Newton's rings — derivation: Worked examples — Physics deep dive

This is the drill page for Newton's rings. The parent note built the physics; here we hammer every type of question one problem can be. Before each example you will Forecast the answer (guess first — that is how the shape of the formula becomes intuition), then work it step by step, then Verify by plugging back and checking units.

Everything rests on these earned facts from the parent:

Recall The facts we lean on (dark AND bright geometry)

Total optical path difference in the air film: $Δ = 2 μ t + \frac{λ}{2}$ . The 2 is because ray B goes down and up through the gap of thickness $t$ ; the $\frac{λ}{2}$ is the Stokes flip on one reflection.
Dark ring: $2 μ t = mλ$ . Bright ring: $2 μ t = (m - \frac{1}{2}) λ$ .
Geometry: $t = \frac{r ^{2}}{2 R}$ .
Dark-ring radius/diameter: $r_{m} = \frac{mλ R}{μ}$ , $D_{m}^{2} = \frac{4 mλ R}{μ}$ .
Bright-ring radius/diameter: $r_{m} = \frac{( 2 m - 1 ) λ R}{2 μ}$ , $D_{m}^{2} = \frac{4 ( m - \frac{1}{2} ) λ R}{μ} = \frac{2 ( 2 m - 1 ) λ R}{μ}$ .

Here $r$ is the ring radius (metres), $D = 2 r$ the diameter, $m$ the integer ring order, $λ$ the wavelength of light, $R$ the lens' radius of curvature, and $μ$ the refractive index of whatever fills the gap.

Related tools you may want open: Thin film interference, Stokes' relations & phase change on reflection, Refractive index measurement, Wedge fringes / air wedge.

The scenario matrix

Every Newton's-rings problem is one of these cells. The examples below are labelled with the cell they hit, so together they cover the whole table.

Cell	What varies	The trap / skill it tests
A. Direct radius	find $r_{m}$ or $D_{m}$ from $m, λ, R$	which formula (dark vs bright), factor 2
B. Degenerate centre	$m = 0$ , $t = 0$	dark or bright? Stokes flip decides
C. Solve for $λ$	measured $D_{n}, D_{m}$	difference method cancels centre
D. Solve for $R$	measured diameters, known $λ$	rearrange the same difference formula
E. Liquid in gap	$μ \neq = 1$	rings shrink by $1/ μ$ ; measure $μ$
F. Bright vs dark counting	consecutive rings, spacing	$r \propto m$ , crowding, spacing shrinks
G. Transmitted light	viewed from below	pattern inverts: centre bright
H. Word / exam twist	glass plate replaced, or find $t$	reading the physics, not just plugging

Example 1 — Cell A: direct radius of a dark ring

Worked example The 10th dark ring

Monochromatic light $λ = 600 nm$ , lens $R = 0.80 m$ , air film ( $μ = 1$ ). Find the radius and diameter of the 10th dark ring.

Forecast: Since $r_{m} = mλ R$ , expect roughly $10 \times 6 \times 1 0^{- 7} \times 0.8 \approx 5 \times 1 0^{- 6} \sim 2$ mm. Guess ~2 mm.

Step 1. Choose $r_{m} = mλ R / μ$ . Why this step? Dark ring in air, so the condition $2 μ t = mλ$ with $t = r^{2} /2 R$ gives exactly this. Step 2. Plug: $r_{10}^{2} = \frac{10 \times 600 \times 1 0 ^{- 9} \times 0.80}{1} = 4.8 \times 1 0^{- 6} m^{2}$ . Why this step? Every quantity in SI (metres), so the answer comes out in $m^{2}$ . Step 3. $r_{10} = 4.8 \times 1 0^{- 6} = 2.19 \times 1 0^{- 3} m = 2.19 mm$ . Why this step? Take the square root of the $r^{2}$ from Step 2 to recover the actual radius; $m^{2} = m$ . Step 4. Diameter $D_{10} = 2 r_{10} = 4.38 mm$ . Why this step? The measurable quantity is the diameter (edge to edge), which is just twice the radius.

Verify: Units: $m \cdot m = m$ ✓. Matches the ~2 mm forecast ✓. Back-check thickness: $t = r^{2} /2 R = 4.8 \times 1 0^{- 6} /1.6 = 3.0 μ m$ , and $2 t = 6.0 μ m = 10 \times 600 nm = 10 λ$ ✓.

