Problem 32 - University Physics with Modern Physics

Question

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 nm? The index of refraction of the film is 1.33, and there is air on both sides of the film.

Step-by-Step Solution

Answer

The thinnest soap film is approximately 90.23 nm thick.

Step 1: Understanding the problem

We are asked to find the thinnest thickness of a soap film that appears black when light of wavelength 480 nm is shone upon it. The film has an index of refraction of 1.33 and is surrounded by air.

Step 2: Conceptualizing Thin Film Interference

For a film to appear black, destructive interference occurs at the reflected light. For a soap film, the phase change upon reflection happens at both interfaces: when light enters and exits the film, since the index of refraction is higher than the air.

Step 3: Applying the Condition for Destructive Interference

Destructive interference for light reflected from thin films occurs when the optical path difference is a half integer multiple of the wavelength: \[ 2nt = (m+\frac{1}{2})\lambda \]where \( n \) is the refractive index, \( t \) is the thickness, and \( \lambda \) is the wavelength of the light in a vacuum.

Step 4: Determining the Appropriate Order of Destructive Interference

To find the thinnest film, we want the smallest non-zero thickness, i.e., the first-order minimum reflection \( (m=0) \).

Step 5: Substituting Values Into the Formula

For \( m = 0 \), we have: \[ 2(1.33)t = (0+\frac{1}{2})(480 \, \text{nm}) \]\[ 2.66t = 240 \, \text{nm} \]

Step 6: Solving for Thickness \( t \)

Solving for \( t \) gives: \[ t = \frac{240}{2.66} \, \text{nm} \]Thus, by calculating, we get: \[ t \approx 90.23 \, \text{nm} \]

Key Concepts

Destructive Interference

Destructive interference is a fascinating phenomenon that occurs when two or more waves overlap in such a way that they cancel each other out. In the context of thin films, like a soap film, destructive interference happens when the path difference between the reflecting waves results in a phase shift that leads them to cancel.
When light waves strike a thin film, some of the light reflects off the top surface and some transmits through, reflects off the bottom surface, and then comes back through the film to interfere with the light reflecting from the top surface.

• For destructive interference, the waves must combine 180 degrees out of phase, effectively
• This creates a dark spot, such as when the soap film appears black under specific conditions

To achieve this setup in our problem, the light must undergo a path difference characterized by an integer multiple of half wavelengths, leading to the cancellation of the light: \[ 2nt = (m+\frac{1}{2})\lambda \] Here, \( n \) is the refractive index of the film, \( t \) is the film's thickness, and \( \lambda \) is the light's wavelength.

Optical Path Difference

Optical path difference plays a key role in determining interference patterns in thin films. It refers to the difference in the paths traveled by two light waves reflecting internally and externally from a thin film.
When light enters from air into a medium like a soap film, it experiences a change because the film has a higher index of refraction.

• The amount of path the light "seems" to travel is not only the physical distance but is modified by the refractive index
• This modification is essential for interference, whether constructive or destructive

In our scenario with the black-appearing soap film, the optical path difference needs to ensure destructive interference. This means the optical path length related to the thickness and refractive index leads to the cancellation of certain light wavelengths: \[ 2nt = (m+\frac{1}{2})\lambda \]This formula captures the essence of how thickness \( t \), the refractive index \( n \), and the wavelength \( \lambda \) interact to produce the color or lack thereof perceived in thin films.

Index of Refraction

The index of refraction, often simply called refractive index, is a measure of how much the speed of light is reduced inside a medium compared to vacuum. It is a critical component in understanding how light behaves in different materials.
For instance, in our problem, the index of refraction for the soap film is 1.33, which means light travels slower in the film than in air.

• This slower speed results in bending or refracting the light, a key mechanism in interference phenomena
• The index directly contributes to the optical path difference responsible for interference patterns

In the formula for destructive interference, the index \( n \) amplifies the actual distance light travels in the medium, impacting how light wavelengths interfere after reflecting off the film's surfaces. An index of refraction greater than 1 indicates a medium where light experiences a significant level of slowing down and characteristic behavior change. Thus, understanding the precise value of this index is vital for calculating the effect of thin films on light interference.