## DP IB Physics: HL

### Revision Notes

Syllabus Edition

First teaching 2014

Last exams 2024

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# 9.3.5 Thin Film Interference

## Thin Film Interference

• Thin film interference causes the iridescence seen in:
• Nature on peacock feathers
• Glossy flower petals
• Soap bubbles
• The shiny side of a CD
• Thin layers - or films - of oil on water
• This phenomenon occurs when light waves reflecting off the top and bottom surfaces of a thin film interfere with one another The colourful pattern observed on a CD is a result of thin film interference

#### Conditions for Thin Film Interference

• To see the interference light must be incident on a material which:
• Is very thin
• Has a higher refractive index than the medium surrounding it
• Also transmits light
• The effect is caused by the reflection of waves from the top and bottom surfaces of the thin film Phase changes at the top and bottom of a thin film

Light incident on the top surface of the thin film (Part A of the process)

• Part of the light wave reflects at a boundary between a less-dense and a more-dense medium (e.g. Air to Oil where nair < noil ), and a phase change is seen, such that:

A wave reflected at a boundary with a medium of a higher refractive index undergoes a phase change of half a wavelength ( ), 180° or π rad A wave incident on the boundary with a more dense material is both reflected and transmitted. The transmitted wave continues without a phase change and the reflected wave has a 180° phase change. What was the peak on the incident wave becomes the trough on the reflected wave.

• As shown above, the light is reflected and transmitted from a boundary between a less dense and a more dense material
• But the light is transmitted only from a boundary between a more dense to less dense material

Hence, the wave is reflected and transmitted between:

Air and Water

Oil and Water

and transmitted only between:

Oil and Air

• A less dense material has a lower refractive index (n)
• The refractive index of air is 1
• The refractive index of water is 1.33
• The refractive index of oil is 1.4-1.5
• Normally in thin-film interference situations: nair < nwater < noil Light is reflected and transmitted at the boundary from a less dense to a more dense material. Light is transmitted only at the boundary from a more dense to a less dense material. Hence, in this diagram P and Q exist but the third unlabelled ray does not.

Why does this happen?

• Imagine the ray of light is a football
• When kicked at a MORE DENSE solid brick wall the ball is reflected back from the wall into the same side as it was kicked from
• When kicked at a LESS DENSE glass window, well depending on how hard the ball is kicked and how strong the glass the ball could pass through the window to the other side into the house or bounce back off the window and return to the same side it was kicked from
• So, the object could pass through or be reflected back from a less dense boundary but reflected only from a more dense boundary

Light transmitting through the thin film (Part B of the process)

• Part of the light wave also refracts as it enters the thin film and is transmitted through the material to the bottom surface
• The light wave is now travelling through a denser medium, so it travels at a slower speed and has a shorter wavelength When the light ray enters a denser medium the wavelength becomes shorter.

Light incident on the bottom surface of the thin film (C)

• If the bottom surface is at a boundary with a more dense material (a higher refractive index e.g. oil-water where  noil < nwater) then both of the following will occur:
• Reflection with a π phase change back into the thin film
• Refraction of the transmitted light into the next medium
• This is the same situation as the light wave incident on the top surface of thin film Light is transmitted with no phase change between the boundary of a less dense to a more dense material and reflected with a phase change.

• If the bottom surface is at a boundary with a less dense material (a lower refractive index e.g. oil-air where nair < noil ) then the following will occur:
• Transmission with refraction will occur out of the material
• No reflection will occur at the boundary

Light transmitting through the thin film and refracting back into the first medium (D)

• Light travelling through the thin film undergoes a wavelength shift due to the density of the plastic
• The wavelength shift is a multiple (normally labelled m) of a half-integer multiple of wavelengths
• The thinner the film, the smaller the value of Light reflecting off the thin-film air boundary undergoes a wavelength shift related to the density of the thin-film. Light reflecting off the air-thin film boundary undergoes a phase change of half a wavelength. This results in constructive interference and a bright light seen by the observer.

Observing Constructive Interference

• Constructive interference will be seen as brighter colours by the observer
• The light intensity has increased making the colour appear brighter
• The total path difference of the two waves reflected and refracted off the top surface of the thin film must be a multiple of wavelengths for constructive interference to be seen by the observer The combination of the phase change due to the reflection off the air-film surface and the phase change within the film here, results in constructive interference seen by an observer because the path difference is a multiple number of wavelengths.