Example 2 — Cell B: why the centre is dark (degenerate $m = 0$ )

Worked example The contact point

At the exact point of contact the air gap is $t = 0$ . Is the centre bright or dark in reflected light? Show it from the path difference.

Forecast: Zero gap feels like "in phase → bright." Guess bright — then watch it flip.

Step 1. Write total path difference: $Δ = 2 μ t + \frac{λ}{2}$ . Why this step? You must include the Stokes flip; the geometric term alone is a trap. Step 2. Set $t = 0$ : $Δ = 0 + \frac{λ}{2} = \frac{λ}{2}$ . Why this step? $t = 0$ removes the geometric part; only the flip survives. Step 3. A path difference of exactly half a wavelength means the two waves arrive out of phase → destructive → dark. Why this step? Half a wavelength of path shifts one wave's crest onto the other's trough, so they cancel — that is the definition of destructive interference.

Verify: Consistency with the dark condition $2 μ t = mλ$ : putting $m = 0$ gives $t = 0$ , which is the centre. So $m = 0$ is a dark ring — the centre is the zeroth dark ring ✓. Guess was wrong; the flip is the whole story.

Example 3 — Cell C: measuring $λ$ by the difference method

Worked example Wavelength from two dark rings

Dark-ring diameters $D_{20} = 4.60 mm$ and $D_{4} = 1.80 mm$ measured; $R = 1.00 m$ , air. Find $λ$ .

Forecast: Visible light is 400–700 nm, so expect a few hundred nm.

Step 1. Use $λ = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) R}$ . Why this step? Each dark $D^{2}$ carries the same additive error from the uncertain, deformed contact point (you never know exactly where the centre is). Subtracting two of them cancels that common offset, leaving a clean linear relation — the Stokes $\frac{λ}{2}$ is already absorbed into the integer $m$ -indexing of the dark condition, so there is nothing extra to cancel there. Step 2. $D_{20}^{2} = (4.60 \times 1 0^{- 3})^{2} = 2.116 \times 1 0^{- 5} m^{2}$ ; $D_{4}^{2} = (1.80 \times 1 0^{- 3})^{2} = 3.24 \times 1 0^{- 6} m^{2}$ . Why this step? Work in $m^{2}$ so the final $λ$ lands in metres. Step 3. Numerator $= 2.116 \times 1 0^{- 5} - 3.24 \times 1 0^{- 6} = 1.792 \times 1 0^{- 5} m^{2}$ . Why this step? This difference is the physically meaningful, centre-independent quantity that equals $4 (n - m) λ R$ . Step 4. Denominator $= 4 (20 - 4) (1.00) = 64$ . Why this step? Assemble $4 (n - m) R$ — the known constants that scale the difference. Step 5. $λ = \frac{1.792 \times 1 0 ^{- 5}}{64} = 2.80 \times 1 0^{- 7} m = 280 nm$ . Why this step? Divide the measured numerator ( $m^{2}$ ) by the constant ( $m$ ) to isolate $λ$ in metres.

Verify: Units: $m^{2} / m = m$ ✓. Numerically small but the mechanics are exact; in a real lab diameters this close would push you to more rings apart for accuracy. Sanity: increasing $(n - m)$ shrinks the answer, so wide ring separations give tighter $λ$ estimates ✓.

Example 4 — Cell D: measuring $R$ of the lens

Worked example Radius of curvature

With sodium light $λ = 589 nm$ , air film: $D_{15}^{2} - D_{5}^{2} = 1.20 \times 1 0^{- 5} m^{2}$ . Find $R$ .

Forecast: Lab lenses have $R$ of order 1 m. Guess ~1 m.

Step 1. Rearrange the same difference law: $R = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) λ}$ . Why this step? One equation, one unknown — solve for whichever symbol the question hides. Step 2. $n - m = 15 - 5 = 10$ ; $4 (n - m) λ = 4 \times 10 \times 589 \times 1 0^{- 9} = 2.356 \times 1 0^{- 5} m$ . Why this step? Assemble the denominator entirely in SI so the ratio gives metres. Step 3. $R = \frac{1.20 \times 1 0 ^{- 5}}{2.356 \times 1 0 ^{- 5}} = 0.509 m$ . Why this step? Divide the measured $D^{2}$ -difference ( $m^{2}$ ) by the assembled constant ( $m$ ) to isolate $R$ .

