Problem 12 - Physics

Question

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00)\). A thin film of liquid floats on the gasoline and has a refractive index that is between \(1.00\) and \(1.40\). Light that has a wavelength of \(625 \mathrm{~nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is \(242 \mathrm{~nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

Step-by-Step Solution

Answer

The refractive index of the film is approximately 1.29.

Step 1: Understand the Problem

We need to find the refractive index of a film on gasoline, knowing that constructive interference of light occurs at a minimum thickness of 242 nm. The light wavelength in vacuum is 625 nm, and the refractive index must be between the indices of gasoline (1.40) and air (1.00).

Step 2: Determine Conditions for Constructive Interference

For constructive interference, the condition is that the optical path difference (OPD) should be an integer multiple of the wavelength in the film:\[ 2nt = m \lambda_f \]where \(n\) is the refractive index of the film, \(t\) is the thickness of the film, and \(\lambda_f\) is the wavelength in the film. Since the film gives the minimum thickness for interference to occur, the minimum integer \(m\) is used.

Step 3: Calculate Wavelength in the Film

The wavelength of light in the film, \(\lambda_f\), is related to the wavelength in vacuum, \(\lambda_0\), by the expression:\[ \lambda_f = \frac{\lambda_0}{n} \]where \(n\) is the refractive index we want to find.

Step 4: Utilize Minimum Thickness Condition

Given the minimum nonzero thickness \(t = 242\,\mathrm{nm}\), the simplest case is when \(m = 1\). Plugging this into the equation for constructive interference,\[ 2nt = \lambda_0 \]substitute \(\lambda_0 = 625\,\mathrm{nm}\) and solve for \(n\).

Step 5: Solve for the Refractive Index

Substitute the known values into the simplified equation:\[ n = \frac{\lambda_0}{2t} = \frac{625\,\mathrm{nm}}{2 \times 242\,\mathrm{nm}} \]Calculate to find the refractive index of the thin film.

Step 6: Calculate the Result

Perform the calculation:\[ n = \frac{625}{484} \approx 1.29 \]Thus, the refractive index of the film is approximately 1.29.

Key Concepts

Refractive Index

The refractive index is a crucial property in optics, representing how much light slows down as it passes through a material compared to its speed in a vacuum. This slowing down happens because the oscillating electric field of light interacts with the electric charges in the material.

• A higher refractive index means light travels slower through the material.
• Refractive indices vary between different materials; for example, air has a refractive index of approximately 1.00, while glass can have values around 1.5.

In the given exercise, a thin film has a refractive index between 1.00 (air) and 1.40 (gasoline). Knowing the refractive index is essential for understanding how light behaves as it transitions between different media.

Interference

Interference is a phenomenon that happens when two or more waves overlap and combine. The resulting wave can have greater, lower, or unchanged amplitude, depending on how the waves line up.

• Interference can be constructive or destructive.
• This phenomenon is observed in various settings, from sound waves to light waves.

In optics, light interference is visible as patterns of bright and dark regions, which occur because light waves from different paths recombine. This recombination can amplify or cancel out wave amplitudes, creating visible patterns.

Wavelength

Wavelength is a fundamental property of waves, defining the distance between consecutive wave peaks. It determines the color of light when visible light is considered. In the equation for wavelength, \[ ext{wavelength} = rac{ ext{speed of light}}{ ext{frequency}} \]The wavelength of light changes depending on the medium due to the change in speed. In a vacuum, the given wavelength was 625 nm, but it changes when light passes through materials because its speed is reduced.

• It's crucial to know the medium to calculate the new wavelength using its refractive index.
• This alteration affects the conditions needed to create or observe interference patterns.

Constructive Interference

Constructive interference occurs when waves add together because their peaks align, resulting in a wave of greater amplitude than the individual waves. In optics, constructive interference typically leads to bright regions, as the intensity of light increases.

• The condition for constructive interference in thin films is when the path difference between waves is an integer multiple of the wavelength.
• In the given problem, it is represented by the equation \( 2nt = m \lambda_f \), where \( m \) is an integer.

This means that, for light reflected from the thin film to be bright, the path difference must satisfy this integer condition, which is met with the minimum possible film thickness of 242 nm in the given exercise.