AP Physics 2- 14.9 Thin-Film Interference- Study Notes- New Syllabus
AP Physics 2- 14.9 Thin-Film Interference – Study Notes
AP Physics 2- 14.9 Thin-Film Interference – Study Notes – per latest Syllabus.
Key Concepts:
- Thin-Film Interference
- Practical Examples of Thin-Film Interference
Thin-Film Interference
Thin-film interference occurs when light reflects from the upper and lower boundaries of a thin transparent layer, producing a resultant wave due to the superposition of the reflected waves. This phenomenon is most significant when the thickness of the film is comparable to the wavelength of light.
Phase Changes at Interfaces
Reflection from a higher refractive index:
A phase change of 180° (\(\pi\) radians) occurs when light reflects from a medium with a higher refractive index than the medium it is coming from.
Example: Air (\(n = 1.0\)) → Water (\(n = 1.33\))
Reflection from a lower refractive index:
No phase change occurs when light reflects from a medium with a lower refractive index than the incident medium.
Example: Water (\(n = 1.33\)) → Air (\(n = 1.0\))
Refraction:
The phase of a wave does not change when it is refracted into another medium; only its speed and wavelength change:
$ \mathrm{v = \frac{c}{n}, \quad \lambda_{medium} = \frac{\lambda_0}{n}} $
Formation of Interference
- Light is partially reflected at the top surface of the film and partially transmitted.
- The transmitted portion is reflected at the bottom surface of the film.
- The two reflected waves interfere, forming a resultant wave, the nature of which (constructive or destructive) depends on the phase difference and path difference.
Path Difference
The effective path difference between the two reflected waves is given by:
$ \mathrm{\Delta = 2 n t \cos \theta_t + \phi} $
- \(\mathrm{n}\) = refractive index of the film
- \(\mathrm{t}\) = thickness of the film
- \(\theta_t\) = angle of refraction inside the film
- \(\phi\) = phase shift due to reflection (0 or \(\pi\))
Conditions for Interference
The interference of light in a thin film depends on the phase difference between the light reflected from the top and bottom surfaces of the film. This phase difference arises from both the extra path traveled inside the film and any phase changes upon reflection. Depending on this phase difference, the reflected waves can interfere constructively (bright) or destructively (dark).
Constructive Interference:
Occurs when the reflected waves reinforce each other to produce maximum intensity.
$\mathrm{2 n t \cos \theta_t + \phi = m \lambda_{medium}}, \quad m = 0, 1, 2, \dots$
Destructive Interference:
Occurs when the reflected waves cancel each other to produce minimum intensity.
$\mathrm{2 n t \cos \theta_t + \phi = (m + \frac{1}{2}) \lambda_{medium}}, \quad m = 0, 1, 2, \dots $
Factors Affecting Interference
- Thickness of the film (\(\mathrm{t}\))
- Wavelength of light (\(\mathrm{\lambda}\))
- Phase shifts due to reflection (\(\phi\))
- Angle of incidence (\(\theta_i\)) and refraction (\(\theta_t\))
Applications
- Colorful patterns in soap bubbles and oil slicks
- Anti-reflection coatings on lenses and glasses
- Optical coatings in cameras, microscopes, and other instruments
Example :
Light of wavelength \(\mathrm{600 \, nm}\) strikes a thin oil film (\(n = 1.25\)) on water at normal incidence. The film thickness is \(\mathrm{250 \, nm}\). Determine whether the reflected light will interfere constructively or destructively.
▶️ Answer/Explanation
Step 1: Phase changes: Top surface (air → oil, \(n_2 > n_1\)) → \(\pi\), Bottom surface (oil → water, \(n_2 < n_1\)) → 0
Step 2: Path difference: \(\mathrm{\Delta = 2 n t = 2 \times 1.25 \times 250 \, nm = 625 \, nm}\)
Step 3: Condition for constructive interference with one \(\pi\) phase shift: \(\mathrm{2 n t = (m + 1/2) \lambda = (0 + 1/2) \times 600 = 300 \, nm}\)
Step 4: Since \(\Delta = 625 \neq 300\), light will partially interfere; closer to destructive interference for \(m = 0\).
Example :
Light of wavelength \(\mathrm{550 \, nm}\) is incident normally on a thin anti-reflection coating (\(n = 1.38\)) of thickness \(\mathrm{t = 100 \, nm}\). Determine the order of the first constructive interference for reflected light.
▶️ Answer/Explanation
Step 1: One phase change occurs (air → coating), so constructive interference: \(\mathrm{2 n t = (m + 1/2) \lambda}\)
Step 2: Substitute values: \(\mathrm{2 \times 1.38 \times 100 = (m + 1/2) \times 550}\)
\(\mathrm{276 = 550 (m + 1/2)} \Rightarrow m + 1/2 \approx 0.502 \approx 0\)
Step 3: First constructive interference occurs at \(m = 0\).
Practical Examples of Thin-Film Interference
Soap Bubbles and Oil Films
- Color variations are observed due to interference of light reflected from the top and bottom surfaces of thin films of varying thickness.
- The spectrum of colors arises because different thicknesses produce constructive interference for different wavelengths (\(\lambda\)) of light.
- Example: A soap bubble may appear red in one region and blue in another depending on the local film thickness.
Anti-Reflection Coatings
- Used on lenses, glasses, and camera optics to reduce glare and unwanted reflections.
- Principle: Reflected light from the top and bottom surfaces of the coating interferes destructively, canceling the reflection.
The simplest anti-reflection coating has thickness:
$ \mathrm{t = \frac{\lambda}{4 n}} $ where \(\mathrm{n}\) is the refractive index of the coating and \(\lambda\) is the wavelength of light in air.
- Requirements for the coating: \(\mathrm{n_{air} < n_{coating} < n_{surface}}\) and incident light is assumed normal to the surface.
Summary of Effects
- Variation in film thickness → different colors in soap bubbles/oil films.
- Quarter-wavelength coating → destructive interference for reflected light → reduced glare.
Example:
A camera lens is coated with a thin film of refractive index \(\mathrm{n = 1.38}\) to reduce reflection of light with wavelength \(\mathrm{550 \, nm}\) (green light). Find the required thickness of the coating for maximum destructive interference.
▶️ Answer/Explanation
Step 1: Use the quarter-wavelength formula: \(\mathrm{t = \frac{\lambda}{4 n}}\)
Step 2: Substitute values: \(\mathrm{t = \frac{550 \, nm}{4 \times 1.38} \approx 99.6 \, nm \approx 100 \, nm}\)
Step 3: Therefore, the coating thickness should be approximately \(\mathrm{100 \, nm}\) to minimize reflection.