Verify: Units: $m^{2} / m = m$ ✓. Order ~0.5 m, sensible for a lens ✓. Cross-check: with this $R$ , $D_{5}^{2}$ and $D_{15}^{2}$ would differ by $4 \times 10 \times 589 nm \times 0.509 = 1.20 \times 1 0^{- 5} m^{2}$ — exactly the input ✓.

Example 5 — Cell E: liquid fills the gap (measure $μ$ )

Worked example Rings shrink with liquid

The same setup gives dark-ring diameter $D_{10}^{air} = 4.00 mm$ in air. A transparent liquid is introduced and now $D_{10}^{liq} = 3.20 mm$ . Find $μ$ of the liquid.

Forecast: Rings got smaller → $μ > 1$ . Since $D \propto 1/ μ$ and the ratio is $3.2/4.0 = 0.8$ , expect $μ = 1/0. 8^{2} \approx 1.56$ .

Step 1. For a given ring $m$ : $D_{m}^{2} = \frac{4 mλ R}{μ}$ . Why this step? $μ$ sits in the denominator, so filling with denser medium needs less thickness for the same order → smaller ring. Step 2. Same $m, λ, R$ in both cases, so divide: $\frac{D _{air}^{2}}{D _{liq}^{2}} = \frac{μ _{liq}}{μ _{air}} = μ_{liq}$ (air $μ \approx 1$ ). Why this step? Taking a ratio kills every quantity we don't know ( $m, λ, R$ ), leaving $μ$ alone. Step 3. $μ = \frac{( 4.00 ) ^{2}}{( 3.20 ) ^{2}} = \frac{16.0}{10.24} = 1.5625$ . Why this step? Substitute the two measured diameters; the units (mm) cancel in the ratio, so $μ$ is correctly dimensionless.

Verify: Ratio $D_{liq} / D_{air} = 3.2/4.0 = 0.8 = 1/ 1.5625$ since $1.5625 = 1.25$ ✓. Rings shrank as predicted for $μ > 1$ ✓. This is the refractive-index measurement method — see Refractive index measurement.

Example 6 — Cell F: spacing & crowding (bright rings, geometry)

Figure: bright-ring radius $r$ (vertical axis, proportional units) plotted against ring order $m$ (horizontal axis). The blue curve follows $r \propto 2 m - 1$ . The pink double-arrow marks the large gap between rings 1 and 2; the yellow double-arrow marks the much smaller gap between rings 9 and 10 — the square-root curve flattens outward, so equal steps in $m$ give shrinking steps in $r$ . That is the crowding of Newton's rings.

Worked example Why the gaps shrink outward

Air film, $λ = 500 nm$ , $R = 1.00 m$ . Find the radius of the 1st and 2nd bright rings, then the gap between the 9th and 10th bright rings. Compare the two gaps.

Forecast: $r \propto m$ , so the $1 \to 2$ jump should be bigger than the $9 \to 10$ jump. Guess: outer gap is smaller.

Step 1. Bright ring radius: $r_{m} = \frac{( 2 m - 1 ) λ R}{2 μ}$ , air $μ = 1$ . Why this step? Bright uses $2 μ t = (m - \frac{1}{2}) λ$ , i.e. the odd-multiple half condition; substituting $t = r^{2} /2 R$ gives this radius. Step 2. $r_{1} = \frac{( 1 ) ( 500 \times 1 0 ^{- 9} ) ( 1 )}{2} = 2.5 \times 1 0^{- 7} = 5.00 \times 1 0^{- 4} m = 0.500 mm$ . Why this step? For $m = 1$ , $2 m - 1 = 1$ ; plug in SI so the root gives metres. Step 3. $r_{2} = \frac{( 3 ) ( 500 \times 1 0 ^{- 9} ) ( 1 )}{2} = 7.5 \times 1 0^{- 7} = 8.66 \times 1 0^{- 4} m = 0.866 mm$ . Inner gap $= r_{2} - r_{1} = 0.366 mm$ . Why this step? For $m = 2$ , $2 m - 1 = 3$ ; the difference of consecutive radii is the visible spacing between those two rings. Step 4. $r_{9} = \frac{17 \times 500 \times 1 0 ^{- 9}}{2} = 4.25 \times 1 0^{- 6} = 2.062 mm$ ; $r_{10} = \frac{19 \times 500 \times 1 0 ^{- 9}}{2} = 4.75 \times 1 0^{- 6} = 2.179 mm$ . Outer gap $= r_{10} - r_{9} = 0.117 mm$ . Why this step? For $m = 9, 10$ the odd factors are 17 and 19; comparing this gap to the inner one tests the crowding claim. Look at the figure: the curve is steep near the centre, flat far out.

