Bubbles in the air, peacock feathers in bloom and oily puddles all reveal a light phenomenon: thin film interference.
It's a tricky concept involving reflection, refraction and phase changes at interfaces.
Key Concepts
Use quizzes to practise application of theory.
START QUIZ!
The image, taken from a GeoGebra simulation of thin film interference, shows a wave passing through a thin film.
The refractive index of the film is:
\(n = {c_1\over c_2} ={λ_1\over λ_2} = {6\over 1.5}\)
The image, taken from a GeoGebra simulation of thin film interference, shows a wave passing through a thin film.
If the wavelength of the wave in air is 400 nm, the thickness of the film is
The wavelength in the film is \(1\over 4\) the wavelength in air = 100 nm.
Film is \(2λ\) thick = 200 nm = 200 x 10-9 = 0.2 x 10-6.
The image, taken from a GeoGebra simulation of thin film interference, shows a wave passing through a thin film.
Which sequence gives the colours of the waves in this order:
incident on 1st face - reflected off first face - reflected off second face
We see the green wave reflect off the left face out of phase.
The red wave reflects off the right face in phase.
The black line is incident on the left face and emergent from the right.
NB: Sorry if you are colour blind - the exam would find a mechanism for distinguishing in black and white!
The image, taken from a GeoGebra simulation of thin film interference, shows a wave passing through a thin film. The wavelength in the film is λ.
The path difference between the red and green waves is:
The path difference = 2 x thickness.
In this case, the thickness coincides with being 2λ.
The image, taken from a GeoGebra simulation of thin film interference, shows a wave passing through a thin film.
The phase difference between the green and red waves is:
At the left boundary, the green wave reflects with \(\pi\) phase difference to the black wave that enters. The red wave has travelled two wavelengths towards the right boundary and two wavelengths back to the left. Each wavelength contributes a further \(2\pi\) phase difference. The total is \(9\pi\).
A parallel sided soap film of thickness 0.01 μm is illuminated with light of wavelength 700 nm.
The appearance of the film is:
The film thickness (\(10^{-8} \text { m}\)) is much less than the wavelength of the light (almost \(10^{-6} \text{ m}\)) so the path difference is negligible. The phase difference is therefore \(\approx \pi\) so the waves cancel.
Lines are formed by wedges of film with continuously changing thickness.
The thickness of a soap bubble reduces as the soap drains from it. What is the minimum value of the product of thickness and refractive index for constructive interference of light of wavelength 600 nm?
Constructive interference occurs when \(2dn=(m+{1\over 2})\lambda\). Reducing the right hand side using integer \(m=1\):
\(dn={1\over4}λ\)
The image represents a wave passing into a more dense medium.
The reflected wave that has been produced in the same time frame is best represented by:
What was a trough is reflected as a peak.
The image represents a wave passing into a less dense medium.
The reflected wave that has been produced in the same time frame is best represented by:
What was a trough is reflected as a trough.
How much of Thin film interference have you understood?
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