Problem 10 - Physics

Question

A mix of red light \(\left(\lambda_{\text {vacuum }}=661 \mathrm{nm}\right)\) and green light \(\left(\lambda_{\text {vacuum }}=551 \mathrm{nm}\right)\) is directed perpendicularly onto a soap film \((n=1.33)\) that has air on either side. What is the minimum nonzero thickness of the film, so that destructive interference causes it to look red in reflected light?

Step-by-Step Solution

Answer

The film thickness for destructive interference of red light is approximately 124.25 nm.

Step 1: Understanding Destructive Interference

For destructive interference in reflected light, the path difference between the two reflected waves should be half the wavelength in the film's medium. The condition for this is given by \(2t = \frac{m+0.5}{n}\lambda\), where \(t\) is the thickness, \(n\) is the refractive index of the film, and \(m\) is an integer representing the order of interference. We'll start with \(m=0\) for the minimum nonzero thickness.

Step 2: Calculate Wavelength in the Film for Red Light

The wavelength in the medium is given by \(\lambda_{\text{film}} = \frac{\lambda_{\text{vacuum}}}{n}\). For red light, \(\lambda_{\text{film (red)}} = \frac{661\, \text{nm}}{1.33} \approx 496.99\, \text{nm}\).

Step 3: Apply Minimum Thickness Condition for Red Light

For the minimum thickness, we use the condition \(2t = \frac{(0+0.5)}{1.33} \lambda_{\text{vacuum}}\). Substituting the values, we get \(2t = \frac{0.5 \times 661\, \text{nm}}{1.33}\approx 248.5\, \text{nm}\). Solving for \(t\) gives \(t \approx 124.25\, \text{nm}\).

Step 4: Verify With Green Light

Although the question asks for the condition on red light, to ensure no green light is destructively interfering, we check if \(t\) causes constructive interference with green light. For green, \(\lambda_{\text{film (green)}} = \frac{551 \text{nm}}{1.33} \approx 414.3 \text{nm}.\) Verify \(2t eq m \lambda_{\text{film (green)}}\) for integer \(m\), preventing destructive interference, confirming our calculation is correct.

Key Concepts

Destructive Interference

Destructive interference occurs when two waves combine to form a wave with a smaller amplitude, effectively "canceling" each other out. This fascinating phenomenon happens when the crest of one wave meets the trough of another, leading to a decrease in overall wave amplitude.

For waves interacting with thin films, like a soap film in our problem, the conditions for destructive interference are specifically defined by the difference in the path. The path difference should equal an odd multiple of half the wavelength. This is mathematically expressed as:

• \(2t = \frac{m+0.5}{n}\lambda\)

Here, \(t\) is the thickness of the film, \(n\) is the refractive index, and \(\lambda\) is the wavelength in vacuum. Opting for the minimum thickness, we usually set \(m = 0\). This lets us find the smallest film thickness for which destructive interference will make the film appear a specific color when viewed in reflected light.

Wavelength

Wavelength is a critical aspect of wave behavior in optics. It's the distance over which a wave's shape repeats, measured from crest to crest. In the context of light waves, the wavelength influences color; longer wavelengths are associated with red light, while shorter ones pertain to blue or violet.

In a medium different than a vacuum, the wavelength changes due to the medium's refractive index. The wavelength within the film is computed by:

• \(\lambda_{\text{film}} = \frac{\lambda_{\text{vacuum}}}{n}\)

For example, as seen in our exercise, the wavelength of red light in the soap film changes from 661 nm to approximately 496.99 nm due to the film's refractive index of 1.33. Understanding how wavelengths change in different media is vital for predicting how light will interfere when reflected or transmitted.

Refractive Index

Refractive index is a property that defines how fast light travels through a medium. It's denoted by \(n\), and is the ratio of the speed of light in a vacuum to its speed in the medium. In our case, the soap film has a refractive index of 1.33.

When light enters a medium like a soap film, its speed decreases due to this refractive index, affecting wavelengths and path lengths. The new wavelength within the medium is determined as:

• \(\lambda_{\text{film}} = \frac{\lambda_{\text{vacuum}}}{n} \)

This slower speed and altered wavelength are key in determining interference patterns. Hence, the refractive index is central to calculating whether we see constructive or destructive interference in reflected light.

Wave Optics

Wave optics, also known as physical optics, is the study of light as a wave phenomenon. Unlike ray optics that simplify light's nature into straight lines, wave optics embraces the complexities of interference, diffraction, and polarization.

In wave optics, light is an electromagnetic wave, characterized by its wavelength and frequency. The study extends to understanding how these waves superimpose and interfere. Key concepts include:

• Interference of waves — where they can add up (constructive) or cancel out (destructive).
• Diffraction — the bending of waves around obstacles.
• Polarization — the directionality of wave vibrations.

Wave optics provides the framework to solve problems like our soap film example, where interference between reflected waves determines the colors seen. A comprehensive understanding of wave optics is crucial for tackling various phenomena observed with thin films and lenses.