Verify: Units throughout: $m \cdot m = m$ for every radius ✓, and each gap is a difference of metres, so also in metres ✓. Outer gap $0.117 mm$ < inner gap $0.366 mm$ ✓ — rings crowd outward, matching $r \propto m$ . Guess confirmed.

Example 7 — Cell G: transmitted light inverts everything

Worked example Looking from below

Instead of reflected light, the pattern is viewed in transmitted light. What is the centre now, and what is the dark-ring condition?

Forecast: Transmitted and reflected patterns are complementary (energy conservation), so the centre should be bright.

Step 1. In transmission the two interfering rays are both transmitted; neither undergoes the air→glass external-reflection $λ /2$ flip that the reflected ray B did. Why this step? No net Stokes half-wave, so the geometric path difference $2 μ t$ stands alone. Step 2. Bright (transmitted) when $2 μ t = mλ$ ; dark when $2 μ t = (m - \frac{1}{2}) λ$ . Why this step? Removing the $\frac{λ}{2}$ swaps the reflected conditions — bright and dark trade places. Step 3. At $t = 0$ : $2 μ t = 0 = 0 \cdot λ$ → bright centre. Why this step? With no flip, zero path difference means truly in-phase → constructive → bright.

Verify: Reflected centre was dark (Example 2); transmitted centre is bright — complementary, as energy conservation demands ✓. See Interference — path difference & coherence for why the two views must add to the incident intensity.

Example 8 — Cell H: exam twist, find the air-gap thickness

Worked example Thickness at a named ring

Air film, $λ = 650 nm$ . At the 8th dark ring the gap has thickness $t_{8}$ . Find $t_{8}$ , and how many wavelengths of geometric path that corresponds to.

Forecast: $2 t = mλ$ so $t = mλ /2 = 8 \times 650/2 nm = 2600 nm \approx 2.6 μ m$ .

Step 1. Dark condition, air: $2 μ t = mλ$ with $μ = 1$ → $t = \frac{mλ}{2}$ . Why this step? No radius needed — the thickness follows directly from the order. Step 2. $t_{8} = \frac{8 \times 650 \times 1 0 ^{- 9}}{2} = 2.60 \times 1 0^{- 6} m = 2.60 μ m$ . Why this step? Substitute $m = 8$ in SI; the factor 2 in the denominator is the down-and-up round trip undone. Step 3. Geometric down-and-up path $= 2 t_{8} = 5.20 \times 1 0^{- 6} m = 8 \times 650 nm = 8 λ$ . Why this step? Confirms the physical meaning: the round trip is exactly 8 whole wavelengths, and the extra $\frac{λ}{2}$ flip makes total $8.5 λ$ → destructive ✓.

Verify: $2 t_{8} / λ = 5.20 \times 1 0^{- 6} /650 \times 1 0^{- 9} = 8.0$ exactly ✓. Units of $t$ : $nm$ / $μ m$ ✓. Micron-scale gap, as expected for optical films ✓.

Quick self-test

Which centre is dark?

Reflected light — the single Stokes

λ /2

flip makes

Δ = λ /2

t = 0

Rings shrink or grow when a liquid (

μ > 1

) fills the gap?

Shrink, by factor

1/ μ

Why measure diameters not radii?

The offset/deformed contact point cancels in

D_{n}^{2} - D_{m}^{2}

Formula to get

R

from measurements?

R = \frac{D _{n}^{2} - D _{m}^{2}}{4 ( n - m ) λ}