Problem 17 - Physics

Question

Consult Interactive Solution \(\underline{27} .17\) at to review a model for solving this problem. 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 film will appear bright due to constructive interference at 640 nm.

Step 1: Understand the Concept of Interference

Interference of light occurs when two or more waves overlap, leading to regions of constructive and destructive interference. Constructive interference leads to brightness, while destructive interference leads to darkness. For thin films, these conditions are affected by the thickness of the film and the wavelength of light.

Step 2: Review the Condition for Destructive Interference

For destructive interference in thin films, we use the formula for the film's minimum thickness that results in a dark appearance: \( 2nt = (m+\frac{1}{2})\lambda \). Given that \( \lambda = 640 \ \mathrm{nm} \) results in darkness, we know \( = 1 \) \( t \) corresponds to the minimum nonzero thickness.

Step 3: Determine Minimum Thickness Corresponding to Darkness

Using the condition for destructive interference, solve for thickness, \( t \): \( 2nt = (m+\frac{1}{2})\lambda \).Given that the wavelength in air \( \lambda = 640 \ \mathrm{nm} \), the refractive index of oil \( n \) (higher than water), and \( m = 0 \) since it's the minimal nonzero thickness, we calculate \( t = \frac{(0 + \frac{1}{2}) \cdot 640}{2n} = \frac{320}{n} \).

Step 4: Identify Condition for Constructive Interference

Constructive interference for the thin film occurs when \( 2nt = m\lambda' \), where \( m \) is an integer and \( \lambda' \) is the wavelength in air. Substitute \( t = \frac{320}{n} \) to solve for \( \lambda' \).

Step 5: Solve for Wavelengths of Constructive Interference

Using \( t = \frac{320}{n} \) in the constructive formula: \[ 2n \left(\frac{320}{n}\right) = m\lambda' \Rightarrow 640 = m\lambda' \Rightarrow \lambda' = \frac{640}{m} \]. Identify values of \( m \) that ensure \( \lambda' \) is within the visible spectrum (380 to 750 nm).

Step 6: Check Values of m for Visible Spectrum

Calculate possible \( m \) values:1. \( m = 1 \Rightarrow \lambda' = 640/1 = 640 \ \mathrm{nm} \) (within visible spectrum, already considered)2. \( m = 2 \Rightarrow \lambda' = 640/2 = 320 \ \mathrm{nm} \) (not visible)3. \( m = 3 \Rightarrow \lambda' = 640/3 \approx 213.3 \ \mathrm{nm} \) (not visible)4. Continue until finding \( m \) giving \( \lambda' \leq 750 \ \mathrm{nm} \). Only consider visible values.

Key Concepts

Constructive Interference

When light waves meet, they can combine or interfere in various ways. Constructive interference occurs when those waves align with each other in such a manner that their crests (or peaks) overlap, leading to a reinforcement of their combined intensity and manifesting as a brighter region.
In thin film interference, like the oil film on water, the condition for constructive interference is that the thickness of the film allows the outgoing waves to constructively interfere after reflection. For this to happen, the effective path difference between the two interfering waves needs to be a whole multiple of the wavelength of the light, expressed mathematically as:

• \[ 2nt = m \lambda' \]

where:

• \( n \) is the refractive index of the medium,
• \( t \) is the thickness of the film, and,
• \( \lambda' \) is the wavelength of light in air.

This means the film appears bright at specific wavelengths due to constructive interference.

Destructive Interference

Destructive interference is the phenomenon that occurs when two light waves overlap in a way that their crests and troughs are out of step, effectively canceling each other out. This results in a darker appearance due to reduced light intensity in that region.
In the context of thin films, destructive interference depends on the thickness of the film, the wavelength of the incident light, and the refractive index of the film. For minimum nonzero thickness leading to darkness, the condition for destructive interference can be modeled as:

• \[ 2nt = (m + \frac{1}{2})\lambda \]

where:

• \( m \) is an integer,
• \( \lambda \) is the wavelength in vacuum,

When these conditions are satisfied, waves emerging from the film interfere destructively, leading to a darkened appearance at those specific wavelengths.

Visible Spectrum

The visible spectrum is the small portion of the electromagnetic spectrum that is visible to the human eye, spanning approximately from 380 nm to 750 nm. This range includes all the colors that we typically see, from violet to red. The perception of an oil film's color is due to the light waves in this visible spectrum region being selectively interfered with.
When performing calculations for the wavelengths at which an oil film appears to be bright or dark, it is essential to ensure that the wavelengths fall within the visible spectrum range. Any wavelength that results from constructive interference but lies outside of this range will not contribute to the observable brightness and thus needs to be disregarded for the purpose of visible observation. When determining the wavelengths of light for constructive interference, it is constrained between 380 nm and 750 nm to ensure the colors can be seen.

Refractive Index

The refractive index, denoted as \( n \), is a dimensionless number that describes how light propagates through a medium compared to the vacuum of space. It illustrates how much the light slows down when passing through different materials. A higher refractive index means that light travels slower through that medium.
In thin film interference, the refractive index plays a crucial role in determining the path difference and the effective wavelength of light inside the film. For a film of oil on water, since the refractive index of oil is greater than that of water, light traveling through the oil experiences a decrease in speed, thus modifying the interference conditions. Calculating the phase changes and the condition for both destructive and constructive interference requires using the refractive index to factor how much the light "bends" or slows down within the film. This ultimately affects the color the film reflects, contributing to the rich and beautiful patterns observed in oil films.