Additional: Thin Films Interference

Interference due to Reflection in Thin Films

When light interacts with a thin transparent film (like a soap bubble, an oil slick on water, or anti-reflective coatings on lenses), the light waves reflected from the top and bottom surfaces of the film interfere. This interference can lead to beautiful coloured patterns when viewed under white light.

Mechanism of Thin Film Interference

Consider a thin film of a transparent material with thickness $t$ and refractive index $n_f$, placed between two other media (e.g., air above and below, or air above and water below). When a ray of light (say, from air, $n_a$) is incident on the top surface of the film, a portion of the light is reflected (ray 1), and the rest is refracted into the film. The refracted ray travels through the film and strikes the bottom surface. Here, another portion of the light is reflected back into the film, and the rest is transmitted. The ray reflected from the bottom surface travels back up through the film and emerges into the top medium, where it is refracted and travels in the same direction as the ray reflected from the top surface (ray 2).

Diagram illustrating interference of light reflected from the top and bottom surfaces of a thin film.

(Image Placeholder: A ray of light incident on a thin film (refractive index n_f, thickness t) with air above (n_a) and below (n_s). Show the ray incident at angle i. Show a ray reflected from the top surface (ray 1). Show a ray refracted into the film at angle r, travels to the bottom surface, reflects back, refracts out at the top surface, travelling parallel to ray 1 (ray 2). Indicate the path difference between ray 1 and ray 2.)

Ray 1 and Ray 2 are derived from the same incident ray, so they are coherent. They travel in the same direction and overlap. Their superposition results in interference.

The observed interference pattern (whether the region appears bright or dark for a particular wavelength) depends on the path difference between ray 1 and ray 2, and any phase changes upon reflection.

Path Difference

Assuming the film is in air ($n_a \approx n_s \approx 1$), the path difference arises primarily because ray 2 travels an extra distance within the film (down and back up) compared to ray 1, which reflects directly from the top surface. For a ray incident at an angle $i$ and refracting at angle $r$ inside the film, and film thickness $t$, the optical path difference between ray 1 and ray 2 is approximately:

$ \text{Optical Path Difference} = n_f (2t \cos r) $

where $r$ is the angle of refraction inside the film. For normal incidence, $i=0$, $r=0$, and the optical path difference is $2n_f t$.

Phase Change upon Reflection

In addition to the path difference, a phase change of $\pi$ radians (180°) occurs when light reflects from the boundary of an optical medium where the light is coming from a rarer medium and reflecting from a denser medium. This is analogous to a wave on a string reflecting from a fixed end.

Reflection at the top surface (Air-Film): If $n_a < n_f$, there is a phase change of $\pi$ radians for ray 1 upon reflection.
Reflection at the bottom surface (Film-Substrate): If $n_f < n_s$, there is a phase change of $\pi$ radians for the ray reflected at the bottom surface. If $n_f > n_s$, there is no phase change (0 radians).

For a thin film of soap solution in air ($n_a < n_f > n_s$, where $n_s$ is the refractive index of the air/medium below the film, which is also air), reflection at the top surface (air to soap solution, rarer to denser) introduces a phase change of $\pi$. Reflection at the bottom surface (soap solution to air, denser to rarer) introduces no phase change. So, there is an effective phase difference of $\pi$ (corresponding to $\lambda/2$ optical path difference) between ray 1 and ray 2 due to the reflections themselves.

The total optical path difference between ray 1 and ray 2 is the sum of the geometrical path difference ($2n_f t \cos r$) and any optical path difference due to phase changes upon reflection. The $\pi$ phase change at a boundary is equivalent to an extra path of $\lambda/2$.

