Problem 61 - Physics

Question

(a) What, if any, phase change occurs when light, traveling in air, reflects from the interface between the air and a soap film \((n=1.33) ?\) (b) What, if any, phase change occurs when light, traveling in a soap film, reflects from the interface between the soap film and a glass plate \((n=1.52) ?(\mathrm{c})\) Is the wavelength of the light in a soap film greater than, smaller than, or equal to the wavelength in a vacuum? A soap film \((n=1.33)\) is \(465 \mathrm{~nm}\) thick and lies on a glass plate \((n=1.52)\). Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{~nm}\), travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?

Step-by-Step Solution

Answer

Phase changes occur with both reflections. Wavelength in soap film is smaller. Destructive interference occurs for 372 nm, causing darkness in reflected light.

Step 1: Understand Phase Change on Reflection (Air to Soap Film)

In optics, when light reflects from a medium with a lower refractive index (like air at \(n=1\)) to a medium with a higher refractive index (such as the soap film at \(n=1.33\)), a phase change of \(\pi\) (or 180 degrees) occurs.

Step 2: Phase Change on Reflection (Soap Film to Glass)

When light reflects from a medium with a refractive index of \(n=1.33\) (soap film) to a medium with a higher refractive index of \(n=1.52\) (glass), another phase change of \(\pi\) occurs.

Step 3: Wavelength in Soap Film

The wavelength of light in a medium is given by \(\lambda = \frac{\lambda_0}{n}\) where \(\lambda_0\) is the wavelength in vacuum, and \(n\) is the refractive index of the medium. Since \(n=1.33\) for the soap film, the wavelength in the film is smaller than in vacuum.

Step 4: Condition for Destructive Interference

For destructive interference in thin films, the condition \(2nt = (m+\frac{1}{2})\lambda\) is used, where \(t\) is the thickness of the film, \(n\) is the refractive index, and \(m = 0, 1, 2, \ldots\). The phase change from both reflections causes destructive interference for odd multiples of half the wavelength inside the film.

Step 5: Calculate Specific Wavelengths

Given the film thickness \(t = 465\, \mathrm{nm}\) and refractive index \(n = 1.33\), solve \(2(1.33)(465) = (m+\frac{1}{2})\lambda\) for \(\lambda\), staying within the range of 380 - 750 nm. Solve this equation for different values of \(m\) until values within this wavelength range are found.

Key Concepts

Phase Change in Optics

When light encounters a boundary between two different media, it can be reflected, transmitted, or both. However, the behavior of light during reflection at such an interface is very specific. When light travels from a medium with a lower refractive index to one with a higher refractive index, a phase change of \(\pi\) (equivalent to a 180-degree shift) occurs.
A classic example of this is when light reflects from air (with a refractive index \(n = 1\)) into a soap film (with a refractive index \(n = 1.33\)). In this situation, the light undergoes a phase shift of \(\pi\).
On the contrary, if the light were reflecting from a higher index medium to a lower index one, like a soap film to air, no phase change would take place.

To consider another scenario, when light reflects off a soap film and enters glass \((n = 1.52)\), the greater refractive index difference once again results in a \(\pi\) phase change.
The principle of phase change is vital in understanding phenomena such as thin film interference, where these shifts in phase interfere with waves reflecting from other surfaces.

Thin Film Interference

Thin film interference occurs due to light waves reflecting from the top and bottom surfaces of a thin layer. These reflections can interfere with each other, enhancing or reducing certain wavelengths upon reflection.
For this interference to occur perceptibly, the thickness of the thin film must be comparable to the wavelength of incident light.

The interference is called "destructive" when the waves are out of phase, leading to cancellation (dark spots). This happens when the condition \[2nt = (m+\frac{1}{2})\lambda\] is met. Here, \(t\) is the thickness of the thin film, \(n\) is the refractive index, and \(\lambda\) is the wavelength of light within the film. The variable \(m\) is an integer.
"Constructive interference" results in enhancement (bright spots), occurring when the waves are in phase.

Thin film interference is responsible for the colorful patterns seen in soap bubbles, oil slicks, and more. It's a fascinating optical phenomenon that's richly full of colors depending on the viewing angle and lighting conditions.

Wavelength in Medium

The wavelength of light changes when it enters a medium other than a vacuum. As light enters a medium with a refractive index \(n\), its speed is reduced, impacting its wavelength but not its frequency. The relationship is given by: \[\lambda = \frac{\lambda_0}{n}\] where \(\lambda_0\) is the original wavelength in a vacuum and \(\lambda\) is the wavelength in the medium.
This means that light entering a medium with a refractive index greater than 1 (like water, glass, or a soap film) will have a shorter wavelength compared to its wavelength in a vacuum.

In our example, the wavelength in the soap film (with \(n = 1.33\)) would be shorter than the same wavelength in air or vacuum. Understanding this concept is crucial because it affects how light interacts within the medium, leading to phenomena like refraction and interference.
Remember, while the wavelength changes, the frequency of light remains constant across different media. This invariant frequency is a fundamental property of wave behavior in physics.