Difficulty: Intermediate | Prerequisites: Double-slit interference (Ch. 17.3), basic wave optics, index of refraction
Thin-film interference produces colourful patterns (oil slicks, soap bubbles) when light reflects off the top and bottom surfaces of a thin transparent layer and the two reflected waves interfere. Single-slit diffraction spreads light into a broad central bright band flanked by weaker fringes, governed by the slit width rather than slit spacing. Circular aperture diffraction sets the fundamental resolution limit of lenses and telescopes.
Thin film
A transparent layer (oil, soap, anti-reflective coating) with a thickness comparable to the wavelength of light. Light reflects from both surfaces, and the two reflected waves interfere.
Phase change on reflection
When light reflects off a medium with a higher index of refraction (going from low n to high n), the reflected wave undergoes a half-wavelength (λ/2) phase shift. When reflecting off a lower-n medium, there is no phase change. Think of it as: the wave flips when it bounces off something denser.
Wavelength in a medium (λ_film)
Light slows down inside a medium with index n, so its wavelength shrinks: λ_film = λ / n. The frequency stays the same.
Huygens' principle
Every point on a wavefront acts as a source of a tiny spherical wavelet, all moving at the wave speed. The new wavefront is the surface tangent to all those wavelets. This explains how a wave spreads after passing through a slit.
Central maximum (single slit)
The broad, bright band at the centre of a single-slit diffraction pattern. It is twice as wide as any other bright fringe in the pattern.
Circular aperture diffraction
When light passes through a circular opening (a lens, a pupil), it produces a circular diffraction pattern: a bright central disc (the Airy disc) surrounded by faint rings. The angular radius of the first dark ring is θ₁ = 1.22λ / D.
Light hits a thin transparent film (index n) sitting on a substrate. Part of the light reflects off the top surface, and part enters the film, reflects off the bottom surface, and exits back through the top. These two reflected beams overlap and interfere.
The critical rule:
If n₁ (the medium the light is coming from) is less than n₂ (the medium it reflects off), the reflected wave picks up a phase change equivalent to λ/2.
If n₁ is greater than n₂, there is no phase change.
For a typical thin film in air (air–film–glass), the top reflection goes from low n (air) to high n (film), so it gets a phase change. The bottom reflection goes from the film to the glass. Whether that one gets a phase change depends on whether the glass has a higher or lower n than the film.
The path-length difference for light travelling through a film of thickness t is 2t (down and back up). Using the wavelength inside the film (λ_film = λ/n):
When only one reflection has a phase change (the usual case for a film in air on glass):
Constructive: 2t = (m + 1/2)(λ/n)
Destructive: 2t = m(λ/n)
When both reflections have a phase change, or neither does:
Constructive: 2t = m(λ/n)
Destructive: 2t = (m + 1/2)(λ/n)
The key is to check each surface for phase changes first, then apply the correct condition.
Two key ideas:
Each point on a wavefront is a source of a spherical wavelet, moving at the wave speed.
The shape of the new wavefront is the curve tangent to all those wavelets.
This explains why light spreads out after passing through a single narrow slit rather than travelling in a straight line.
For a slit of width a, destructive interference (dark fringes) occurs at angles where:
a \sin\theta_p = p\lambdawhere p = ±1, ±2, ±3, ... (p = 0 is excluded because that is the central maximum).
The physical reasoning: at these angles, every point in the top half of the slit can be paired with a point in the bottom half that is exactly λ/2 out of phase, so all light cancels.
The central bright fringe extends between the first dark fringes on either side (p = ±1). Its angular width is:
\theta_p = p\frac{\lambda}{a}The full width of the central maximum on a screen at distance L is:
w = \frac{2\lambda L}{a}A narrower slit produces a wider central maximum. This is the opposite of what many students expect.
The path-length difference between waves from the edges of the slit to the first minimum is:
\Delta r_{1,2} = \frac{a}{2}\sin\theta_r = \frac{\lambda}{2}When light passes through a circular opening of diameter D (a lens, a telescope aperture, the pupil of your eye), it does not focus to a perfect point. Instead, it forms a circular diffraction pattern: a bright central disc surrounded by faint rings.
The angular radius of the first dark ring (the edge of the central disc) is:
\theta_1 = 1.22\frac{\lambda}{D}The factor 1.22 comes from the circular geometry (it replaces the factor of 1 that appears in the single-slit case).
On a screen at distance L, the width (diameter) of the central bright disc is:
w = 2L\tan\theta_1 \approx \frac{2.44\lambda L}{D}A larger aperture D produces a smaller central disc, giving sharper images. This is why telescopes need large mirrors or lenses.
Two point sources (e.g. two stars) can only be distinguished if their Airy discs do not overlap too much. The Rayleigh criterion states that two sources are just resolved when the central maximum of one falls on the first dark ring of the other. This sets the minimum angular separation:
\theta_{\text{min}} = 1.22\frac{\lambda}{D}Formula | Meaning | When to use |
|---|---|---|
2t = m(λ/n) | Constructive (both or neither reflection has phase change) | Thin-film problems after checking phase changes |
2t = (m + 1/2)(λ/n) | Constructive (one reflection has phase change) | Thin-film problems, the more common case |
λ_film = λ/n | Wavelength inside a medium | Any thin-film calculation |
a sin θ_p = pλ | Dark fringes, single slit | Finding minima in single-slit diffraction |
w = 2λL / a | Width of central maximum, single slit | Finding the physical width of the bright centre band |
θ₁ = 1.22λ / D | First dark ring, circular aperture | Circular aperture diffraction and resolution |
w = 2.44λL / D | Width of central disc, circular aperture | Physical size of the Airy disc on a screen |
Thin-film interference is why oil slicks on wet roads display rainbow colours, and why soap bubbles shimmer. Anti-reflective coatings on camera lenses and glasses use thin films tuned to cause destructive interference for reflected light, so more light passes through.
