In short

A thin film of transparent material with refractive index n and thickness t has two parallel reflecting surfaces. Light falling on the film splits into a ray reflected from the top surface and a ray reflected from the bottom surface. The two reflected rays interfere. Their optical path difference, for a ray refracted into the film at angle r, is

\boxed{\;\Delta_\text{geometric} \;=\; 2 n t \cos r\;}

On top of this, reflection at a boundary with a denser medium introduces an extra half-wavelength (\lambda/2, or equivalently a \pi phase shift). For a film sitting in air, the top reflection (air → film) picks up this shift while the bottom reflection (film → air) does not. The net optical path difference in the reflected light is therefore

\Delta_\text{net} \;=\; 2 n t \cos r \;-\; \tfrac{\lambda}{2}

Bright fringe in reflected light (constructive):

2 n t \cos r \;=\; \left(m + \tfrac{1}{2}\right)\lambda \qquad (m = 0, 1, 2, \ldots)

Dark fringe in reflected light (destructive):

2 n t \cos r \;=\; m\lambda \qquad (m = 0, 1, 2, \ldots)

In transmitted light the conditions swap — a dark reflection means a bright transmission. When white light falls on a film whose thickness varies across its surface, different wavelengths go bright at different places, and the film shows the swirling colours of a soap bubble or an oil slick.

Newton's rings — a plano-convex lens resting on an optical flat — make the "varying thickness" controllable. The resulting concentric dark rings obey

r_m^2 \;=\; m\lambda R \qquad (m = 0, 1, 2, \ldots)

where r_m is the radius of the mth dark ring and R is the lens's radius of curvature.

Walk home through a Delhi by-lane on the morning after Diwali and you will find a rainbow clinging to every puddle. The monsoon has left water lying between the bricks; an autorickshaw has leaked a teaspoon of two-stroke oil into one of them; and where the oil has spread out into a film thinner than a human hair, the puddle has bloomed into bands of green and magenta and amber. Lean over it and tilt your head — the pattern slides across the water, greens becoming blues, magentas becoming oranges. Lean the other way and it changes again. No pigment in the oil is doing this. The oil itself, if you scooped it into a jar, would look like a featureless brown. The colour is not in the oil. It is in the geometry — a sheet of oil so thin that light reflecting from its top surface meets light reflecting from its bottom surface and either reinforces or cancels, wavelength by wavelength.

The same physics paints every soap bubble you have ever seen, every peacock feather, every beetle's iridescent shell, and — less romantically — every anti-reflection coating on every spectacle lens and every camera. If you wear glasses, look at the edges of your lenses in bright sunlight: the faint purple or green tinge you see is a quarter-wavelength coating of magnesium fluoride engineered to make the reflection dark at the middle of the visible spectrum so that more light reaches your eye. The physics of the oil slick and the physics of the coating that makes your camera sharper are the same physics. This article is how it works.

What a "thin film" means

A thin film is a transparent layer with two parallel (or near-parallel) reflecting surfaces, and a thickness comparable to — or at least not enormously larger than — the wavelength of light. The oil slick is perhaps a few hundred nanometres thick. A soap bubble wall is of similar thickness. A camera lens coating is deliberately a quarter-wavelength thick, about 100 nm.

When a light ray falls on such a film, something both reflects off the top and transmits through to the bottom. The transmitted portion refracts at the top surface, crosses the film, reflects off the bottom surface, refracts again on the way back out, and emerges parallel to the first reflected ray. You now have two rays heading toward your eye from almost the same direction. They are also coherent — they come from the same incident wave train, split by partial reflection, so their phases are locked together. (This is the division of amplitude strategy introduced in coherence and conditions for interference.) Coherent waves arriving at the same detector interfere.

Ray geometry for reflection from a thin film A horizontal transparent film of thickness t sits in air. An incident ray strikes the top surface at angle i. Part of it reflects as ray 1. The transmitted ray refracts to angle r inside the film, travels to the lower surface, reflects, and refracts back out as ray 2, parallel to ray 1. The extra path of ray 2 inside the film is AB + BC where angle r is shown at both surfaces; the geometric optical length difference is 2 n t cos r. film, index $n$, thickness $t$ incident ray 1 ray 2 $i$ $r$ $A$ $B$ $C$ $t$ air, index $1$ air, index $1$
Ray 1 reflects at the top surface; ray 2 refracts into the film at angle $r$, reflects at the bottom, and refracts back out. The two emerging rays are parallel and coherent. Their optical path difference is set by the extra travel of ray 2 inside the film and by a possible half-wavelength shift at one of the reflections.

