Thin film interference — reflected and transmitted: Worked examples

Before we count anything, we need three plain-word definitions.

Recall Check you have the vocabulary before the examples

Three quick reveals — do them, then read on. What decides whether we use the standard or swapped reflection condition? ::: The parity (odd/even) of the flip count $F$ . Which wavelength goes on the right-hand side, film or vacuum? ::: The vacuum/air wavelength $λ$ — the film's index $n$ is already inside $2 n t cos θ_{r}$ . In $n_{i} sin θ_{i} = n_{t} sin θ_{r}$ , which index is which for air→film? ::: $n_{i} = 1$ (air, coming from), $n_{t} = n_{film}$ (going into). What does $t$ mean here? ::: The film thickness measured along the normal.

The scenario matrix

Every film problem lives in one of these cells. The worked examples below each name their cell.

Cell	What varies	Flip count $F$	Extra feature
A	Air–film–air (soap)	1 (odd)	normal incidence
B	Air–coat–glass (AR coating)	2 (even)	we want dark reflection
C	Air–oil–water (index in between)	1 (odd)	angled view, $cos θ_{r}$
D	Degenerate $t \to 0$	any	limiting behaviour
E	Transmitted pattern	complement	energy conservation check
F	Non-integer result	—	classify nearest condition
G	Angled light + Snell	1 (odd)	must find $θ_{r}$ first
H	Exam twist: find $λ$ that is missing	1 (odd)	solve for colour
I	Total internal reflection edge case	—	$θ_{i}$ beyond critical angle

A — Air–film–air, normal incidence (soap bubble)

Figure 1 (Cell A geometry). Incoming orange ray splits into ray 1 (reflects off the top, magenta, carries a flip) and ray 2 (dives in, bounces off the bottom with no flip, magenta, returns having travelled extra path $2 t$ ). The two magenta rays leaving the top are what your eye combines. Study the labelled flips before the steps.

Figure — Thin film interference — reflected and transmitted

Caption: air–soap–air, normal incidence — one flip up top, none below, so $F = 1$ .

Worked example Cell A: thinnest bright-red soap film

Soap $n = 1.33$ , light hits straight on ( $cos θ_{r} = 1$ ), red $λ = 650$ nm. What is the thinnest film that reflects bright red?

Forecast: guess — will "thinnest bright" use $mλ$ or $(m + \frac{1}{2}) λ$ ? Jot your guess.

Count flips. Top: air→soap, $1.00 \to 1.33$ , rarer→denser → flip. Bottom: soap→air, $1.33 \to 1.00$ , denser→rarer → no flip. So $F = 1$ (odd). (This is the flip labelling in Figure 1.) Why this step? $F$ decides which master line we use — do it first, always.
Pick the bright line. $F$ odd → bright is $2 n t = (m + \frac{1}{2}) λ$ . Why this step? Odd flip swaps constructive to the half-integer condition.
Smallest $t$ ⇒ $m = 0$ . $t = \frac{λ}{4 n} = \frac{650}{4 ( 1.33 )} = 122.2 nm$ Why this step? $m = 0$ is the smallest non-negative integer, so it gives the thinnest film.

Verify: put $t = 122.2$ nm back: $2 (1.33) (122.2) = 325 = \frac{1}{2} (650)$ ✓ — exactly a half wavelength, matching $m = 0$ bright. Units: nm in, nm on both sides. ✓

B — Air–coat–glass (AR coating), two flips

Worked example Cell B: MgF₂ anti-reflection coating

Coating $n_{c} = 1.38$ between air ( $1.00$ ) and glass ( $1.52$ ). Find the thinnest coating that makes reflection cancel for $λ = 550$ nm (so the lens transmits more). See Anti-reflection coatings.

Forecast: two flips — do the conditions swap or stay standard?

Count flips. Top: air→coat, $1.00 \to 1.38$ → flip. Bottom: coat→glass, $1.38 \to 1.52$ → flip. $F = 2$ (even). Why this step? Both interfaces are rarer→denser because $1.00 < 1.38 < 1.52$ climbs the whole way.
Pick the dark line. $F$ even → dark reflection is $2 n_{c} t = (m + \frac{1}{2}) λ$ . Why this step? Even flips keep standard conditions; standard dark is the half-integer one.
Thinnest ⇒ $m = 0$ . $t = \frac{λ}{4 n _{c}} = \frac{550}{4 ( 1.38 )} = 99.6 nm$ Why this step? Same reasoning as Cell A: smallest integer, thinnest layer.

