A soap bubble is essentially a very thin film of water (n=...

A soap bubble is essentially a very thin film of water $(n=$ 1.33) surrounded by air. The colors that you see in soap bubbles are produced by interference, much like the colors of dichroic glass. a. Derive an expression for the wavelengths $\lambda_{\mathrm{c}}$ for which constructive interference causes a strong reflection from a soap bubble of thickness $d$ Hint: Think about the reflection phase shifts at both boundaries. b. What visible wavelengths of light are strongly reflected from a 390 -nm-thick soap bubble? What color would such a soap bubble appear to be?

A soap bubble is essentially a very thin film of water $(n=$ 1.33) surrounded by air. The colors that you see in soap bubbles are produced by interference, much like the colors of dichroic glass.
a. Derive an expression for the wavelengths $\lambda_{\mathrm{c}}$ for which constructive interference causes a strong reflection from a soap bubble of thickness $d$ Hint: Think about the reflection phase shifts at both boundaries.
b. What visible wavelengths of light are strongly reflected from a 390 -nm-thick soap bubble? What color would such a soap bubble appear to be?

Key Concepts

Visible Wavelengths and Color Appearance

Once the condition for constructive interference is established, it can be applied to determine which wavelengths in the visible spectrum are strongly reflected for a given film thickness. The detected color of the soap bubble is the dominant color from the reflected wavelengths that constructively interfere. In the practical situation described, calculating the wavelengths leads to the identification of a particular visible color, which is then perceived as the color of the soap bubble.

Constructive Interference Condition

The constructive interference condition in thin film interference is derived by equating the optical path difference between the two reflected beams to an integral multiple of the wavelength, adjusted for any phase shifts due to reflection. For a thin film, this typically involves the film’s thickness, its refractive index, and the wavelength of light in the medium, leading to an expression of the form 2nd plus any necessary phase correction equals an integer multiple of the wavelength. This condition determines the specific wavelengths at which reflected light is reinforced.

Thin Film Interference

This concept describes the phenomenon where light waves reflected from the upper and lower boundaries of a very thin film interfere with each other. In the context of a soap bubble, the film has a thickness that is comparable to the wavelength of visible light, leading to constructive or destructive interference effects which determine which colors are enhanced or diminished in the reflected light.

Reflection Phase Shifts

When light reflects off a boundary between materials of different refractive indices, it can undergo a phase shift. Specifically, when light reflects from a medium of lower refractive index to a higher refractive index—such as from air to the soap bubble's water film—it experiences a half-wavelength (? radian) phase shift. In contrast, reflection from a higher refractive index to a lower refractive index does not produce such a shift. These phase shifts are essential in setting up the conditions for constructive and destructive interference in thin films.

Transcript

00:02 I am writing delta theta equals 2 pi delta x over lambda plus delta theta zero, and that's going to equal m times 2 pi for constructive interference.

00:26 The reasoning for that is because the new amplitude of combined waves is 2a cosine of one half delta phi.

00:47 So when you take this delta phi and you put m2 pi, m2 ,000, 2 pi in for it, then the 1 half and the 2 canceled out, and the cosine of m pi is 1, or negative 1.

01:17 So that's where this comes from.

01:21 Now, i'm going to put pi in here for the phase shift, and the main reason for that is that when we were doing thin film onto glass is higher.

01:56 Again, the refractive index for glass is higher than that of the film.

02:03 But the refractive index for air is lower than that of the film.

02:13 So this needs to be the opposite of what we did for the glass.

02:17 And the opposite of what we did is adding pi.

02:23 Okay, 2 pi.

02:27 It needs to go 2d, it needs to go across the length of the film and then back again for the reflection.

02:38 And then the wavelength within the film is lambda over n.

02:44 And then i'm just going to subtract pi on both sides, and i'm going to get 2m minus 1 times pi.

02:59 2m minus 1 times pi.

03:05 Okay? and i'm just trying to look here.