Two questions, then, determine everything about the colours you see in the reflected light.

  1. Geometric question. How much farther does ray 2 travel than ray 1?
  2. Phase-shift question. Does either ray pick up a half-wavelength kick at its reflection that has nothing to do with distance travelled?

Answer both and you have the interference condition.

The geometric path difference — 2nt\cos r

Follow ray 2 carefully. It enters the film at A, travels down to B on the lower surface, reflects, and travels up to C on the upper surface, where it refracts and exits. Inside the film, it travels the distance AB + BC.

Assumptions. The film's two surfaces are flat and parallel. The surrounding medium is air (refractive index 1). The film's refractive index n is real (non-absorbing). The incident light is monochromatic of wavelength \lambda in vacuum.

Step 1. Each internal leg has length t/\cos r.

Look at the triangle formed by A, B, and the foot of the perpendicular from A onto the bottom surface. The vertical leg has length t (the film thickness); the angle at A between the refracted ray and the vertical (downward) normal is r. The hypotenuse is AB.

\cos r \;=\; \frac{t}{AB} \quad\Longrightarrow\quad AB \;=\; \frac{t}{\cos r}

By the symmetry of reflection, BC has the same length as AB:

BC \;=\; \frac{t}{\cos r}

Why: the angle at which ray 2 leaves the bottom surface equals the angle at which it arrived (law of reflection). Both legs make the same angle r with the normal, and both span the same vertical thickness t.

Step 2. Add the two legs to get the total geometric length inside the film.

AB + BC \;=\; \frac{2t}{\cos r}

Step 3. Subtract the correction for the part of ray 1's path that has "already been travelled" before ray 2 exits at C.

Here is the subtle step. At the moment ray 2 emerges from C, ray 1 has already been travelling (above the film) for the same amount of time as ray 2 was inside the film. Over that time, ray 1 has moved from A along its reflected direction. Drop a perpendicular from C to ray 1's line — call the foot D. Only the portion of ray 2's path beyond the point where it and ray 1 are finally side-by-side-on-the-same-wavefront counts as a path difference.

From the geometry, the extra-path in the same medium (air) that ray 1 loses, compared to ray 2's in-film run, works out to AC \sin i = 2 t \tan r \sin i. Using Snell's law \sin i = n \sin r,

AD \;=\; 2 t \tan r \sin i \;=\; 2 t \tan r \cdot n \sin r \;=\; 2 n t \,\frac{\sin^2 r}{\cos r}

Why: AC, the distance between where ray 1 reflected and where ray 2 exits, equals 2 t \tan r (two horizontal legs of length t \tan r each). The component of AC along ray 1 is AC \sin i. Snell's law lets you rewrite \sin i in terms of \sin r and n.

Step 4. Put it together. The geometric path ray 2 travels inside the film has optical length n \times (its geometric length) — light in a medium of index n advances phase as if it had gone n times farther in vacuum. So the optical-path contribution of ray 2 is

n \cdot \frac{2t}{\cos r} \;=\; \frac{2nt}{\cos r}

The optical path difference is ray 2's contribution minus ray 1's "already travelled in air" part:

\Delta_\text{geometric} \;=\; \frac{2nt}{\cos r} \;-\; 2nt \,\frac{\sin^2 r}{\cos r} \;=\; \frac{2nt(1 - \sin^2 r)}{\cos r} \;=\; \frac{2nt \cos^2 r}{\cos r}
\boxed{\;\Delta_\text{geometric} \;=\; 2 n t \cos r\;}

Why: the denominator \cos r cancels one power of \cos r in the numerator, leaving exactly 2nt \cos r. This remarkable piece of cancellation — the reason the formula is so compact — is the first of several moments in wave optics where Snell's law conspires with the Pythagorean identity to produce something clean. For a ray at normal incidence (i = r = 0), \cos r = 1 and \Delta = 2nt — the path difference is just twice the film thickness in optical units, which is intuitively exactly what you expect.

