Problem 17 - Physics

Question

A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light (wavelength \(=640.0 \mathrm{~nm}\) in vacuum). Assuming that the visible spectrum extends from 380 to \(750 \mathrm{~nm}\), for which visible wavelength(s) (in vacuum) will the film appear bright due to constructive interference?

Step-by-Step Solution

Answer

The wavelengths are approximately 480 nm and 960 nm, but only 480 nm is visible.

Step 1: Understanding the Problem

We need to determine the wavelengths for which the oil film appears bright due to constructive interference. Given the film appears dark for red light (640 nm) due to destructive interference, we are asked to find the conditions for constructive interference for other visible wavelengths.

Step 2: Condition for Destructive Interference

The condition for destructive interference at the minimum thickness for the oil film is given by: \(2nt = (m + \frac{1}{2}) \lambda\), where \(n\) is the refractive index of oil, \(t\) is the thickness, and \(\lambda\) is the wavelength in a vacuum. For the given condition, \(\lambda = 640\ nm\).

Step 3: Condition for Constructive Interference

The condition for constructive interference is \(2nt = m \lambda\). We use the fact that both constructive and destructive interference must be occurring for the same thickness \(t\), allowing us to relate this to the previous equation. When substituting, the factor \(\frac{1}{2}\) difference helps us find other wavelengths.

Step 4: Calculating Thickness

Using \(2nt = (m + \frac{1}{2}) 640\), solve for thickness for the first order (\(m=0\)). This sets the instance for constructing the solution.

Step 5: Finding Wavelength for Constructive Interference

Use \(2nt = k \lambda_{bright}\). From the thickness found and knowing \(\lambda_{bright}\), identify integers \(k\) that satisfy this equation within the visible spectrum \(380-750\) nm.

Step 6: Integers for Constructive Interference

Calculate possible integer values for \(k\) from the equation: \( \lambda_{bright} = \frac{640 \times 2}{2m + 1} \), which needs evaluation for \(m=0, 1, 2, \text{etc.}\), constrained by 380 nm ≤ \(\lambda_{bright}\) ≤ 750 nm, to identify visible wavelengths satisfying constructive interference.

Key Concepts

Destructive Interference

Destructive interference in thin films occurs when waves combine to cancel each other out. This happens because of the difference in the path length traveled by the light waves within the film. In specific terms, waves are out of phase by half a wavelength, leading to cancellation. In the case of the oil film on water, destructive interference makes the film appear dark for red light of wavelength 640 nm. The formula that governs this phenomenon is \(2nt = (m + \frac{1}{2}) \lambda \), where:

• \(n\) is the refractive index of the film
• \(t\) is the thickness of the film
• \(\lambda\) is the wavelength in vacuum
• \(m\) is an integer (called the order of interference)

This condition specifies that the film thickness and wavelength are just at the right values to let light waves traveling in opposite directions cancel each other.

Constructive Interference

Constructive interference occurs when waves add together, enhancing each other's effects. In thin films, constructive interference causes certain colors to stand out, making the film appear bright. This happens when the waves remain in phase, thereby reinforcing each other.
The equation for constructive interference is \(2nt = m \lambda\), where the terms maintain their previous meanings:

• \(2nt\) involves the film's thickness and refractive index, as it describes the path difference
• \(\lambda\) is the wavelength of light in vacuum
• \(m\) is an integer that represents interference order

For constructive interference, the thickness \(t\) must satisfy wavelengths that reinforce one another. It stands in contrast with the destructive interference condition, focusing on where waves align perfectly in phase. For this particular exercise, we find other visible wavelengths that meet this formula, providing the film's colorful appearance.

Wavelength

Wavelength refers to the distance between two consecutive peaks (or troughs) of a wave. In the context of this exercise, we work with the visible spectrum range from 380 nm to 750 nm.
When light waves reflect off surfaces with varying thickness like a thin film, different wavelengths will experience different degrees of interference. Since each wavelength corresponds to a color, this interference creates a spectrum of colors seen—like an oil slick on pavement.
For the given problem, the 640 nm wavelength correlates with red light, which appears dark due to destructive interference. The challenge lies in pinpointing other wavelengths within the 380 nm to 750 nm range that meet the conditions for constructive interference, resulting in a bright appearance.

Refractive Index

The refractive index, denoted as \(n\), is a measure of how much a medium slows down light compared to a vacuum. It tells us how light bends as it passes through different materials.
In this exercise, we use the refractive index to calculate how light waves interact with the oil film. This interaction alters wavelengths for constructive or destructive interference. Specifically, the refractive index let us calculate the correct film thickness to achieve different interference patterns.
The refractive index impacts the path difference as light reflects off surfaces, affecting how the light waves add or cancel each other. The refractive index is crucial in determining the oils' optical behavior on wet pavement, contributing to the color patterns observed in this exercise.