Newton's rings — derivation — Physics notes

WHAT is happening?

WHY rings (circles)? The air film has the same thickness $t$ at all points equidistant from the contact point. A locus of constant $t$ is a circle → fringes of equal thickness are circles.

WHY a dark centre? At the exact point of contact $t=0$ , yet there is still a sudden phase flip on one reflection (explained below). Net effect → destructive interference → central dark spot.

HOW the optical path difference arises (first principles)

Two rays matter:

Ray A reflects off the bottom of the glass lens (glass→air boundary).
Ray B transmits, crosses the air gap of thickness $t$ , reflects off the top of the flat plate (air→glass boundary), and comes back.

The geometric extra path travelled by B is down and up through the film. For near-normal incidence in a film of refractive index $\mu$ :

$\text{geometric path difference} = 2\mu t$

The crucial half-wavelength

So the total optical path difference is:

$\boxed{\Delta = 2\mu t \pm \frac{\lambda}{2}}$

The $\pm$ is just bookkeeping; what matters is the extra $\lambda/2$ exists.

Conditions for bright and dark rings

Constructive (bright) needs $\Delta = m\lambda$ ; destructive (dark) needs $\Delta = (m+\tfrac12)\lambda$ . Substituting $\Delta = 2\mu t + \lambda/2$ :

Geometry: relating $t$ to the radius $r$

Derivation (circle theorem / Pythagoras):

Place the centre of the spherical surface at height $R$ above the contact point. A point on the lens surface at horizontal distance $r$ sits at height $t$ above the plate. By the chord (intersecting-chords) theorem for the circle of radius $R$ :

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

Why this step? The vertical chord through the air gap has segments $t$ and $(2R-t)$ ; the perpendicular half-chord is $r$ , and their product equals $r^2$ .

Since $t \ll R$ (the gap is microns, $R$ is ~metres), $2R - t \approx 2R$ :

$r^2 \approx 2Rt \quad\Longrightarrow\quad t = \frac{r^2}{2R}$

Final ring-radius formulas (the payoff)

Dark rings — put $t = \dfrac{r^2}{2R}$ into $2\mu t = m\lambda$ :

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

Bright rings — using $2\mu t = (m-\tfrac12)\lambda$ :

In terms of diameter $D=2r$ (what you actually measure):

$D_m^2 = 4m\lambda R \quad (\text{dark, air})$

Measuring $\lambda$ or $R$ — eliminating the unknown centre

WHY subtract? If the true centre is offset, every $D^2$ carries the same additive error; subtracting two cancels it and cancels the awkward " $+\lambda/2$ " constant. Clean linear relation.

Worked examples

Worked example Example 2 — find

$\lambda$ from measured diameters $D_{15}=4.30$ mm, $D_5=2.48$ mm, $R=1.20$ m, air. Find $\lambda$ .

Step 1: Use $\lambda=\dfrac{D_n^2-D_m^2}{4(n-m)R}$ . Why? Difference method kills centre error. Step 2: $D_{15}^2=1.849\times10^{-5}$ , $D_5^2=6.150\times10^{-6}\ \text{m}^2$ . Step 3: Numerator $=1.234\times10^{-5}$ ; $4(15-5)(1.20)=48$ . Step 4: $\lambda = 1.234\times10^{-5}/48 = 2.57\times10^{-7}$ m. Hmm, that's 257 nm — recheck inputs in a real lab; here it shows the mechanics.

Common mistakes (steel-manned)

Active recall flashcards

Why is the centre of Newton's rings dark in reflected light?

t=0

geometric path is 0, but one reflection (air→glass) adds a

\lambda/2

phase flip → net

\Delta=\lambda/2

→ destructive.

What is the total optical path difference for the air film?

\Delta = 2\mu t + \lambda/2

(factor 2 = down-and-up;

\lambda/2

= Stokes flip).

Dark ring condition in terms of

t

2\mu t = m\lambda

Bright ring condition in terms of

t

2\mu t = (m-\tfrac12)\lambda

Relation between film thickness and radius?

r^2 = t(2R-t)\approx 2Rt

, so

t=r^2/2R

Radius of

m

-th dark ring (air)?

r_m=\sqrt{m\lambda R}

Why use diameter difference

D_n^2-D_m^2

Cancels the uncertain centre and the constant

\lambda/2

term, giving

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

Formula for

\lambda

from the experiment?

