Thin Film Interference

Thin film interference is a phenomenon that occurs when light waves reflect off the surfaces of a thin film, such as soap bubbles or oil slicks, and create a pattern of bright and dark fringes. This effect results from the constructive and destructive interference of light waves, depending on the film's thickness and the wavelength of light. In real life, thin film interference is responsible for the vibrant colours seen in soap bubbles, oil spills, and certain coatings on lenses and glasses. By studying thin-film interference, we gain insights into wave behaviour and can develop advanced technologies in optics and material science. This article explores the principles, experimental setup, and real-life applications of thin film interference.

Thin Film Interference

Thin film interference is a captivating optical phenomenon that occurs when light waves reflect off the different surfaces of a thin film, such as oil on water or a soap bubble. This effect results in a series of colourful patterns caused by the constructive and destructive interference of light waves. The varying thickness of the film and the wavelength of light influence the resulting colours and patterns.

Interference effects are commonly observed in thin films when their thickness is comparable to the wavelength of incident light ( if it is too thin as compared to the wavelength of light it appears dark and if it is too thick, this will result in uniform illumination of the film). A thin layer of oil on the water surface and soap bubbles show various colours in white light due to the interference of waves reflected from the two surfaces of the film.

In thin films, interference takes place between the waves reflected from its two surfaces and waves refracted through it

Interference in Reflected Light

Interference in reflected light is a phenomenon observed when light waves reflected from the different interfaces of a thin film interfere with each other. This type of interference can be either constructive or destructive, depending on the phase difference between the reflected waves. The phase difference arises due to the variation in the optical path length and the potential phase shift upon reflection.

Net path difference between two consecutive waves in the reflected system $\Delta x=2 \mu t \cos r-\frac{\lambda}{2}$

(As the ray suffers reflection at the surface of a denser medium an additional phase difference of $\pi$ or a path difference of $\frac{\lambda}{2}$ is introduced.)

1. Condition of constructive interference (maximum intensity):

The condition for constructive interference (maximum intensity) in reflected light in a thin film is achieved when the path difference between the two reflected waves is an integer multiple of the wavelength of the light. This can be mathematically expressed as

$\begin{aligned} & \Delta x=n \lambda \\ \Rightarrow & 2 \mu t \cos r+\frac{\lambda}{2}=n \lambda \\ \Rightarrow & 2 \mu t \cos r=\left(n-\frac{1}{2}\right) \lambda\end{aligned}$

For normal incidence, i.e r = 0, so $2 \mu t=(2 n-1) \frac{\lambda}{2}$

2. Condition of destructive interference (minimum intensity)

The condition for destructive interference (minimum intensity) in reflected light in a thin film is achieved when the path difference between the two reflected waves is an odd multiple of half the wavelength of the light. This can be mathematically expressed as:

$
\Delta x=2 \mu t \cos r=(2 n) \frac{\lambda}{2} .
$

And For normal incidence $2 \mu t=n \lambda$

Interference in Refracted Light

Net path difference between two consecutive waves in the refracted system = $\Delta x=2 \mu t \cos r$

1. Condition of constructive interference (maximum intensity)

$
\Delta x=2 \mu t \cos r=(2 n) \frac{\lambda}{2} \text {. }
$
and For normal incidence $2 \mu t=n \lambda$

2. Condition of destructive interference (minimum intensity):

$
\Delta x=2 \mu t \cos r=(2 n-1) \frac{\lambda}{2}
$

For normal incidence : $2 \mu t=(2 n-1) \frac{\lambda}{2}$

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Solved Examples Based on Thin Film Interference

Example 1: To demonstrate the phenomenon of interference, we require two sources which emit radiation of

1)nearly the same frequency

2)the same frequency

3)different wavelength

4)the same frequency and have a definite phase relationship.

Solution:

For the interference phenomenon, two sources should emit radiation of the same frequency and have a definite phase relationship.

Hence, the answer is the option (4).

Example 2: To produce a minimum reflection of wavelength near the middle of the visible spectrum (550nm) how thick should a coating of mgf₂ $(\mu=1.38)$ be vacuum coated on a glass surface?

1) $10^{-10} \mathrm{~m}$
2) $10^{-8} \mathrm{~m}$
3) $10^{-7} \mathrm{~m}$
4) $10^{-11} \mathrm{~m}$

Solution:

Thin Film Interference in reflected light

For normal incidence

For destructive interference

$2 \mu t=(2 n+1) \frac{\lambda}{2}$ where $n=1,2,3 \ldots$.

To produce a minimum reflection, destructive interference should happen

So use $2 \mu t=(2 n+1) \frac{\lambda}{2}$ where $n=1,2,3 \ldots$

at n=0

$\begin{aligned} & 2 \mu t=\frac{\lambda}{2} \\ & t=\frac{\lambda}{4 \mu}=\frac{550 * 10^{-9}}{4 * 1.38}=100 \mathrm{~nm} \\ & t=10^{-7} \mathrm{~m}\end{aligned}$

Hence, the answer is the option (3).

Example 3: What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident on the film is of 750 nm? Assume that the index for the film is $\mu=1.33$ :

1)282 nm

2)70.5 nm

3) 141 nm

4)387 nm

Solution:

Here, $2 \mu \mathrm{t}=\frac{\lambda}{2}$
$
\therefore \quad \mathrm{t}_{\min }=\frac{\lambda}{4 \mu}=141 \mathrm{~nm}
$

Hence, the answer is the option (3).

Example 4: What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident on the film is 750nm? Assume that the index for the film is $\mathrm{n}=1.33$

1)282 nm

2)70.5 nm

3) 141 nm

4)387 nm

Solution:

$\begin{aligned} & 2 \mu \mathrm{f}=\frac{\lambda}{2} \\ & \therefore \quad \mathrm{t}_{\min }=\frac{\lambda}{4 \mu}=141 \mathrm{~nm}\end{aligned}$

Hence, the answer is the option (3).

Example 5: What is the minimum thickness of a soap bubble needed for constructive interference in reflected height, if the light incident on the film is 1500 nm ? Assume that the refractive index for the film is $\mathrm{n}=1.33$ :

1)282 nm

2)70.5 nm

3)282 nm

4)387 nm

Solution:

We have, $2 \mu \mathrm{t}=\frac{\lambda}{2}$ $\Rightarrow$ Minimum thickness, $\mathrm{t}=\frac{\lambda}{4 \mu}=\frac{1500 \times 10^{-9}}{4 \times 1.33}=282 \mathrm{~nm}$

Hence, the answer is the option (3).

Summary

Thin film interference occurs when light waves reflect off the surfaces of a thin film, such as soap bubbles or oil slicks, leading to patterns of bright and dark fringes due to constructive and destructive interference. Constructive interference happens when the path difference between reflected waves is an integer multiple of the wavelength, while destructive interference occurs when the path difference is an odd multiple of half the wavelength. This phenomenon explains the vibrant colours seen in thin films and is utilized in various applications, including coatings and optics. Understanding these principles helps in designing advanced optical devices and materials.