The half-wavelength kick — \pi phase shift at reflection

The geometric path difference 2nt\cos r is only half the story. The other half is a purely quantum-mechanical-but-classically-derivable fact about how electromagnetic waves reflect at a boundary:

When light in a medium of lower refractive index reflects off a boundary with a medium of higher refractive index, the reflected wave undergoes a phase shift of \pi — equivalently, an extra optical-path gain of \lambda/2.

When light reflects off a boundary with a medium of lower refractive index, there is no phase shift.

This is the same rule you met when a wave on a string reflects off a fixed end (phase-inverted, analogous to reflection from a denser medium) versus a free end (no inversion, analogous to a less-dense medium). It follows from Maxwell's equations with the boundary conditions on \vec E and \vec B — the "denser medium is like a fixed end" rule is the electromagnetic version.

For a soap film or an oil slick sitting in air, this means:

The net optical path difference between the two reflected rays, once you include the \lambda/2 from the top reflection, is

\Delta_\text{net} \;=\; 2 n t \cos r \;-\; \frac{\lambda}{2}

(The minus sign is a bookkeeping convention; what matters is that one of the two rays is shifted by half a wavelength relative to the other. You can equally well write 2nt\cos r + \lambda/2 because half-wavelength shifts are equivalent mod \lambda.)

Conditions for bright and dark fringes

Constructive interference requires the net optical path difference to be an integer number of wavelengths. Destructive interference requires an odd half-integer number of wavelengths.

Bright (constructive) in reflection:

2 n t \cos r - \frac{\lambda}{2} = m\lambda \qquad\Longrightarrow\qquad 2 n t \cos r = \left(m + \tfrac{1}{2}\right)\lambda

Dark (destructive) in reflection:

2 n t \cos r - \frac{\lambda}{2} = \left(m + \tfrac{1}{2}\right)\lambda \qquad\Longrightarrow\qquad 2 n t \cos r = (m+1)\lambda

with m = 0, 1, 2, \ldots. Absorbing the "+1" into the counting, dark in reflection happens at 2nt \cos r = m\lambda for m = 1, 2, \ldots.

A telling special case: the vanishing soap film

Take a very thin soap film — t \to 0. The formula 2nt\cos r \to 0 gives \Delta_\text{net} \to -\lambda/2, which is the dark-fringe condition. A soap film thinner than a quarter-wavelength reflects no light at all (for the wavelength for which it is tuned). This is directly observable: a vertically held soap film drains over time — the film is thicker at the bottom (gravity pulls the soap solution down) and thinner at the top. As the top drains below a quarter-wavelength, you see a dark band appear at the top of the film. A moment later the whole film pops, because it has become so thin that surface-tension-versus-evaporation wins.

A particle model of light would predict that a thin soap film reflects a lot of light (two surfaces, so twice as much reflection as one). The wave model predicts a thin film reflects no light. The wave model is right.

In transmitted light, the conditions swap

A ray that goes through the film to reach your eye from behind is also the sum of two coherent paths — one transmits directly, the other bounces twice inside the film before emerging. The path-difference geometry is the same (2nt\cos r), but neither reflection in the transmitted-light path is a "low-to-high" reflection — the first goes high-to-low at the bottom, the second goes low-to-high at the top after already having been transmitted (rather than reflected) the first time. Work it out: the net relative phase shift between the two transmitted rays has no extra \lambda/2. So:

Wherever you see a bright fringe in reflection, you see a dark fringe in transmission, and vice versa. Energy is conserved — light that does not reflect at a bright-in-transmission place has gone through.

Why a thin film shows colours in white light

Fix the geometry — a specific t, a specific viewing angle that picks out a specific r — and ask: which wavelength is bright in reflection? The condition

2 n t \cos r = \left(m + \tfrac{1}{2}\right)\lambda

picks out the wavelengths \lambda = 4nt\cos r / (2m+1) for m = 0, 1, 2, \ldots. For typical thin-film thicknesses (t of a few hundred nanometres) and n \approx 1.3 to 1.5 and near-normal incidence, the m = 0 and m = 1 wavelengths fall inside the visible range. So out of the broadband white light falling on the film, one or two narrow bands of wavelength reflect constructively — a colour — and the rest are suppressed.