Total Optical Path Difference

For a soap film in air (phase change $\pi$ at top, 0 at bottom):

$ \text{Total Optical Path Difference} = (2n_f t \cos r) + \frac{\lambda_{film}}{2} $ (from phase change at top surface, $\lambda_{film}$ is wavelength in film)

No, the phase change effect is usually added as a path difference of $\lambda_{medium}/2$. For ray 1, phase change is $\pi$ at air-film boundary. For ray 2 reflected inside the film, phase change at film-air boundary is 0. So, ray 1 has a relative phase shift of $\pi$ compared to ray 2's internal reflection. The total optical path difference contributing to interference is the geometrical path difference plus/minus any path difference equivalent to phase changes.

Correct approach: The total optical path difference between ray 1 and ray 2 when measured in the medium from which light is incident (air) is $2n_f t \cos r$. We need to add an equivalent path difference corresponding to the phase changes at the reflection points. For a soap film in air ($n_{air} < n_{film} > n_{air}$), reflection from air to film adds a phase shift of $\pi$ (equiv. $\lambda_{air}/2$ path difference). Reflection from film to air adds 0 phase shift. So, the total optical path difference between ray 1 and ray 2 is $2n_f t \cos r$. The effective path difference causing interference is $2n_f t \cos r$ plus/minus the path difference equivalent to the phase change difference. The phase change difference between ray 1 (reflected from air-film) and ray 2 (reflected from film-air) is $\pi - 0 = \pi$. A phase difference of $\pi$ is equivalent to an optical path difference of $\lambda/2$.

So, the total effective path difference is $ (2n_f t \cos r) + \frac{\lambda_{air}}{2}$ ? No, it should be relative to the wavelength in the film $\lambda_f = \lambda_{air}/n_f$. The optical path difference is measured in terms of path difference in vacuum or air, so $2n_f t \cos r$ is correct. The phase change upon reflection is a $\pi$ phase shift. This is equivalent to an extra path of $\lambda/2$ where $\lambda$ is the wavelength in the medium the wave is travelling in BEFORE reflection. So, for reflection at the top surface (air to film), $\lambda$ is in air. For reflection at the bottom surface (film to air), $\lambda$ is in the film. This is also complex.

Simpler approach for phase change: A $\pi$ phase shift upon reflection is equivalent to adding or subtracting $\lambda/2$ to the optical path difference. For reflection from an optically denser medium, there's a $\pi$ phase shift. From an optically rarer medium, there's no phase shift. For a soap film in air ($n_{air} < n_{film} > n_{air}$): - Ray 1 reflects from air ($n_a$) to film ($n_f$): reflection from a denser medium. Phase shift $\pi$. - Ray 2 reflects from film ($n_f$) to air ($n_a$): reflection from a rarer medium. Phase shift 0. The net phase difference due to reflections is $\pi - 0 = \pi$, which is equivalent to a path difference of $\lambda/2$. This equivalent path difference must be added to the geometrical path difference. The optical path difference is measured in the medium of incidence or vacuum. The geometrical path difference is $2t/\cos r$. The optical path difference is $n_f (2t \cos r)$. No, the geometrical path in the film is $2t/\cos r$. The optical path is $n_f \times (2t/\cos r)$. Let's recheck standard texts on path difference. Optical path difference is indeed $n_f (2t \cos r)$.

Total effective optical path difference $\Delta X_{eff}$ for interference is $n_f (2t \cos r)$ plus/minus the path difference due to reflection phase changes.

For reflection from denser medium: $\pi$ phase change, adds $\lambda/2$ optical path difference. For reflection from rarer medium: $0$ phase change, adds $0$ optical path difference.

For a soap film in air ($n_{air} < n_{film} > n_{air}$): - Ray 1 reflects from air ($n_{air}$) to film ($n_{film}$), $n_{air} < n_{film}$. Reflection from denser. Phase change $\pi$. Equivalent path $\lambda_{air}/2$. - Ray 2 reflects from film ($n_{film}$) to air ($n_{air}$), $n_{film} > n_{air}$. Reflection from rarer. Phase change 0. Net effect of reflections: One ray gets a $\pi$ phase shift, the other doesn't. The optical path difference $2n_f t \cos r$ is the difference in path lengths traveled in the film. Interference depends on the total phase difference, which is $(2\pi/\lambda_{film}) \times (\text{geometrical path diff}) + (\text{phase diff from reflections})$. Optical path diff is $\lambda_{air} / \lambda_{film} \times \text{geometrical path diff} = n_f \times \text{geometrical path diff}$.