\lambda=(D_n^2-D_m^2)/[4(n-m)R]

What happens to ring diameters if a liquid

\mu

fills the gap?

They shrink by

1/\sqrt\mu

;

D^2\propto 1/\mu

Why do rings crowd outward?

Because

r_m\propto\sqrt m

, consecutive spacing

\Delta r

decreases.

Recall Feynman: explain to a 12-year-old

Press a magnifying-glass lens onto a flat piece of glass. Right where they touch there's no gap; a little outside there's a tiny air gap that grows thicker like the slope of a hill. When light bounces off the top and the bottom of that thin air gap, the two bounced beams race each other. Sometimes they line up (bright), sometimes they cancel (dark), and that makes glowing circles. There's a sneaky rule: one of the bounces flips the wave upside-down, so right at the touching point the waves cancel — the bullseye is dark. The faster the air gap grows, the closer the circles get squished together.

Connections

Thin film interference — Newton's rings are a special "wedge of air" case.
Stokes' relations & phase change on reflection — source of the $\lambda/2$ flip.
Interference — path difference & coherence — $\Delta=m\lambda$ vs $(m+\tfrac12)\lambda$ .
Michelson interferometer — also uses $D^2$ -difference logic to find $\lambda$ .
Refractive index measurement — liquid-in-gap method here.
Wedge fringes / air wedge — straight-line analogue of these circular fringes.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Newton's rings ka funda simple hai. Ek plano-convex lens (curved side neeche) ko ek flat glass plate par rakho. Lens aur plate ke beech mein ek patli air ki layer ban jaati hai — contact point par thickness zero, aur jaise-jaise bahar jaate ho, gap badhta jaata hai. Light is air film ke upar aur neeche se reflect hoti hai, aur do reflected rays aapas mein interfere karti hain. Same thickness wale points ek circle banate hain, isliye pattern circular rings ka aata hai.

Ab ek important baat — Stokes ka phase flip. Jab light low index se high index par reflect hoti hai (air se glass, plate wali side), toh ek extra $\lambda/2$ ka jhatka lagta hai. Isliye exact contact point par, jahan thickness zero hai, phir bhi waves cancel ho jaati hain aur centre dark dikhta hai. Total path difference banta hai $\Delta = 2\mu t + \lambda/2$ . Yahan $2t$ isliye kyunki ray neeche jaati hai aur wapas upar aati hai — do baar thickness travel.

Geometry ka magic: circle theorem se $r^2 = t(2R-t)$ , aur kyunki $t$ bahut chhota hai compared to lens radius $R$ , hum likhte hain $t = r^2/2R$ . Isko condition mein daalo toh dark ring ka radius nikalta hai $r_m=\sqrt{m\lambda R}$ . Iska matlab radius $\sqrt{m}$ ke proportional hai — isliye rings bahar jaate-jaate paas-paas (crowded) hote jaate hain, equally spaced nahi.

Practical use: hum diameter measure karte hain, radius nahi, kyunki centre point thoda uncertain hota hai. Do rings ka difference lo: $D_n^2 - D_m^2 = 4(n-m)\lambda R$ . Isse $\lambda$ ya $R$ nikal sakte ho, aur centre ki galti automatically cancel ho jaati hai. Agar gap mein koi liquid bhar do toh rings $1/\sqrt{\mu}$ se chhoti ho jaati hain — yeh liquid ka refractive index maapne ka tareeka hai. Yahi exam aur lab dono mein kaam aata hai.

Newton's rings — derivation

WHAT is happening?

HOW the optical path difference arises (first principles)

The crucial half-wavelength

Conditions for bright and dark rings

Geometry: relating $t$ to the radius $r$

Final ring-radius formulas (the payoff)

Measuring $\lambda$ or $R$ — eliminating the unknown centre

Worked examples

Common mistakes (steel-manned)

Active recall flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Optics

Connections