Question

What is the thinnest film of a coating with \(n\) = 1.42 on glass (\(n\) = 1.52) for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection?

Short answer

The thinnest film thickness is approximately 114.44 nm.

Step-by-step solution

Understand the concept of destructive interference

Destructive interference occurs when two waves combine to produce a resultant wave with reduced amplitude. For thin films, this happens when the path difference between the two reflected waves is equal to half the wavelength of the light in the film.

Set up the condition for destructive interference

The condition for destructive interference for a thin film is given by \(2nt = (m + \frac{1}{2})\frac{\lambda_0}{n_{air}}\), where \(t\) is the thickness of the film, \(n\) is the refractive index of the film, \(\lambda_0\) is the wavelength of light in the air, \(m\) is an integer, and \(n_{air}\) is the refractive index of air (approximately 1). Since we want the thinnest film, we start with \(m = 0\).

Convert the wavelength of light in the film

The wavelength of light in the film, \(\lambda_f\), is related to its wavelength in air, \(\lambda_0 = 650 \text{ nm}\), by \(\lambda_f = \frac{\lambda_0}{n}\), where \(n = 1.42\). Thus, \(\lambda_f = \frac{650 \text{ nm}}{1.42}\).

Calculate the film thickness

Using the condition for destructive interference, \(2nt = \frac{\lambda_0}{2}\), and substituting for \(\lambda_f\), we have:\[2 \times 1.42 \times t = \frac{650 \text{ nm}}{2}\]Solving for \(t\), \[t = \frac{650 \text{ nm}}{2 \times 2 \times 1.42}\].

Solve for the thickness

Calculate \(t\) using the expression from Step 4:\[t = \frac{650}{4 \times 1.42} = \frac{650}{5.68} \approx 114.44 \text{ nm}\]

Verify the calculation

Double-check the calculations and ensure all values, especially the wavelength and refractive indices, match the problem requirements.

Key concepts

Thin Film Interference

Thin film interference is a fascinating phenomenon that occurs when light waves are reflected off the surfaces of a thin layer and interfere with one another. This types of interference can create colorful patterns, like those seen in soap bubbles or on a thin layer of oil on water.
Thin film interference is primarily associated with the varying path lengths that light waves travel. As light encounters the top and bottom surfaces of the film, some of it gets reflected back. These reflected waves can then interfere with each other, depending on their phase differences and the film's thickness.
One of the most important aspects of thin film interference is whether it results in constructive or destructive interference, which depends on the path difference of the waves. Destructive interference, specifically, occurs when the path difference causes the waves to be out of phase, leading to a reduction or cancellation of the light's intensity, often making certain wavelengths less visible.

Refractive Index

The refractive index (often represented by the symbol "n") is a measure of how much a material slows down light passing through it compared to its speed in a vacuum.
It is an important property of materials because it affects how light bends, or refracts, when entering a different medium.

• A higher refractive index means the material slows down light more, causing more bending.
• The refractive index varies with the wavelength of light, which is why different colors bend differently, leading to phenomena such as dispersion.

In the context of thin film interference, knowing the refractive index of each layer is crucial because it determines the phase change upon reflection and helps calculate the wavelength of light within the film. For instance, in this exercise, the film with a refractive index of 1.42 compared to the glass of 1.52 affects the interference pattern seen and must be accounted for in calculations.

Wavelength in Medium

The wavelength of light is the distance between successive crests of a wave. However, when light travels through a medium other than a vacuum (or air, approximately), its wavelength changes depending on the material's refractive index.
This is because the speed of light is slower in a medium than in a vacuum, making the wavelengths shorter.
The formula to find the wavelength of light in a medium is \[\lambda_{medium} = \frac{\lambda_{air}}{n},\]where \(\lambda_{medium}\) is the wavelength in the medium, \(\lambda_{air}\) is the wavelength in air, and \(n\) is the refractive index.
For the problem given, when the red light with an air wavelength of 650 nm enters a film with a refractive index of 1.42, its wavelength changes. This shorter wavelength in the medium plays a critical role in the interference pattern and is used in calculating the thin film's conditions for destructive interference.

Path Difference

Path difference refers to the difference in distance traveled by two waves reflecting off different surfaces of the thin film. This concept is vital in understanding how interference patterns form in thin films.

• If the path difference is a whole number multiple of the wavelength, constructive interference occurs.
• If it's a half number multiple, destructive interference occurs, causing the light to potentially cancel out.

For thin films, the path difference depends on both the thickness of the film and the change in wavelength within the medium. The calculation involves a factor of 2nt, where \(t\) is the film thickness and \(n\) is the refractive index. This factor ensures that all aspects of light behavior, including phase changes due to reflections, are considered, which is essential for accurately predicting destructive interference conditions. In practice, understanding path difference helps in designing coatings with specific reflective or anti-reflective properties, as explored in this exercise.