Thin film interference — reflected and transmitted: Question bank

Before the drills, let's pin down every symbol and term you'll meet below, so nothing appears unexplained.

Definition The symbols used on this page

$t$ ::: the film thickness — how deep the transparent layer is, measured straight down (perpendicular to its surfaces).
$n$ ::: the Refractive index of the film — how much light slows inside it; wavelength shrinks by this factor inside the medium.
$n_{c}$ ::: the refractive index of a coating layer specifically (used in the anti-reflection item); it plays the same role as $n$ but names the coating rather than a generic film.
$λ$ ::: the wavelength of the light, always measured in vacuum/air (the value you'd quote for "green light", e.g. $550$ nm).
$θ_{i}$ ::: the angle of incidence, the angle the incoming ray makes with the surface normal outside the film — the angle you measure in the air above.
$θ_{r}$ ::: the refracted angle, the angle the ray makes with the surface normal inside the film; smaller than $θ_{i}$ for a denser film. The two are linked by Snell's law: $sin θ_{i} = n sin θ_{r}$ .
$m$ ::: the interference order, a whole number $0, 1, 2, \dots$ counting how many full wavelengths fit into the path difference.
$Δ$ ::: the total path difference between the two interfering rays, measured in optical (wavelength-counting) terms. It bundles the geometric path $2 n t cos θ_{r}$ together with any half-wavelength flip: $Δ = 2 n t cos θ_{r} \pm \frac{λ}{2}$ when there is one flip.

The two figures below are the pictures the rest of this page keeps pointing back to — one for the path geometry, one for the flip rule. Study them before the drills.

Figure — Thin film interference — reflected and transmitted

Now the checklist of the three levers every thin-film question pulls:

Recall The three levers every thin-film question pulls

Path ::: The geometric extra path the lower ray travels is $2 t cos θ_{r}$ (first figure), made optical as $2 n t cos θ_{r}$ . Flip ::: A ==phase change of $π$ == ( $= λ /2$ ) happens only when reflecting off a denser medium (low $n \to$ high $n$ ) — see the second figure. Count ::: Odd number of flips (1) ⇒ conditions swap; even (0 or 2) ⇒ standard conditions.

True or false — justify

Recall A soap film about to pop looks black in reflected light.

T/F? True. As $t \to 0$ the geometric path vanishes, but the single air→film flip still contributes $λ /2$ . Two rays exactly out of step ⇒ perfect destructive interference ⇒ dark.

Recall The

$n$ in $2 n t cos θ_{r}$ means we should also use the in-film wavelength $λ / n$ on the right side. T/F? False. The factor $n$ already converts geometric path to optical path. The $λ$ on the right stays the vacuum/air wavelength — using $λ / n$ double-counts the medium.

Recall For an oil film on water, viewed in reflection, "bright when

$2 n t cos θ_{r} = mλ$ " is the correct constructive rule. T/F? False. Air→oil flips, oil→water (denser→rarer) does not: one net flip. So the conditions swap and bright becomes $2 n t cos θ_{r} = (m + \frac{1}{2}) λ$ .

Recall Where the film is bright in reflection, it is also bright when viewed through in transmission.

T/F? False. Reflection and transmission are complementary (photographic negatives). Energy not reflected is transmitted, so a reflection-bright band is transmission-dark. See Conservation of energy.

Recall Increasing the angle of viewing always increases the effective path difference

$2 n t cos θ_{r}$ . T/F? False. As the angle grows, $θ_{r}$ grows and $cos θ_{r}$ decreases, so the path difference shrinks. That's why film colors shift as you tilt a soap bubble.

Recall In the formula we should use the external angle of incidence

$θ_{i}$ . T/F? False. The extra path is travelled inside the film, so the relevant angle is the refracted one, $θ_{r}$ , linked to $θ_{i}$ by Snell's law: $sin θ_{i} = n sin θ_{r}$ .

Recall A film with

zero net flips (0 or 2) and $t \to 0$ looks bright in reflection. T/F? True. With no flip term, $Δ = 2 n t cos θ_{r} \to 0$ , the two rays are in step ⇒ constructive ⇒ bright. This is the exact opposite of the soap film, purely because of flip count.

Recall Thin-film interference needs two separate light sources like

Young's double slit. T/F? False. Both interfering rays come from the same incident ray splitting at the surfaces (division of amplitude), so they are automatically coherent. Young's double slit instead splits one wavefront through two slits (division of wavefront).

Recall White light on a thick (

$t \sim 1$ mm) glass plate shows vivid interference colors. T/F? False. Two reasons stack up. First, white light has a very short coherence length (only microns): the front-surface and back-surface reflections are separated by far more than that, so they can no longer keep a fixed phase relationship and simply do not interfere. Second, even if they could, so many wavelengths satisfy bright/dark conditions at once that all colors would overlap and wash to white. Vivid color needs $t$ within the coherence length, comparable to $λ$ .

Spot the error

Recall "Both surfaces reflect, so both give a

$λ /2$ flip; add $λ$ ." — where's the slip? A flip happens only when going into a denser medium (second figure). The bottom reflection is often denser→rarer (no flip). You must check each interface's index ordering separately, not assume both flip.

Recall "The extra path is the full slant length

$2 t / cos θ_{r}$ ." — fix it. The two emerging rays must be compared along a common wavefront (perpendicular to the rays), as in the first figure. Projecting the slant travel onto the film normal cuts it to $2 t cos θ_{r}$ ; the secant does not survive.