Verify: $2 (1.38) (99.6) = 274.9 \approx \frac{1}{2} (550) = 275$ ✓ — half a wavelength, so the two reflected waves cancel. Note this looks like Cell A's formula but arrives via a totally different flip count — the coincidence is because both are " $λ /4$ " cases.

C — Air–oil–water, angled view (index sits in between)

Worked example Cell C: oil on water at an angle

Oil $n = 1.40$ on water ( $n = 1.33$ ), $λ = 600$ nm, internal angle gives $cos θ_{r} = 0.80$ , thickness $t = 500$ nm. Bright or dark in reflection?

Forecast: oil is denser than water — does the bottom reflection flip?

Count flips. Top: air→oil, $1.00 \to 1.40$ → flip. Bottom: oil→water, $1.40 \to 1.33$ , denser→rarer → no flip. $F = 1$ (odd). Why this step? Water being less dense than oil kills the second flip — a common trap.
Compute the path. $Δ = 2 n t cos θ_{r} = 2 (1.40) (500) (0.80) = 1120 nm .$ Why this step? This is the raw optical path we compare against $λ$ .
Express in wavelengths. $Δ/ λ = 1120/600 = 1.867$ . Why this step? Interference cares only about how many wavelengths fit.
Classify. $F$ odd → bright at $(m + \frac{1}{2}) = 1.5, 2.5, \dots$ ; dark at $m = 1, 2, \dots$ . $1.867$ is nearest to $2$ (distance $0.133$ ) versus $1.5$ (distance $0.367$ ). So it is closest to a dark condition. Why this step? Real thicknesses rarely hit exact maxima; we report the nearest band.

Verify: $∣1.867 - 2∣ = 0.133 < ∣1.867 - 1.5∣ = 0.367$ ✓ — dark wins. Units: all nm cancel in $Δ/ λ$ . ✓

D — Degenerate limit $t \to 0$

Worked example Cell D: why a bubble about to pop turns black

Air–soap–air, $F = 1$ . As the top of a draining soap bubble thins toward $t \to 0$ , what does reflection do?

Forecast: zero path — surely it's bright (waves in step)? Watch out.

Take the limit of the path term. $Δ = 2 n t cos θ_{r} \to 0$ as $t \to 0$ . Why this step? Geometric extra path vanishes when there is no film to cross.
But apply the flip. $F = 1$ is odd, so bright needs $Δ = (m + \frac{1}{2}) λ$ and dark needs $Δ = mλ$ . With $Δ = 0 = 0 \cdot λ$ we hit $m = 0$ dark. Why this step? The half-wave flip supplies a $λ /2$ mismatch even with zero path — the two waves are exactly out of step.
Conclusion. Reflection $\to$ dark (black) for all colours at once. Why this step? The condition $Δ = 0$ is wavelength-independent, so every colour cancels → black, not coloured.

Verify: at $t = 0$ , $2 n t = 0$ and $m = 0$ dark predicts extinction; the single flip alone is $λ /2$ — perfect destructive interference. This is the famous black top on a dying bubble. ✓

E — Transmitted pattern (complement + energy)

Worked example Cell E: what the

same Cell C oil film looks like from below Same oil film as Cell C ( $Δ/ λ = 1.867$ , $F = 1$ , nearest to $m = 2$ ). Is the transmitted light bright or dark there?

Forecast: if reflection is dark, transmission is...?

Use the transmitted condition. From the master transmission formula, transmitted bright is $Δ = mλ$ (the plain condition, opposite to this film's swapped reflection rule). Why this step? The transmitted beam that interferes is built from two internal reflections at the bottom surface; those are the same type of reflection, so their net flip is even — no $λ /2$ is introduced for transmission. Reflection here had one flip (odd), transmission has none (even); Phase change on reflection guarantees these differ by exactly $λ /2$ , which is why the two patterns are opposite.
Evaluate. $Δ/ λ = 1.867$ nearest to $m = 2$ → bright in transmission. Why this step? Same number, but now the integer condition is the bright one.
Cross-check with energy. Reflection was dark ⇒ almost no light bounced ⇒ that light transmitted ⇒ transmission bright. Why this step? Conservation of energy forbids both being dark; the patterns are negatives.