Change the thickness a little (move to a different spot on a soap bubble, where the film is thicker or thinner because of gravity or evaporation) and the reflected colour changes. Tilt the bubble so your viewing angle shifts — \cos r changes, the constructive wavelength changes, the colour changes. This is why a soap bubble's colour is never uniform: it shifts in space because t varies across the bubble's surface, and it shifts in time because the bubble drains.

Explore the thin-film colour yourself

The interactive below lets you drag a film-thickness slider and watch the m = 0 constructive wavelength \lambda_0 = 4nt/(2m+1) = 4nt (at normal incidence, \cos r = 1) sweep through the visible spectrum. For an oil-on-water film with n = 1.5, thicknesses from about 60 nm to 400 nm span the full visible range. The coloured bar in the readout shows roughly what you would see.

Interactive: thin-film constructive wavelength vs thickness A rising line showing the m = 0 constructive-reflection wavelength λ₀ = 4 n t for n = 1.5 as a function of film thickness t in nanometres. A draggable red point lets you pick t; readouts show t in nm and the constructive wavelength in nm, plus an approximate colour label. film thickness $t$ (nm) constructive $\lambda_0$ (nm) 50 150 250 350 450 300 500 700 900 visible band $\lambda_0 = 4 n t$ (n=1.5) drag the red point
Drag the red point to change the film thickness. The first-order constructive-reflection wavelength $\lambda_0 = 4nt$ (setting $\cos r = 1$ for near-normal incidence and $n = 1.5$) is plotted against $t$. For $t \approx 100$ nm the bright-reflection wavelength is in the blue; for $t \approx 150$ nm it is in the green; for $t \approx 200$ nm it is in the red. Above the visible band ($> 700$ nm), the eye sees a mix of higher-order colours instead of a single tint.

Anti-reflection coatings — thin films put to work

Spectacle lenses, camera lenses, the front glass of LCD screens and binoculars — essentially every optical surface that matters in modern India — carries a thin-film coating engineered to make the reflection dark at the middle of the visible spectrum. This is thin-film interference used as a tool, not studied as a phenomenon.

The design: deposit a transparent film with refractive index n_f intermediate between air (n = 1) and glass (n_g \approx 1.5), with thickness t chosen so that 2 n_f t = \lambda/2 at normal incidence for the middle of the visible range (\lambda \approx 550 nm). Magnesium fluoride (n_f \approx 1.38) is the standard. A quarter-wavelength film is about 100 nm thick.

Why both reflections now invert. For an anti-reflection coating, both the air→film interface and the film→glass interface are low-to-high refractive-index jumps (air 1.0 → film 1.38 → glass 1.5). Both reflections pick up a \pi shift. The two \pi shifts cancel: their relative phase shift is zero. So the net path difference is just 2n_f t (no \lambda/2 correction), and destructive interference in reflection requires

2 n_f t \;=\; \left(m + \tfrac{1}{2}\right)\lambda \qquad(m = 0 \text{ gives the thinnest working film})

For m = 0: t = \lambda/(4 n_f), the famous quarter-wavelength coating. At \lambda = 550 nm and n_f = 1.38, this is t = 550/(4 \cdot 1.38) \approx 99.6 nm. At the edges of the visible band (400 nm and 700 nm) the coating is not perfectly destructive, so a residual purple or green tinge remains. This is the faint colour you see at a glancing angle on your spectacles, and the reason camera lenses (which stack several different coatings) look dark-bluish-green when tilted.

This is not an incidental application. Without anti-reflection coatings, a modern zoom lens with 15 glass elements would lose about half of all incoming light to surface reflections and drown the image in internal flare. Coatings are why you can take a backlit photo on your phone outside IIT Delhi at noon without the Sun ghosting across the frame.

Worked examples

Example 1: A soap film in sunlight

A vertically held soap film (refractive index n = 1.33) is illuminated by white sunlight at near-normal incidence. At a certain horizontal position, its thickness is t = 400 nm. Which visible wavelength is most strongly reflected at that position? What colour will you see there?

Soap film with a ray reflecting from top and bottom surfaces A vertical soap film of thickness 400 nm and refractive index 1.33. An incident white-light ray strikes near normal. Ray 1 reflects off the front surface; ray 2 refracts in, reflects off the back surface, and exits parallel to ray 1. The two rays emerge as one beam that is bright in green-yellow light. soap film $n = 1.33$, $t = 400$ nm incident (white) ray 1 ray 2
At this spot the film is 400 nm thick. Both reflected rays (red, from the two surfaces) combine. The net path difference is $2nt - \lambda/2$; the wavelength for which this equals an integer multiple of $\lambda$ is the bright one.