Recall "For anti-reflection, we want reflection bright, so

$2 n_{c} t = mλ$ ." — spot both errors. An AR coating wants reflection to cancel (destructive), so it's minimizing, not maximizing. And with two flips the phases add to $π + π = 2 π$ ; since a full $2 π$ turn brings a wave back to its starting phase, that is equivalent to no shift at all, giving destructive reflection at $2 n_{c} t cos θ_{r} = (m + \frac{1}{2}) λ$ (which is $2 n_{c} t = (m + \frac{1}{2}) λ$ at normal incidence, where $cos θ_{r} = 1$ ).

Recall "Very thin film always looks dark in reflection." — when does this fail?

Only true for an odd flip count (like air–soap–air). For a coating between air and glass (two flips), a vanishingly thin film looks bright. The blanket claim ignores flip parity.

Recall "Since light is inside the film, use

$c / n$ and get a different $λ$ , then also multiply the path by $n$ ." — what's wrong? That applies the medium correction twice. Either work with in-film wavelength $λ / n$ and geometric path $2 t cos θ_{r}$ , or with vacuum $λ$ and optical path $2 n t cos θ_{r}$ — never both.

Recall "

Newton's rings have a bright center because the path difference is zero there." — correct it. At the contact point the geometric path $\to 0$ , so $Δ \to λ /2$ from the single flip: the air-gap has one net flip (glass→air→glass gives a flip only at the second, air→glass, interface). So the center is dark, not bright — same $t \to 0$ logic as the soap film.

Why questions

Recall Why does the derivation use

$cos θ_{r}$ and not $sec θ_{r}$ ? Because the two exiting rays must be compared along a common wavefront perpendicular to them (first figure). As the lower ray zig-zags through the film it also drifts sideways by $2 t sin θ_{r}$ ; that sideways offset means the upper ray's wavefront reaches a point $2 t sin θ_{r} \cdot tan θ_{r}$ further along its slant, and we must subtract that already-counted lateral piece. The algebra $\frac{2 t}{c o s θ _{r}} - 2 t sin θ_{r} tan θ_{r} = 2 t cos θ_{r}$ then collapses the secant to a cosine.

Recall Why must reflection and transmission patterns be complementary?

By Conservation of energy, every photon either reflects or transmits. If a wavelength interferes destructively in reflection, that missing energy has to appear in the transmitted beam, and vice versa.

Recall Why does a soap film show a

spectrum of colors rather than one? Different wavelengths satisfy the bright condition $2 n t cos θ_{r} = (m + \frac{1}{2}) λ$ at different thicknesses. As the film thickness varies across its surface, each region reflects a different color brightly.

Recall Why do we invoke

Refractive index at all, instead of just measuring distances? Phase is counted in wavelengths, and wavelength shrinks by $n$ inside the film. Two waves with equal geometric paths but different media can still be out of step, so optical path ( $n \times$ distance) is what governs interference.

Recall Why does the flip come from reflection off a

denser medium specifically? It follows from the boundary conditions on the electromagnetic wave (see Phase change on reflection): reflecting off a higher-index medium forces the reflected wave to invert, exactly like a wave on a rope reflecting off a fixed (heavier) end.

Recall Why is

Young's double slit a useful cousin to remember here? Both rely on Interference of light of two coherent beams, and both use "path difference = whole $λ$ ⇒ bright." Thin films just add the reflection-flip twist that double-slit interference doesn't have.

Edge cases

Recall What happens to the reflected condition at exactly

$t = 0$ (perfect contact)? The geometric path is zero, so only the flip term survives. Odd flips ⇒ $Δ = λ /2$ ⇒ dark; even flips ⇒ $Δ = 0$ ⇒ bright. The film's ordering of indices decides the outcome.

Recall What if the film index equals the medium above it (

$n_{f i l m} = n_{ab o v e}$ )? There is no index step at the top surface, so no reflection (and no flip) there — the "film" is optically invisible from the top and produces no two-ray interference.

Recall At grazing incidence (

$θ_{r} \to 9 0^{\circ}$ ), what does the path difference approach? $cos θ_{r} \to 0$ , so the geometric part $2 n t cos θ_{r} \to 0$ and $Δ$ collapses toward just the flip term. Only the flip matters, so the reflected fringe behaviour approaches the $t \to 0$ (flip-dominated) case.

Recall At oblique incidence, does the simple "flip on reflection off a denser medium" rule still hold for both polarizations?

Only approximately. The reflection phase actually depends on polarization: the component with the electric field in the plane of incidence (p) and the one perpendicular to it (s) pick up different phase shifts at oblique angles, and near Brewster's angle the p-reflection can even flip the other way. At near-normal incidence the s/p difference vanishes and the simple $π$ -flip rule is exact — which is why introductory treatments assume it.

Recall For an air wedge between two identical glass plates, is the thin edge bright or dark in reflection?

Dark. Glass→air (denser→rarer) gives no flip; air→glass (rarer→denser) gives one flip: one net flip. At the thin edge $t \to 0$ , one flip ⇒ $Δ \to λ /2$ ⇒ destructive ⇒ dark line.

Recall What if a film is illuminated by two colors and one is bright while the other is dark at the same spot?

Entirely possible: the bright/dark condition depends on $λ$ , so at a fixed thickness $2 n t cos θ_{r}$ may equal $(m + \frac{1}{2}) λ_{1}$ (bright) and $m^{'} λ_{2}$ (dark) simultaneously. This is why films look tinted.

Recall If you slowly increase

$t$ from zero for a soap film in monochromatic light, what sequence of reflected brightness do you see? Starting dark (flip only), it brightens to the first maximum at $t = λ /4 n$ , dims to dark at $t = λ /2 n$ , brightens again at $t = 3 λ /4 n$ , cycling every $λ /2 n$ increase in thickness.