Let's use the optical path difference $2 n_f t \cos r$. And add phase differences. For a soap film in air (reflection from air to film: $\pi$ phase change; reflection from film to air: $0$ phase change): Effective path difference for interference: $(2 n_f t \cos r) + \lambda_{air}/2$ (due to the $\pi$ phase difference between the two reflected rays).

For a film on a substrate (e.g., anti-reflective coating on glass, $n_{air} < n_{film} < n_{glass}$): - Ray 1 reflects from air ($n_{air}$) to film ($n_{film}$), $n_{air} < n_{film}$. Phase change $\pi$. - Ray 2 reflects from film ($n_{film}$) to substrate ($n_{glass}$), $n_{film} < n_{glass}$. Phase change $\pi$. Net effect of reflections: $\pi - \pi = 0$ phase difference. Effective path difference for interference: $2 n_f t \cos r$.

Conditions for Constructive and Destructive Interference

The conditions for constructive and destructive interference in thin films are determined by the total effective optical path difference between the interfering rays.

General Conditions (Considering Phase Changes)

Let $\Delta X_{opt} = 2n_f t \cos r$ be the optical path difference due to travel in the film. Let $\Delta\phi_{ref}$ be the phase difference introduced due to reflections at the two interfaces. The total phase difference is $\Delta\Phi = \frac{2\pi}{\lambda_{air}} \Delta X_{opt} + \Delta\phi_{ref}$.

Conditions for interference:

Constructive Interference (Bright): Total phase difference is an even multiple of $\pi$ ($0, 2\pi, 4\pi, ...$). $ \Delta\Phi = 2 m \pi $, where $m$ is an integer (0, 1, 2, ...).
Destructive Interference (Dark): Total phase difference is an odd multiple of $\pi$ ($\pi, 3\pi, 5\pi, ...$). $ \Delta\Phi = (2 m + 1) \pi $, where $m$ is an integer (0, 1, 2, ...).

Conditions for Thin Film in Air (or when $n_{air} < n_{film} > n_{air}$ type interfaces)

As discussed, for a soap film in air, there is a $\pi$ phase difference between ray 1 and ray 2 due to reflections. This adds an effective path difference of $\lambda_{air}/2$. The total optical path difference is $2n_f t \cos r$. The effective path difference that matters for interference comparing the two rays is $2n_f t \cos r$. The condition for interference includes the phase change difference. Total phase difference = Phase diff from path + Phase diff from reflections.

Total phase difference = $\frac{2\pi}{\lambda_{air}} (2n_f t \cos r) + \pi$ (from reflections)

Bright Fringes: $\frac{2\pi}{\lambda_{air}} (2n_f t \cos r) + \pi = 2m\pi$ $ \frac{2n_f t \cos r}{\lambda_{air}} + \frac{1}{2} = m $ $ 2n_f t \cos r = (m - \frac{1}{2})\lambda_{air} = (2m-1)\frac{\lambda_{air}}{2} $ where $m = 1, 2, 3, ...$ (if $m=0$, right side is negative, not physically meaningful path difference for bright fringe).
Dark Fringes: $\frac{2\pi}{\lambda_{air}} (2n_f t \cos r) + \pi = (2m+1)\pi$ $ \frac{2n_f t \cos r}{\lambda_{air}} + \frac{1}{2} = m + \frac{1}{2} $ $ \frac{2n_f t \cos r}{\lambda_{air}} = m $ $ 2n_f t \cos r = m\lambda_{air} $ where $m = 0, 1, 2, 3, ...$