Verify: reflection nearest $m = 2$ dark, transmission nearest $m = 2$ bright — complementary ✓. Numerically $1.867$ classified identically ( $m = 2$ ) but flipped bright/dark label. ✓

F — Non-integer result, classify carefully

Worked example Cell F: is this film maximally bright, maximally dark, or in between?

Air–soap–air ( $n = 1.33$ , $F = 1$ ), normal incidence, $λ = 500$ nm, $t = 300$ nm. Classify the reflection.

Forecast: guess bright, dark, or mid-grey.

Path. $Δ = 2 (1.33) (300) = 798$ nm.
In wavelengths. $798/500 = 1.596$ . Why this step? Fractional part decides how close to a band we are.
Classify ( $F$ odd). Bright targets $1.5, 2.5$ ; dark targets $1.0, 2.0$ . $d_{b} = ∣1.596 - 1.5∣ = 0.096$ ; $d_{d} = ∣1.596 - 2.0∣ = 0.404$ . Neither is $\leq 0.05$ , so by the rule above this is partial intensity leaning bright (near the $m = 1$ maximum). Why this step? We apply the explicit cutoff rule so "near" is not a vague guess.

Verify: $d_{b} = 0.096 < d_{d} = 0.404$ ✓ — leans bright; and $0.096 > 0.05$ so it is partially bright, not a perfect maximum, exactly as the rule classifies. ✓

G — Angled light: find $θ_{r}$ first via Snell

Figure 2 (Cell G geometry). The orange incident ray at $θ_{i} = 30°$ bends to the smaller refracted angle $θ_{r}$ inside the film (magenta). The dotted line is the normal; the vertical double arrow marks the thickness $t$ . Note how only the projection onto the normal enters the path, which is why $cos θ_{r}$ (not $sec θ_{r}$ ) appears.

Caption: angled incidence — find $θ_{r}$ from Snell, then $t = λ / (4 n cos θ_{r})$ .

Worked example Cell G: you are given the

outside angle Air–soap–air, $n = 1.33$ , $λ = 550$ nm. Light arrives at angle of incidence $θ_{i} = 30°$ . Find $cos θ_{r}$ , then the thinnest bright-reflecting thickness. Uses Snell's law and Refractive index.

Forecast: will the angled path make the film thicker or thinner than the normal-incidence answer?

Snell's law for the internal angle. With $n_{i} = 1$ (air) and $n_{t} = 1.33$ (film): $n_{i} sin θ_{i} = n_{t} sin θ_{r} \Rightarrow sin θ_{r} = \frac{sin 30°}{1.33} = \frac{0.5}{1.33} = 0.3759$ . Why this step? The extra path is traversed inside the film (the magenta ray in Figure 2), so we need $θ_{r}$ , not $θ_{i}$ .
Cosine of refracted angle. $cos θ_{r} = 1 - 0.375 9^{2} = 1 - 0.1413 = 0.9268$ . Why this step? The path formula uses $cos θ_{r}$ ; get it from the Pythagorean identity, no calculator angle needed.
Thinnest bright ( $F = 1$ , $m = 0$ ). $2 n t cos θ_{r} = \frac{1}{2} λ \Rightarrow$ $t = \frac{λ}{4 n c o s θ _{r}} = \frac{550}{4 ( 1.33 ) ( 0.9268 )} = 111.5 nm .$ Why this step? Rearranged the odd- $F$ bright condition for $t$ .

Verify: at normal incidence ( $cos θ_{r} = 1$ ) we'd get $103.4$ nm; tilting needs a thicker film ( $111.5$ nm) because $cos θ_{r} < 1$ shrinks the path, so more thickness restores it ✓. Check: $2 (1.33) (111.5) (0.9268) = 274.9 \approx \frac{1}{2} (550)$ ✓.

H — Exam twist: which colour is missing from reflection?

Worked example Cell H: reverse-engineer the colour

Air–soap–air, $n = 1.33$ , $t = 400$ nm, normal incidence. Which visible wavelength (400–700 nm) is darkest (destroyed) in reflection?

