab: bada $D$ ya chhota $\lambda$ → chhota $\theta_R$ → behtar resolution. Isi liye telescope bade banaye jaate hain — sirf zyada light ke liye nahi, balki do paas-paas ke stars ko alag dekhne ke liye. Microscope mein cheez paas hoti hai, to wahan numerical aperture (N.A. $=n\sin\beta$ ) kaam aata hai: $d_{\min}=0.61\lambda/\text{N.A.}$ , aur isi liye oil-immersion ( $n>1$ ) se finer detail dikhta hai.

Ek common galti: resolving power ko magnification samajh lena. Zyada zoom karne se blur bada ho jaata hai, detail nahi aata — ise "empty magnification" kehte hain. Asli detail diffraction limit se decide hota hai. Doosri galti: $\theta$ ko degrees mein use karna — formula radians deta hai, aur physical spacing $s=\theta L$ ke liye radians hi chahiye.

Resolving power — Rayleigh criterion

WHY does diffraction limit resolution at all?

The Rayleigh Criterion

HOW: deriving the limiting angle from scratch

Two important special cases

Worked Examples

Forecast-then-Verify

Common Mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Optics

Connections