Problem 30 - University Physics with Modern Physics

Question

A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction \(1.52 .\) (a) What minimum thickness is required if light with wavelength 550 \(\mathrm{nm}\) in air reflected from the two sides of the film is to interfere constructively? (b) It is found to be difficult to manufacture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference?

Step-by-Step Solution

Answer

(a) Minimum thickness is 74.32 nm. (b) Next greatest thickness is 223 nm.

Step 1: Understanding the Problem

We need to determine the thickness of a thin plastic film that causes constructive interference of light reflecting from its surfaces. The index of refraction for the film is 1.85 and for the glass is 1.52. We need to use the information that the wavelength of light in air is 550 nm. We will use the condition for constructive interference in thin films.

Step 2: Condition for Constructive Interference

Constructive interference occurs when the path difference between the two waves leads to an interference of type \(m \lambda\), where \(m\) is any non-negative integer (0, 1, 2,...). For thin films, constructive interference occurs at: \(2nt = (m + \frac{1}{2}) \lambda_0\), where \(n\) is the index of refraction, \(t\) is the film thickness, and \(\lambda_0\) is the wavelength in air.

Step 3: Calculate Wavelength in the Film

Calculate the wavelength of light inside the film using \(\lambda = \frac{\lambda_0}{n}\). Substitute \(\lambda_0 = 550\ \text{nm}\) and \(n = 1.85\):\[\lambda = \frac{550\ \text{nm}}{1.85} \approx 297.30\ \text{nm}\]

Step 4: Finding the Minimum Thickness for Constructive Interference

Substitute the wavelength in the film into the constructive interference condition equation. For \(m = 0\), we solve: \(2nt = (0 + \frac{1}{2}) \lambda_0\). Rewriting and solving for \(t\), we get: \(t = \frac{\lambda}{4} = \frac{297.30\ \text{nm}}{4} = 74.32\ \text{nm}\).

Step 5: Finding the Next Greatest Thickness

For the next thickness, use \(m = 1\), so the equation becomes \(2nt = (1 + \frac{1}{2}) \lambda_0\). Rewriting, \(t = \frac{3 \lambda}{4} = \frac{3 \times 297.30\ \text{nm}}{4} \approx 223\ \text{nm}\).

Key Concepts

Index of Refraction

The index of refraction, often symbolized by the letter 'n', is a critical concept in understanding how light travels through different materials. When light passes from one medium to another, its speed and direction can change. This is because different materials have different optical densities, which affect the speed of light traveling through them.

• If a material has a higher index of refraction, it means that light travels slower through it compared to a material with a lower index.
• In the problem, the plastic film has an index of refraction of 1.85, while the glass has a lower index of 1.52, showing that light slows down more in the plastic film than in the glass.
• Knowing the index of refraction allows us to calculate the speed and bending (or refraction) of light as it moves from air into the film or glass.

This concept is very important in predicting how light will behave as it transitions between different layers, which is fundamental for applications like reducing glare through anti-reflective coatings.

Constructive Interference

Constructive interference is a phenomenon that occurs when two or more light waves overlap in such a way that they combine to form a brighter light. This happens when the path difference between the waves is a whole number multiple of the wavelength plus a half.

• In the context of thin films, such as the plastic film coating on a car window, constructive interference is used to increase reflectivity.
• For constructive interference to occur, the path difference must satisfy the formula: \(2nt = (m + \frac{1}{2}) \lambda_0\), where \(n\) is the index of refraction of the film, \(t\) is the thickness, \(m\) is a non-negative integer, and \(\lambda_0\) is the wavelength in air.
• This is why manipulating the thickness of the film can control whether the light waves reinforce each other (constructive interference) or cancel each other out (destructive interference).

Understanding how to achieve constructive interference in thin films can help in designing coatings that enhance brightness or reduce unwanted glare.

Wavelength of Light

The wavelength of light is a fundamental property that describes the distance between consecutive peaks of a wave. Measured in nanometers (nm), it is critical for understanding how light interacts with different materials.

• The exercise provided a wavelength of 550 nm in air, which is typical for visible light.
• When light enters a medium like the plastic film, its wavelength changes according to the material's index of refraction. This new wavelength in the material is calculated as \(\lambda = \frac{\lambda_0}{n}\), where \(\lambda_0\) is the wavelength in air and \(n\) is the medium's index of refraction.
• For the plastic film with \(n = 1.85\), the wavelength becomes approximately 297.30 nm within the material.

This change in wavelength is crucial because it affects how light will interfere inside the medium, influencing the optical properties like reflectivity and color of observed surfaces.