Visual walkthrough — Newton's rings — derivation

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Step 1 — Set the stage: what is actually touching what?

WHAT. Take a plano-convex lens (flat on top, gently curved — convex — on the bottom) and rest it, curved-side-down, on a flat glass plate. They touch at exactly one point. Everywhere else there is a paper-thin gap of air between the curved glass and the flat glass.

WHY. Interference needs two surfaces close enough that light bouncing off both stays "in step" (this closeness idea is coherence). The trapped air layer is our thin film. Nothing else in the setup matters yet.

PICTURE.

• The cyan curve is the bottom of the lens.
• The white line is the flat plate.
• The amber dot is the single contact point — call the vertical line through it the axis.
• $r$ = horizontal distance from the axis (we will use it constantly).
• $t$ = the height of the air gap at distance $r$. At the axis, $t=0$.

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Step 2 — Two rays are born: A and B

WHAT. Shine monochromatic light (one single colour → one wavelength $\lambda$) straight down. At the bottom face of the lens the beam splits into two:

• Ray A reflects straight back off that glass→air surface.
• Ray B keeps going, crosses the air gap, reflects off the top of the flat plate (air→glass), and comes back up.

WHY. These two rays started as one beam, so they are perfectly in step (coherent). When they recombine at your eye they will either reinforce or cancel — that decision is the entire pattern.

PICTURE.

• Ray A (white) — short bounce, off the top of the gap.
• Ray B (amber) — the long detour, down through the gap and up again.
• The shaded band is the air film of thickness $t$.

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Step 3 — Measure the extra distance ray B travelled

WHAT. For light hitting nearly straight down (near-normal), ray B goes down through the film and back up. It crosses the thickness $t$ twice. So its extra geometric travel over ray A is:

$\underset{\text{down and up}}{\underbrace{2}}\times \underset{\text{index of the film}}{\underbrace{\mu }}\times \underset{\text{film thickness}}{\underbrace{t}}=2\mu t$

WHY each symbol.

• The 2 — because B makes a round trip through the gap, not a one-way trip. Forgetting this halves every answer (a classic slip).
• $\mu$ (mu) — the refractive index of whatever fills the gap. Light "counts distance" more slowly in a denser medium, so we weight the physical thickness by $\mu$. For plain air, $\mu \approx 1$, so $2\mu t\to 2t$.
• $t$ — the thickness from Step 1.

PICTURE.

The amber path is drawn slightly tilted so you can see the down-leg and up-leg; each leg is $t$ long, total $2t$ (times $\mu$).

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Step 4 — The hidden half-wavelength (Stokes' flip)

WHAT. There is one more effect that has nothing to do with distance. When light reflects off a surface where the next medium is denser (higher index), the wave flips — as if it lost half a wavelength. This is a Stokes phase change of $\pi$.

• Ray A: glass → air reflection. Air is less dense → no flip.
• Ray B: air → glass reflection at the plate. Glass is more dense → flip, worth $+\frac{\lambda }{2}$.

Exactly one of the two rays flips. So the total optical path difference is:

$\Delta =\underset{\text{distance part (Step 3)}}{\underbrace{2\mu t}}+\underset{\text{Stokes flip on B only}}{\underbrace{\frac{\lambda }{2}}}$

WHY it matters most at the centre. At the contact point $t=0$, the distance part vanishes but the $\frac{\lambda }{2}$ stays. That leftover half-wave is what will make the very centre dark.

PICTURE.

• Left wave: reflects with no flip (crest stays a crest).
• Right wave: reflects off denser glass and comes back inverted — the amber crest became a trough. That inversion is the extra $\lambda /2$.

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Step 5 — Turn "path difference" into bright / dark rules

WHAT. Two waves reinforce (bright) when B lags A by a whole number of wavelengths, and cancel (dark) when the lag is a whole-plus-a-half.

$\text{Bright: }\Delta =m\lambda \text{Dark: }\Delta =\left(m+\frac{1}{2}\right)\lambda$

Now substitute $\Delta =2\mu t+\frac{\lambda }{2}$. The $\frac{\lambda }{2}$ on the left swaps the two conditions:

WHY the swap is the punchline. Put $t=0$ into the dark condition: $0=0\cdot \lambda$ ✓. So the exact centre satisfies the dark rule — the observed dark spot is explained, purely by the Stokes flip.

PICTURE.

Two ripples drawn on top of each other: on the left, crest-on-trough = flat line = dark; on the right, crest-on-crest = tall wave = bright.