• According to the data booklet, the formula for the path difference for constructive interference is:

2dn • Where:
• = thickness of the film
• = refractive index of medium
• = an integer related to the refractive index and thickness of the medium
• The thinnest film exists when = 0
•  λ = wavelength of the light in air
• Constructive interference occurs when two waves are out of phase by an integer number of wavelengths The wave reflected from A has undergone a phase change so now has a path difference of • The wave coming from C inside the film must have a phase difference of a half-integer multiple of wavelengths So, the total phase difference would be + multiple of • In the formula for the path difference for constructive interference: 2dn • So, the thinnest film exists when = 0
• So the multiple of would just be one
• When m = 1 the multiple would be • A common question is to be asked to calculate the thickness of the thinnest film

Observing Destructive Interference

• Darker colours are observed through destructive interference
• The light intensity has decreased making the colour appear darker
• The total path difference of the two waves reflected and refracted off the top surface of the thin film must be a half-integer multiple of wavelengths for destructive interference to be seen by the observer The combination of the phase change due to the reflection off the air-film surface and the phase change within the film here, results in destructive interference seen by an observer because the path difference is a half integer multiple number of wavelengths.

• According to the data booklet, the formula for the path difference for destructive interference is:

2dn = mλ

• Where:
• = thickness of the film
• = refractive index of medium
• = an integer related to the refractive index and thickness of the medium
• λ = wavelength of the light in air
• Destructive interference occurs when two waves are out of phase by a half-integer number of wavelengths The wave reflected from A has undergone a phase change so now has a path difference of • The wave coming from C inside the film must have a phase difference of a half-integer multiple of wavelengths  So, the total phase difference would be + multiple of  • In the formula for the path difference for destructive interference: dn • So, the thinnest film exists when = 1
• Other films exist with destructive interference for only odd values of m
• When m = 2 for example constructive interference and not destructive takes place as = λ

Distance Travelled inside the film

• Light travels the same distance within a thin film whether it undergoes constructive or destructive interference
• Light enters from the top surface of the thin film, passes through the thin film of thickness
• Light is then reflected off the bottom surface travelling a further distance of d to the original surface
• So, the total distance travelled whether the light interferes constructively or destructively is 2d The reflected wave travels a total distance of twice the thickness of the film Model the thin film as a parallel-sided, rectangular ‘slice’ with a thickness of ‘d’

#### Uses of Thin Film Interference

• Thin films can prevent light from being reflected
• Due to the conservation of energy, this increases the light which is transmitted through a medium
• This effect is utilised in:
• Camera lenses - these often have a coating to prevent reflection
• Solar cells - this is to ensure a high proportion of incident light can be captured

## Solving Problems Using Thin Films

#### The Mathematics of Constructive and Destructive Interference

• Consider a thin film, such as a soap bubble, modelled as a very thin, parallel-sided rectangle, with thickness d and refractive index higher than that of the surrounding air, i.e. n > 1
• If the thin film is sitting on top of a different denser medium e.g. water then the following situations will not apply because of the phase change obtained from the reflection at the oil-water boundary A thin film can be modelled as a parallel-sided, rectangular ‘slice’ with a thickness of ‘d’

• Constructive interference occurs when:

The thickness of the coating is • This is because the light from the bottom of the film has travelled an extra distance (there and back)
• Since the first ray has undergone a phase change of π, or half a wavelength, , the condition for constructive interference is:

Path difference, • Where:
• d = thickness of the film (m)
• m = an integer (generally taken as m = 0)
• λ0 = wavelength of light in soap (m)
• The value of λ0 can be obtained using the relationship: • Where:
• λ = wavelength of light in a vacuum (m)
• n = refractive index of the film
• The condition for constructive interference can, therefore, be rewritten in terms of λ as: • Rearranging this gives: • Destructive interference occurs when:

The thickness of the coating is • This is because the light from the bottom of the film has travelled an overall distance of λ
• This time, the condition for destructive interference is:

Path difference, • The value of λ0 can be obtained using the relationship: • The condition for destructive interference can, therefore, be rewritten in terms of λ as: • Rearranging this gives: #### Worked example

A camera lens has a reflective coating applied to ensure that as much of the light falling on the lens is transmitted, with minimal reflection.

The lens has refractive index of 1.72 and the coating a refractive index of 1.31.

Estimate the thickness of coating required to minimise reflection of visible light. You can assume an average wavelength of 540 nm.

Step 1: List the known quantities

• Refractive index of coating, n = 1.31
• Wavelength, λ = 540 nm

Step 2: Use the data booklet to find the conditions for thin-film interference

 Constructive Interference Destructive Interference  Step 3: Consider the phase changes at the boundaries

• Firstly, at the boundary with the coating, which has higher optical density, phase change = π
• Secondly, at the boundary with the lens, which has higher optical density, phase change = π

Step 4: Determine the correct equation to use

• To prevent reflection, this requires destructive interference, so the equation to use is: • For a minimum thickness, m = 0

Step 5: Rearrange to make thickness, d, the subject and calculate  = 100 nm 