Using integer $m = 0, 1, 2, ...$ for both cases, the conditions for a thin film in air are often written as:

Bright Fringes: $2n_f t \cos r = (2m+1)\frac{\lambda_{air}}{2}$
Dark Fringes: $2n_f t \cos r = m\lambda_{air}$

where $m = 0, 1, 2, ...$ and $\lambda_{air}$ is the wavelength in air. For normal incidence, $\cos r = \cos 0^\circ = 1$, and the conditions are $2n_f t = (2m+1)\lambda_{air}/2$ for bright, $2n_f t = m\lambda_{air}$ for dark.

Conditions for Thin Film on a Denser Substrate (or when $n_{air} < n_{film} < n_{substrate}$)

For a film on a denser substrate (e.g., anti-reflective coating of MgF$_2$ on glass, $n_{air} < n_{MgF_2} < n_{glass}$), reflection at the air-film interface ($n_{air} < n_{film}$) gives a $\pi$ phase change. Reflection at the film-substrate interface ($n_{film} < n_{substrate}$) also gives a $\pi$ phase change. The net phase difference from reflections is $\pi - \pi = 0$.

The total phase difference = $\frac{2\pi}{\lambda_{air}} (2n_f t \cos r)$.

Bright Fringes: $\frac{2\pi}{\lambda_{air}} (2n_f t \cos r) = 2m\pi \implies 2n_f t \cos r = m\lambda_{air}$
Dark Fringes: $\frac{2\pi}{\lambda_{air}} (2n_f t \cos r) = (2m+1)\pi \implies 2n_f t \cos r = (2m+1)\frac{\lambda_{air}}{2}$

where $m = 0, 1, 2, ...$ and $\lambda_{air}$ is the wavelength in air. Note that the conditions for bright and dark fringes are swapped compared to the soap film in air case.

Example 1. A thin film of oil (refractive index 1.4) floats on water (refractive index 1.33). White light is incident normally on the oil film. What is the minimum thickness of the oil film that appears bright in yellow light (wavelength 580 nm in air)?

Answer:

Film material: oil, $n_f = 1.4$.

Medium above film: air, $n_{air} = 1$.

Medium below film: water, $n_{water} = 1.33$.

Wavelength of light in air, $\lambda_{air} = 580$ nm $= 580 \times 10^{-9}$ m.

Incidence is normal, so $i=0$, and refraction angle $r=0$, $\cos r = 1$.

Let's check phase changes upon reflection:

Reflection at the top surface (Air-Oil): $n_{air} < n_f$ ($1 < 1.4$). Reflection from denser medium. Phase change $\pi$.
Reflection at the bottom surface (Oil-Water): $n_f > n_{water}$ ($1.4 > 1.33$). Reflection from rarer medium. Phase change 0.

Net phase difference from reflections = $\pi - 0 = \pi$, equivalent to $\lambda_{air}/2$ optical path difference.

The total optical path difference between the two reflected rays is $2n_f t \cos r = 2n_f t (1) = 2n_f t$.

Condition for Bright Fringes (Constructive Interference): Total phase difference = $2m\pi$. $\frac{2\pi}{\lambda_{air}}(2n_f t) + \pi = 2m\pi$. $ \frac{2n_f t}{\lambda_{air}} + \frac{1}{2} = m $ $ 2n_f t = (m - \frac{1}{2})\lambda_{air} = (2m-1)\frac{\lambda_{air}}{2} $

We want the minimum thickness for a bright fringe, which corresponds to the smallest valid value of $m$. Since thickness $t$ must be positive, and $n_f$ and $\lambda_{air}$ are positive, $2n_f t > 0$. $(2m-1)$ must be positive. If $m=0$, $2m-1 = -1$. If $m=1$, $2m-1 = 1$. So the minimum positive value of $(2m-1)$ is 1, which occurs when $m=1$. (Some texts use $m = 0, 1, 2, ...$ for conditions and then adjust the formula form to be $2n_f t \cos r = (m+1/2)\lambda_{air}$ for bright in this case. Let's use our derived formula and check which $m$ values work. The formula $2n_f t \cos r = (2m-1)\lambda_{air}/2$ requires $m \ge 1$ for $2m-1 \ge 1$. Minimum thickness for bright is for $m=1$).