Step 1. State the bright-reflection condition at normal incidence.

At normal incidence, \cos r = 1, so the condition 2nt \cos r = (m + \tfrac12)\lambda becomes

2 n t = \left(m + \tfrac{1}{2}\right)\lambda \qquad\Longrightarrow\qquad \lambda = \frac{2 n t}{m + \tfrac{1}{2}}

Why: near-normal means you can read off the formula without a Snell's-law detour for r. Most everyday soap-film observations (looking roughly straight at a bubble held in front of you) satisfy this.

Step 2. Compute 2nt.

2 n t \;=\; 2 \times 1.33 \times 400 \text{ nm} \;=\; 1064 \text{ nm}

Why: this is the optical path of the round trip inside the film. Every bright-reflection wavelength must divide this (after the \lambda/2 correction).

Step 3. Find the m values that put \lambda in the visible range (400 nm – 700 nm).

\lambda = \frac{1064}{m + \tfrac{1}{2}} \text{ nm}
  • m = 0: \lambda = 1064 / 0.5 = 2128 nm (infrared — invisible).
  • m = 1: \lambda = 1064 / 1.5 = 709 nm (edge of red — visible but very weak to the eye).
  • m = 2: \lambda = 1064 / 2.5 = 425.6 nm (violet — visible).
  • m = 3: \lambda = 1064 / 3.5 = 304 nm (ultraviolet — invisible).

Why: only m = 1 and m = 2 yield wavelengths the eye can detect. Both are near the edges of the visible spectrum, so the reflected light is a mix of deep red and violet — which your brain interprets as magenta.

Step 4. Check which wavelength is dark.

Dark in reflection: 2 n t = k\lambda, so \lambda = 1064/k nm.

  • k = 2: \lambda = 532 nm (green — strongly suppressed).

The green is almost entirely missing. The reflected light is the visible-spectrum minus green, which is exactly the magenta the eye sees.

Result: The film at t = 400 nm reflects predominantly red (709 nm) and violet (426 nm) and suppresses green (532 nm). Your eye sees magenta.

What this shows: Thin-film colours are not pigments. They are the visible spectrum with particular wavelengths either reinforced or cancelled by the film's geometry. As the film drains and t changes, the missing wavelength slides across the spectrum, and the colour you see shifts continuously.

Example 2: Newton's rings

A plano-convex lens of radius of curvature R = 1.20 m rests with its curved side down on an optical flat glass plate. Monochromatic sodium light (\lambda = 589 nm) illuminates the setup from above at near-normal incidence. What is the radius of the 10th dark ring (counting from the central dark spot as m = 0)?

Newton's rings geometry A plano-convex lens of radius of curvature R sits curved-side down on a flat plate. The tiny air film between the lens and the plate has thickness t that increases with horizontal distance r from the point of contact. A ray travelling down reflects off both the bottom of the lens and the top of the plate, producing concentric interference rings. optical flat plano-convex lens, R = 1.20 m contact point $t$ $r$
The thin air film under the lens has thickness $t(r) \approx r^2/(2R)$, so at radius $r$ the optical-path difference on round-trip is $2t = r^2/R$. Rings of constant $r$ are rings of constant $t$, giving the concentric interference pattern.

Step 1. Identify what the film is.

The film is air between the bottom of the lens and the top of the flat — refractive index n = 1.

The top reflection (from the lens's lower surface, glass → air) does not invert — high-to-low. The bottom reflection (air → glass at the flat's upper surface) does invert — low-to-high. So exactly one of the two reflections carries a \lambda/2 shift. The net optical path difference is

\Delta_\text{net} = 2 t - \frac{\lambda}{2}

Why: the system is a thin air film rather than a thin glass film, but the counting of phase shifts is the same principle applied to this geometry. One low-to-high reflection, one high-to-low; one \lambda/2 kick total.

Step 2. Write the dark-ring condition.

Dark (destructive) in reflection: \Delta_\text{net} = (m + \tfrac12)\lambda, i.e.

