(II) Let's explore why only "thin" layers exhibit thin-film...

(II) Let's explore why only "thin" layers exhibit thin-film interference. Assume a layer of water, sitting atop a flat glass surface, is illuminated from the air above by white light (all wavelengths from 400 $\mathrm{nm}$ to 700 $\mathrm{nm}$ ). Further, assume that the water layer's thickness $t$ is much greater than a micron $(=1000 \mathrm{nm}) ;$ in particular, let $t=200 \mu \mathrm{m} .$ Take the index of refraction for water to be $n=1.33$ for all visible wavelengths. $(a)$ Show that a visible color will be reflected from the water layer if its wavelength is $\lambda=2 n t / m,$ where $m$ is an integer. (b) Show that the two extremes in wavelengths $(400 \mathrm{nm}$ and 700 $\mathrm{nm}$ ) of the incident light are both reflected from the water layer and determine the $m$ -value associated with each. $(c)$ How many other visible wavelengths, besides $\lambda=400 \mathrm{nm}$ and $700 \mathrm{nm},$ are reflected from the "thick" layer of water? (d) How does this explain why such a thick layer does not reflect colorfully, but is white or grey?

(II) Let's explore why only "thin" layers exhibit thin-film
interference. Assume a layer of water, sitting atop a flat
glass surface, is illuminated from the air above by
white light (all wavelengths from 400 $\mathrm{nm}$ to 700 $\mathrm{nm}$ ).
Further, assume that the water layer's thickness $t$ is
much greater than a micron $(=1000 \mathrm{nm}) ;$ in particular,
let $t=200 \mu \mathrm{m} .$ Take the index of refraction for
water to be $n=1.33$ for all visible wavelengths. $(a)$ Show
that a visible color will be reflected from the water layer
if its wavelength is $\lambda=2 n t / m,$ where $m$ is an integer.
(b) Show that the two extremes in wavelengths $(400 \mathrm{nm}$
and 700 $\mathrm{nm}$ ) of the incident light are both reflected from
the water layer and determine the $m$ -value associated
with each. $(c)$ How many other visible wavelengths,
besides $\lambda=400 \mathrm{nm}$ and $700 \mathrm{nm},$ are reflected from the
"thick" layer of water? (d) How does this explain why
such a thick layer does not reflect colorfully, but is white
or grey?

Key Concepts

Thin-film interference

Thin-film interference is a phenomenon that occurs when light waves reflected from different surfaces of a thin material layer combine, resulting in constructive or destructive interference. This interference depends on the phase difference between the reflected rays, which in turn is determined by factors such as the thickness of the film, its refractive index, and the wavelength of the incident light.

Optical path difference

The optical path difference is the extra distance that a light wave travels within a medium compared to another wave traveling in a different path. In thin films, this difference—involving the physical thickness of the layer multiplied by its refractive index—determines the phase shift between the reflected waves, which is essential for predicting how they will interfere.

Constructive interference conditions

Constructive interference occurs when the phase difference between two or more light waves is an integer multiple of 2?, leading to an enhanced reflected intensity at specific wavelengths. In thin films, this is typically described by the condition 2nt = m?, where t is the film thickness, n its refractive index, m an integer, and ? the wavelength in vacuum.

Layer thickness impact

The thickness of the film plays a critical role in determining the interference effects. When the layer is thin, only a few wavelengths constructively interfere to produce vivid color effects, whereas a much thicker layer causes many wavelengths to satisfy the interference condition, leading to a mixture that appears white or grey rather than displaying distinct colors.

Interference order (integer m)

The interference order, represented by the integer m, indicates the number of half-wavelengths (or whole wavelengths, depending on the phase changes at the interfaces) that fit into the effective optical path through the film. This quantization is crucial for understanding which wavelengths will be constructively reflected and how the reflection spectrum is structured.

Spectral range of visible light

The visible spectrum, typically ranging from about 400 nm to 700 nm, is significant in thin-film interference because variations within this range can result in different interference conditions. The interplay of these wavelengths in reflective thin films determines whether the reflected light appears as specific colors or a mix that results in a neutral white or grey tone.

Transcript

00:01 In this question, we are going to explore the interference pattern of white light on a thick layer of water on top of glass in order to learn why is it that only thin films will produce, you know, some pretty rainbow colors with their interference.

00:26 So one of the key things i like to start with is a quick sketch of our situation just indicating that we have a layer of air, and then we're going to have this thick layer of water, it will thick as tea, and then we're going to have a layer of glass.

00:54 And we're asked first to show that a visible color will be reflected from the water layer if its wavelength is lambda equals 2nt divided by m where m is an integer.

01:09 So we need visible colors to constructively interfere.

01:17 And we are going to have a phase shift of half a wavelength at the air water interface because we're moving from a lower index of refraction into a higher index of refraction.

01:39 So we'll pick up a half wavelength here.

01:44 And then similarly, because the index of refraction of glass is going to be greater than that of water, we are going to have another phase shift here.

01:55 So what this tells us is we need the path through the glass to be an integer number of wavelengths to maintain the construct of interference.

02:17 So the fact that we have both reflected rays getting a phase shift means they're going to emerge, well the second one's going to emerge in phase and all automatically constructively interfering with the one reflected from the top layer.

02:33 So the path through glass is an integer number of wavelengths.

02:43 And this now lets us set up the fact that 2t, because we're going to go down and then back up, equals m times our wavelength within the film.

02:58 Well our wavelength within the film is equal to the incident wavelength or wavelength in the vacuum divided by n.

03:08 So that lets us say now that 2t equals m lambda over n.

03:15 And now we can rearrange to say that lambda is going to be 2nt t over m as we were asked to show.

03:27 And now show that the two extremes of wavelengths for visible light, 400 nanometers and 700 nanometers, are both reflected from the water layer.

03:39 Fun.

03:40 So what this means is for our red wavelength of 700 nanometers, let's calculate m for it.

03:53 So all i did was multiply both sides of that equation by m and divide by lambda so that we have 2nt divided by lambda r in this case.

04:05 So we have 2 times our index of refraction of 1 .33.

04:10 We were given a thickness of 200 micrometers, so 200 times 10 to the minus 6 meters.

04:17 And then we have to express our wavelength in meters as well...

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