Video: Film Thickness Required for Destructive Interference

A thin film with a refractive index of 1.32 is surrounded by air. What is the minimum thickness of this film that minimizes the reflection of normally incident light of wavelength 500 nm?

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Video Transcript

A thin film with a refractive index of 1.32 is surrounded by air. What is the minimum thickness of this film that minimizes the reflection of normally incident light of wavelength 500 nanometers?

We can start by highlighting some of the important information weโ€™ve been given. Weโ€™re told that the index of refraction of the film is 1.32; weโ€™ll call that ๐‘›. Weโ€™re also told that the wavelength of light involved is 500 nanometers, which weโ€™ll call ๐œ†. We want to know the smallest thickness of the film that minimizes the reflection of normally incident light; weโ€™ll call this thickness ๐‘‘.

As we begin our solution, letโ€™s draw a diagram of whatโ€™s going on. In this example, we have light of wavelength ๐œ† normally or perpendicularly incident on our thin film ๐‘›. When the light hits the surface of the film, some of it reflects off the film and bounces back with a 180-degree phase shift and some transmits into the film, reflects off of the interface at the bottom of the film, travels back upward, and then leaves the film back into the air.

Weโ€™re told we want to minimize the amount of light that reflects back off of the interface. That means that we want to use destructive interference to make that reflected light minimum. For thin films, there is a relationship describing the mathematics of destructive interference. That relationship says that two times the thin film thickness ๐‘‘ times its refractive index ๐‘› is equal to ๐‘š times the wavelength of the light moving through the film, ๐œ†, where ๐‘š is an integer value.

When we apply this relationship to our scenario, weโ€™re told we want to solve for the minimum thickness of the film that allows destructive interference to happen. That means weโ€™ll let ๐‘š be equal to one. When we rearrange to solve for ๐‘‘, the thin film thickness, we find that is equal to ๐œ† divided by two ๐‘›.

We can now plug in the given values for ๐œ† and ๐‘›, being careful to convert ๐œ† into units of meters first. When we calculate this fraction, we find that ๐‘‘ is equal to 189 nanometers. Thatโ€™s the smallest thickness of this thin film that will minimize the reflection of the light off of the air-and-film interface.

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