2 t - \tfrac{\lambda}{2} = (m + \tfrac{1}{2})\lambda \qquad\Longrightarrow\qquad 2 t = (m + 1)\lambda

Relabel m + 1 \to m', so 2t = m'\lambda for m' = 1, 2, 3, \ldots. The central spot (t = 0) is m' = 0, and it is dark — at the point of contact, the only contribution is the \lambda/2 shift, giving total destructive interference. (This dark central spot is a direct experimental confirmation of the \pi-shift rule.)

Step 3. Relate the film thickness t to the ring radius r.

From the geometry of a sphere of radius R touching a flat at its south pole, the height of the sphere above the flat at horizontal distance r from the contact point is

t(r) \;=\; R - \sqrt{R^2 - r^2}

For small r \ll R, Taylor-expand:

\sqrt{R^2 - r^2} = R\sqrt{1 - r^2/R^2} \;\approx\; R\left(1 - \frac{r^2}{2R^2}\right) = R - \frac{r^2}{2R}
t(r) \;\approx\; \frac{r^2}{2R}

Why: in a typical Newton's rings setup, r is a few millimetres and R is of order a metre, so r/R \lesssim 10^{-2}. The small-r approximation is excellent.

Step 4. Substitute into the dark-ring condition.

2 \cdot \frac{r^2}{2R} = m'\lambda \qquad\Longrightarrow\qquad \boxed{\;r_m^2 = m'\lambda R\;}

For m' = 10:

r_{10}^2 = 10 \times 589 \times 10^{-9} \text{ m} \times 1.20 \text{ m} = 7.07 \times 10^{-6} \text{ m}^2
r_{10} = \sqrt{7.07 \times 10^{-6}} \approx 2.66 \times 10^{-3} \text{ m} = 2.66 \text{ mm}

Why: r_m grows as \sqrt{m}, so rings are packed closer together further out. Your lab travelling microscope will read consecutive rings whose squared radii differ by a constant \lambda R — that is the measurement you take to extract \lambda from known R, or R from known \lambda.

Result: The 10th dark ring has radius r_{10} \approx 2.66 mm.

What this shows: Newton's rings turn a thin air film with deliberately varying thickness into a precise wavelength-measuring instrument. In a Class 12 optics lab, you set the lens on the flat, illuminate with a sodium lamp, measure a dozen ring diameters through a travelling microscope, and extract \lambda to within 1 per cent. The same experiment also measures R — used to calibrate lenses, even today.

Example 3: Anti-reflection coating on spectacle glass

A spectacle lens made of crown glass (n_g = 1.52) is to be coated with magnesium fluoride (n_f = 1.38) to minimise reflection at \lambda = 550 nm (the green-yellow middle of daylight). What is the required coating thickness?

Step 1. Identify the reflections and their phase shifts.

  • Air → MgF₂: 1.00 \to 1.38, low-to-high, \pi shift.
  • MgF₂ → glass: 1.38 \to 1.52, low-to-high, \pi shift.

Both reflections invert. The two \pi shifts cancel — they contribute zero relative phase difference between the two reflected rays.

Why: this is the subtle point that distinguishes an anti-reflection coating from a soap film. For soap-in-air, one of the two reflections inverts and the other does not, so there is a net \lambda/2 kick. For MgF₂-on-glass, both reflections invert, so the net kick is zero.

Step 2. Write the destructive condition for zero relative phase shift.

2 n_f t \cos r = \left(m + \tfrac{1}{2}\right)\lambda

At normal incidence (\cos r = 1) and m = 0 for the thinnest working coating,

2 n_f t = \tfrac{\lambda}{2} \qquad\Longrightarrow\qquad t = \frac{\lambda}{4 n_f}

Step 3. Plug in numbers.

t = \frac{550 \text{ nm}}{4 \times 1.38} = \frac{550}{5.52} \approx 99.6 \text{ nm}

Result: A quarter-wavelength coating of MgF₂ about 100 nm thick suppresses reflection at \lambda = 550 nm.

What this shows: Anti-reflection coatings are thin-film interference engineered in reverse. Instead of observing which colour the film makes bright, you choose the film thickness to force a specific colour (the centre of your sensitivity band) to be dark. At 400 nm and 700 nm the cancellation is imperfect, leaving the residual purple-green tinge you see on your glasses in sunlight — the edges of the visible band, no longer lined up with the interference zero.