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Step 6 — The geometry link: why $t$ grows as $r^2$

WHAT. The lens surface is part of a sphere of radius $R$. Drop the sphere onto the plate; at horizontal distance $r$ from the contact point, the surface has risen by height $t$. Pythagoras on the sphere gives:

$r^2=t(2R-t)$

Since the gap is microns and $R$ is metres, $t\ll R$, so $2R-t\approx 2R$:

$\boxed{t=\frac{r^2}{2R}}$

WHY each symbol.

• $R$ — radius of the sphere the lens was ground from; large ($\sim$ metre).
• The full form $r^2=t(2R-t)$ comes from the intersecting-chord theorem: the vertical chord of length $2R$ splits into $t$ and $(2R-t)$, and the half-chord across it is $r$.
• Dropping $t$ next to $2R$ is the key approximation — legal because $t/R\sim 10^{-6}$.

The important shape of the answer: $t$ is proportional to $r^2$, not to $r$. Double the radius → four times the gap.

PICTURE.

The right triangle inside the sphere: horizontal leg $r$, and the little sagitta $t$ climbing up to the surface. Notice how flat the curve is near the centre — that is why $t$ starts tiny and speeds up.

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Step 7 — Put geometry into the ring rule → the payoff formulas

WHAT. Substitute $t=\frac{r^2}{2R}$ into the dark condition $2\mu t=m\lambda$:

$2\mu \cdot \frac{r_m^2}{2R}=m\lambda ⟹\frac{\mu r_m^2}{R}=m\lambda ⟹r_m^2=\frac{m\lambda R}{\mu }$

For air ($\mu =1$):

WHY the shape. $r_m\propto \sqrt{m}$. Going from ring 1 to 2 the radius grows like $\sqrt{1}\to \sqrt{2}$; from 100 to 101 it barely moves. So the rings crowd together outward — never a ruler-even spacing.

PICTURE.

Concentric dark rings with a dark centre; the tick marks show the gaps shrinking as $\sqrt{m}$ climbs — proof by eye that they crowd.

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Step 8 — The centre and the offset-centre problem (degenerate cases)

WHAT — centre, $m=0$. Dark rule with $m=0$ gives $r_0=0$: the very centre is a dark spot, matching Step 4. This is the degenerate ring of zero radius.

WHAT — measuring diameters, not radii. In practice the contact point is squashed and fuzzy, so you cannot trust $r$ from the "centre". Instead measure a full diameter $D$ across a ring, and take differences of two rings $n$ and $m$:

$D_n^2-D_m^2=4n\lambda R-4m\lambda R=4(n-m)\lambda R$

$\boxed{\lambda =\frac{D_n^2-D_m^2}{4(n-m)R}}$

WHY subtract. Any offset of the estimated centre adds the same constant to every $D^2$, and the Stokes $\lambda /2$ also sits inside every ring the same way. Subtracting two kills both nuisances at once — a clean straight line.

PICTURE.

A ring with the true centre offset from the guessed one; the two half-widths differ, but the full diameter (edge to edge) is immune to the offset.

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The one-picture summary

One diagram, whole chain: split into rays A & B → extra path $2\mu t$ → plus Stokes $\lambda /2$ → condition $2\mu t=m\lambda$ → geometry $t=r^2/2R$ → radius $r_m=\sqrt{m\lambda R}$ → crowding rings.

Recall Feynman retelling — say it back in plain words

A round lens sits on a flat sheet of glass; between them is a wisp of air that is zero-thick where they touch and gets thicker the further out you go. Shine one pure colour of light down. Some bounces off the top of the air; some dives through, bounces off the plate, and comes back — this second beam travelled an extra down-and-up, which is $2t$ (times the index $\mu$ of whatever's in the gap). On top of that, the beam that bounced off the denser plate flips over, which is like adding half a wavelength. So even where the gap is zero, the two beams are half a wavelength out — they cancel, and the very middle is dark. Move out: the gap grows like the square of the distance ($t=r^2/2R$, from the sphere's geometry), so every time the extra path clocks up another whole wavelength you get a dark ring, and halfway between, a bright one. Because $r$ goes like the square root of the ring number, the outer rings pile up close together. To measure a wavelength cleanly you don't trust the fuzzy centre — you measure diameters across the rings and subtract two, which cancels both the centre error and the half-wavelength, leaving $\lambda =(D_n^2-D_m^2)/[4(n-m)R]$.

Recall Quick self-test

Why dark centre? ::: At $t=0$ the only difference is the Stokes $\lambda /2$ → destructive. Why factor 2 in $2\mu t$? ::: Ray B crosses the film down and up. Why $t\propto r^2$? ::: Sphere geometry: $r^2\approx 2Rt$. Why crowd outward? ::: $r_m=\sqrt{m\lambda R}\propto \sqrt{m}$, so spacing shrinks. Why subtract diameters? ::: Cancels unknown centre offset and the $\lambda /2$ constant.