Forecast: dark for odd $F$ means $2 n t = mλ$ . How many $m$ land in the visible?

Dark condition ( $F = 1$ odd). $2 n t = mλ \Rightarrow λ = \frac{2 n t}{m} = \frac{2 ( 1.33 ) ( 400 )}{m} = \frac{1064}{m}$ nm. Why this step? Solve the dark equation for $λ$ instead of $t$ — the twist is "find colour."
Scan $m$ for visible answers. $m = 1 : 1064$ nm (infrared, out). $m = 2 : 532$ nm (green, in). $m = 3 : 354.7$ nm (UV, out). Why this step? Only integer $m$ giving $400 \leq λ \leq 700$ counts as a real missing colour.
Answer. $λ = 532$ nm (green) is missing → the film looks its complementary magenta-ish in reflection. Why this step? Removing green from white light shifts the perceived colour.

Verify: $1064/2 = 532$ nm, inside 400–700 ✓; $m = 1$ (1064) and $m = 3$ (354.7) both fall outside the visible ✓, so green is the unique missing colour.

I — Edge case: total internal reflection kills the path

Angled-incidence problems have a hidden trap: if the light tries to pass from a denser medium into a rarer one at too steep an angle, it does not refract at all — it reflects entirely. This is total internal reflection (TIR), and it changes what the bottom surface does.

Worked example Cell I: oil-on-water viewed too steeply

Oil $n = 1.40$ on water $n = 1.33$ . Inside the oil, the down-going ray hits the oil→water bottom surface. At what internal angle does it stop transmitting into the water, and what happens to the interference then?

Forecast: the bottom ray is oil ( $1.40$ ) → water ( $1.33$ ), denser→rarer — so TIR is possible. Guess: above the critical angle, is there still a two-beam interference pattern in transmission?

Critical angle at the oil→water interface. $n_{1} = 1.40$ (oil), $n_{2} = 1.33$ (water): $sin θ_{c} = \frac{1.33}{1.40} = 0.95 \Rightarrow θ_{c} = arcsin (0.95) = 71.8°.$ Why this step? This is the internal angle beyond which the bottom surface reflects everything.
What happens above $θ_{c}$ . For internal angle $θ_{r} > 71.8°$ , the bottom reflection becomes total — no light leaks into the water, so there is no transmitted beam to interfere. The transmitted pattern of Cell E simply vanishes. Why this step? Conservation of energy: if $0%$ transmits, $100%$ reflects; the "transmission negative" has nothing to negate.
The reflected pattern survives — but with a twist. The reflected ray is still there (now at full strength), so a reflection pattern exists, but TIR adds its own phase shift (not the simple $λ /2$ ), so the naive flip count no longer gives clean bright/dark bands. Practical takeaway for exams: check $θ_{r}$ against $θ_{c}$ before trusting the transmitted-light formula. Why this step? It flags exactly when the machinery of Cells A–H stops applying.

Verify: $sin θ_{c} = 1.33/1.40 = 0.95 \leq 1$ ✓ (a real angle exists because oil is denser than water); $θ_{c} = 71.8°$ , and for $θ_{r}$ below this the ordinary formulas hold, above it transmission is zero ✓.

Recall Whole-page self-test

All nine cells, one line each. Cell A thinnest bright, air–soap–air, $λ = 650$ ? ::: $t = λ /4 n = 122.2$ nm. Cell B AR coating thinnest dark, $λ = 550$ , $n_{c} = 1.38$ ? ::: $99.6$ nm. Cell C oil film $Δ/ λ = 1.867$ reflection? ::: nearest $m = 2$ → dark. Cell D bubble $t \to 0$ reflection? ::: black (dark) for all colours. Cell E same oil in transmission? ::: bright (complement of reflection). Cell F soap $t = 300$ , $λ = 500$ : $Δ/ λ = 1.596$ , classify? ::: partial, leans bright (near $m = 1$ max, not exact). Cell G angled soap $θ_{i} = 30°$ thinnest bright? ::: $111.5$ nm. Cell H missing colour, $t = 400$ nm? ::: green, $532$ nm. Cell I oil→water critical angle? ::: $θ_{c} = arcsin (1.33/1.40) = 71.8°$ ; above it transmission vanishes.

Related builds: Interference of light · Young's double slit · Newton's rings · Phase change on reflection.