Common confusions

If you came here to understand why a soap bubble has colours and how an anti-reflection coating works, you have everything you need. What follows is the rigorous treatment of the \pi phase shift and the exact intensity formula that the 2nt\cos r condition only approximates.

The Fresnel reflection coefficients and the \pi-shift rule

Why does reflection off a denser medium invert the wave? Maxwell's equations with the electromagnetic boundary conditions (tangential \vec E and \vec H continuous across the interface) give the Fresnel reflection coefficient for normal incidence:

r \;=\; \frac{n_1 - n_2}{n_1 + n_2}

where n_1 is the index of the incident medium and n_2 that of the second medium. The coefficient is negative when n_2 > n_1 — and a negative reflection coefficient is exactly a \pi phase shift (a sign flip of the wave amplitude).

r < 0 \iff n_2 > n_1 \iff \pi \text{ shift at reflection}

This is not a postulate; it is a consequence of Maxwell's equations. The mechanical analogue — a wave on a string reflecting off a fixed end (heavy rope) or a free end (light rope) — is the same story reframed: the heavy rope resists the wave's displacement and sends back an inverted copy, exactly as a denser optical medium resists the \vec E field and sends back an inverted reflection.

Intensity of the reflected light — beyond "bright" and "dark"

The 2nt\cos r condition tells you where the maxima and minima sit but not how bright they are. The full two-beam interference formula, with reflection coefficients r_1 from the top surface and r_2 from the bottom, gives the reflected intensity as

I_R \;=\; I_0 \left(r_1^2 + r_2^2 + 2 r_1 r_2 \cos \delta\right)

where \delta = 2\pi(2nt\cos r)/\lambda + \pi is the total phase difference (the +\pi is the shift at the top reflection). The amplitude at a bright fringe is r_1 + r_2 (if both are positive, or both negative with the \pi counted); at a dark fringe it is r_1 - r_2. For a soap film (r_1 \approx (1-1.33)/(1+1.33) = -0.14 and |r_2| \approx 0.14), the bright intensity is proportional to (0.14 + 0.14)^2 = 0.078 — about 8 per cent of the incident light — while the dark intensity is zero exactly (perfect cancellation). A soap film reflects at most 8 per cent of any one wavelength; the rest goes through.

Fabry-Pérot and multiple reflections — when two-beam is not enough

Real thin films do not just reflect twice. The ray inside the film also reflects at the top and re-bounces; each round trip contributes a fainter coherent copy to the reflected beam. Summing the infinite geometric series gives the Airy function:

I_R \;=\; I_0 \cdot \frac{F \sin^2(\delta/2)}{1 + F \sin^2(\delta/2)}, \quad F = \frac{4 R}{(1-R)^2}, \quad R = r_1^2

For low-reflectivity films (R \ll 1), F \ll 1 and the Airy function reduces to the two-beam cosine. For high-reflectivity multilayers (R \to 1, as in dielectric mirrors), F \to \infty and the fringes become very sharp — bright peaks sitting on a black background. This is the principle behind Fabry-Pérot interferometers, the filters used in laser cavities and in the narrow-bandpass filters on telescopes like the GMRT at Khodad.

Why the m = 0 fringe in reflection is always dark for a soap film

A soap film at zero thickness would have 2nt\cos r = 0. The bright condition 2nt\cos r = (m + \tfrac12)\lambda has no m \ge 0 solution at t = 0; the dark condition 2nt \cos r = m\lambda is satisfied at m = 0. A soap film in the limit t \to 0 is dark in reflection at every wavelength. As the film drains and thins, the whole visible spectrum collapses to black just before the film pops — a final moment of transparency, then nothing. You can see this on a vertically held film after a few seconds: the top becomes utterly dark, then tears.

Colour charts — the Newton's colour scale

For t less than about 400 nm the film shows a single tint (first-order colour). Between 400 and 800 nm it shows a mixture of two orders. Above about 1 μm the film is so thick that many orders contribute and the eye sees white. The sequence of first-order colours as t grows — black, grey, white, yellow, orange, red, violet, blue, green — is called Newton's colour scale and is used by mineralogists to estimate the thickness of thin mineral sections under polarised light. The same scale explains why soap bubbles, oil slicks, and the thin layer of oxide on a hot metal all show the same colour sequence in the same order.

Where this leads next