Circular aperture diffraction is the reason telescopes and microscopes have resolution limits. The Hubble Space Telescope's 2.4 m mirror gives it far sharper images than a backyard 20 cm telescope, precisely because a larger D means a smaller Airy disc. The same physics limits the resolving power of your eye (pupil diameter roughly 2 to 8 mm).
Students often swap the constructive and destructive conditions for thin films. The conditions flip depending on whether one or both reflections have a phase change. Always check phase changes at each surface first.
A narrower single slit produces a wider diffraction pattern, not a narrower one. This feels counterintuitive, but the maths is clear: w = 2λL/a, so smaller a means larger w.
Students sometimes confuse the single-slit formula (a sin θ = pλ for dark fringes) with the double-slit formula (d sin θ = mλ for bright fringes). Notice: same form, but one gives minima and the other gives maxima.
The 1.22 factor in circular aperture diffraction is not arbitrary. It comes from the first zero of the Bessel function that describes diffraction through a circular opening. You do not need to derive it, but you do need to remember to include it.
⚠️ Thin-film problems almost always require you to determine phase changes at each surface before writing any equation. Missing this step is the most common source of wrong answers.
⚠️ For single-slit diffraction, remember that a sin θ = pλ gives the dark fringes (minima), not the bright ones. This is the opposite convention from the double-slit / grating equation.
⚠️ The 1.22 factor in circular aperture problems is heavily tested. If a problem says "circular aperture" or "lens" or "telescope", use 1.22λ/D, not λ/D.
⚠️ Resolution problems (can you distinguish two stars, two headlights, etc.) use the Rayleigh criterion: θ_min = 1.22λ/D.
True or false: A phase change occurs when light reflects from a medium with a lower index of refraction. (False: phase change occurs when reflecting off a higher-n medium.)
Fill in the blank: The width of the central maximum in single-slit diffraction is w = ______. (2λL / a)
True or false: Making a slit narrower makes the central diffraction maximum narrower. (False: it makes it wider.)
Fill in the blank: For a circular aperture of diameter D, the first dark ring is at angle θ₁ = ______. (1.22λ / D)
True or false: In thin-film interference, 2t is the path-length difference because light crosses the film once. (False: light crosses it twice, down and back up.)
Q: A thin film of oil (n = 1.50) on water (n = 1.33) is 200 nm thick. For which visible wavelength does the film appear brightest in reflected light?
A: Top reflection (air to oil): n increases, so phase change. Bottom reflection (oil to water): n decreases, so no phase change. One phase change, so constructive condition: 2t = (m + 1/2)(λ/n). With m = 0: λ = 2tn / (0.5) = 2(200)(1.50) / 0.5 = 1200 nm (infrared, not visible). With m = 1: λ = 2(200)(1.50) / 1.5 = 400 nm (violet). The film appears violet.
Q: Light of wavelength 600 nm passes through a single slit 0.10 mm wide. A screen is 2.0 m away. How wide is the central maximum?
A: w = 2λL / a = 2(600 × 10⁻⁹)(2.0) / (0.10 × 10⁻³) = 0.024 m = 24 mm.
Q: A telescope has an aperture of 10 cm. What is the minimum angular separation it can resolve for light at 550 nm?
A: θ_min = 1.22λ/D = 1.22(550 × 10⁻⁹) / (0.10) = 6.71 × 10⁻⁶ rad ≈ 6.7 μrad.
Q: Why does the single-slit formula give minima while the double-slit formula gives maxima, even though both look like "something times sin θ equals mλ"?
A: In the double-slit case, d sin θ = mλ means the path difference between the two slits is a whole number of wavelengths, so the waves arrive in phase: a maximum. In the single-slit case, a sin θ = pλ means every point in the upper half of the slit can be paired with a cancelling point in the lower half (separated by a/2, with path difference λ/2), so all light cancels: a minimum.
Thin-film interference ties back to the superposition and phase concepts from earlier chapters. The phase-change-on-reflection rule is the optical analogue of a wave on a string reflecting from a fixed end.
Single-slit diffraction connects directly to the multi-slit and grating work from Section 17.3. In practice, the single-slit diffraction envelope modulates the multi-slit interference pattern, so a real grating pattern shows sharp peaks whose intensities are shaped by the single-slit envelope.
Circular aperture diffraction leads into the resolution limits of optical instruments, a topic that returns in telescope design, microscopy, and even medical imaging.
Thin-film interference, thin film optics, oil slick colours, soap bubble physics, anti-reflective coating, phase change on reflection, wavelength in a medium, single-slit diffraction, Huygens' principle, Huygens wavelets, diffraction minima, central maximum width, circular aperture diffraction, Airy disc, Airy pattern, Rayleigh criterion, angular resolution, telescope resolution limit, 1.22 lambda over D, diffraction-limited, wave optics, physical optics, Chapter 17 physics