For minimum thickness, $m=1$:

$ 2n_f t_{min} = (2(1)-1)\frac{\lambda_{air}}{2} = \frac{\lambda_{air}}{2} $

$ t_{min} = \frac{\lambda_{air}}{4n_f} $

Substitute the values:

$ t_{min} = \frac{580 \times 10^{-9} \text{ m}}{4 \times 1.4} = \frac{580 \times 10^{-9}}{5.6} $ metres.

$ t_{min} \approx 103.57 \times 10^{-9} $ metres.

$ t_{min} \approx 103.57 $ nm.

The minimum thickness of the oil film that appears bright in yellow light is approximately 103.6 nanometres.

Colors in Soap Bubbles and Oil Slicks

Thin film interference is responsible for the vibrant colours often seen in soap bubbles and oil slicks on water. These colours are observed when white light (which is a mixture of all visible wavelengths) is incident on the thin film.

Explanation of Colours

When white light is incident on a thin film, different wavelengths (colours) satisfy the conditions for constructive or destructive interference at different locations on the film, depending on the local thickness of the film and the angle of viewing.

For a particular thickness $t$ and angle of viewing $r$, some wavelengths will satisfy the condition for constructive interference, and these colours will appear bright to the observer.
Other wavelengths will satisfy the condition for destructive interference, and these colours will be missing from the reflected light, making the film appear the complementary colour.

Since the thickness of a soap bubble or an oil slick is often not uniform, different parts of the film will have different thicknesses. Also, as the film evaporates or flows, the thickness changes over time. Therefore, when viewed under white light, different areas of the film appear to have different colours, and these colours can change as the film thickness changes.

The pattern of colours observed in a soap bubble or oil slick is a direct manifestation of the interference of light reflected from its top and bottom surfaces, and the dependence of the interference conditions on the thickness of the film and the wavelength of light.

At locations where the film is very thin (thickness $t$ is very small), the optical path difference $2n_f t \cos r$ is very small. If the phase change upon reflection at both surfaces is the same (e.g., $n_{air} < n_{film} < n_{substrate}$ type boundary), then for $t \approx 0$, the path difference is near zero, leading to destructive interference for all wavelengths, and the film might appear dark. If the phase changes are different (e.g., soap film in air), then for $t \approx 0$, there is an effective path difference of $\lambda/2$ due to reflection phases, leading to destructive interference for $\lambda/2 = \lambda_{air}/2 \implies \lambda_{air} \approx 0$ (blue end) and constructive for larger wavelengths, but the conditions for thicker regions dominate visibility.

As the thickness increases, the path difference $2n_f t \cos r$ increases, and different wavelengths satisfy the constructive/destructive conditions. For instance, a certain thickness might cause constructive interference for blue light, destructive for red light, and partial interference for other colours, resulting in a particular perceived colour. A slightly different thickness will lead to a different set of constructive/destructive wavelengths and thus a different colour.

Near the top of a vertically held soap film, gravity causes the film to thin out. Interference colours appear, with the thinnest region at the very top often appearing black just before the film breaks, because the thickness becomes so small that the optical path difference is almost zero, and destructive interference occurs for all visible wavelengths (due to the $\pi$ phase difference upon reflection from the top surface, cancelling out the ray from the bottom surface which has negligible path difference and no phase change).

The vibrant colours seen in soap bubbles and oil slicks are striking examples of the wave nature of light and the phenomenon of interference.