Problem 28
Question
A plastic film with index of refraction 1.70 is applied to the surface of a car window to increase the reflectivity and thus to keep the car's interior cooler. The window glass has index of refraction 1.52. (a) What minimum thickness is required if light of wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively? (b) Coatings as thin as that calculated in part (a) are difficult to manufacture and install. What is the next greater thickness for which constructive interference will also occur?
Step-by-Step Solution
Verified Answer
Minimum thickness: 80.88 nm; Next thickness: 242.65 nm.
1Step 1: Understanding the Problem
We want to find the thickness of the film that causes constructive interference of reflected light. Constructive interference occurs when the phase difference between the two waves results in them being in phase.
2Step 2: Identify Interference Condition
For constructive interference in a thin film, the path difference must be an integer multiple of the wavelength in the film. However, due to the phase change on reflection, the condition for constructive interference becomes \(2nt = (m + 0.5)\lambda_0\), where \(n\) is the film's refractive index, \(t\) is the thickness, \(m\) is an integer, and \(\lambda_0\) is the wavelength in air.
3Step 3: Calculate Minimum Thickness
Convert the wavelength in air to the wavelength in the film: \(\lambda = \frac{\lambda_0}{n} = \frac{550 \text{ nm}}{1.70}\). Using the condition \(2nt = (m + 0.5)\lambda_0\) with \(m = 0\), solve for \(t\): \(t = \frac{(m + 0.5)\lambda_0}{2n}\). Substituting values, \(t = \frac{(0.5)(550)}{2 \times 1.70} \approx 80.88 \text{ nm}\).
4Step 4: Determine Next Thickness
For the next thickness, use \(m = 1\) in the formula: \(t = \frac{(1.5)(550)}{2 \times 1.70} \approx 242.65 \text{ nm}\).
5Step 5: Conclusion
The minimum thickness for constructive interference is approximately 80.88 nm. The next greater thickness is approximately 242.65 nm.
Key Concepts
Constructive InterferenceIndex of RefractionOptical CoatingsWavelength in Film
Constructive Interference
Constructive interference is a phenomenon that occurs when two or more waves superimpose to produce a new wave of greater amplitude. This happens when the waves are in phase, meaning the crests and troughs of the waves align perfectly. In the context of thin film interference, constructive interference can enhance certain wavelengths of light, making colors more vivid or increasing reflectivity.
In a thin film, light reflects off both the top surface of the film and the bottom surface where it meets the second medium. For constructive interference to occur, the light waves reflecting off these surfaces must be in phase. When the phase difference leads to an integer plus a half integer multiple of the wavelength, the conditions are right for constructive interference, with maxima in reflected light intensity. This is expressed mathematically as:
In a thin film, light reflects off both the top surface of the film and the bottom surface where it meets the second medium. For constructive interference to occur, the light waves reflecting off these surfaces must be in phase. When the phase difference leads to an integer plus a half integer multiple of the wavelength, the conditions are right for constructive interference, with maxima in reflected light intensity. This is expressed mathematically as:
- The condition is given by: \(2nt = (m + 0.5)\lambda_0\)
- Here, \(n\) is the index of refraction of the film, \(t\) is the thickness of the film, and \(\lambda_0\) is the wavelength of light in air.
- \(m\) is an integer representing the order of interference.
Index of Refraction
The index of refraction is a number that describes how fast light travels through a material. It is denoted by \(n\) and is calculated as the speed of light in a vacuum divided by the speed of light in the material. When light enters a medium from air, it slows down, and this is characterized by the index of refraction.
For optical coatings, such as those on a car window, knowing the index of refraction is crucial. It affects how light passes through and reflects off surfaces. The index of refraction determines:
For optical coatings, such as those on a car window, knowing the index of refraction is crucial. It affects how light passes through and reflects off surfaces. The index of refraction determines:
- How much the light bends as it enters the new medium.
- The proportion of light that is reflected vs. transmitted when encountering the boundary.
- The wavelength inside the material: \(\lambda = \frac{\lambda_0}{n}\), where \(\lambda_0\) is the wavelength in air.
Optical Coatings
Optical coatings are thin layers of material applied to surfaces like lenses or windows to control light reflection and transmission. These films can be engineered to enhance or reduce reflections, making them essential in applications like glasses, camera lenses, and automotive windows.
Optical coatings function by causing interference of specific wavelengths of light. By carefully designing the thickness and material of these films, manufacturers can achieve desired optical properties.
Some key aspects include:
Optical coatings function by causing interference of specific wavelengths of light. By carefully designing the thickness and material of these films, manufacturers can achieve desired optical properties.
Some key aspects include:
- Reducing glare by minimizing reflections on surfaces.
- Increasing reflectivity to keep interiors cooler by reflecting sunlight.
- Enhancing colors by maximizing constructive interference at specific wavelengths.
Wavelength in Film
The wavelength of light changes when it moves from one medium to another, depending on the refractive index of the new medium. The wavelength in a film is shorter than in air if the index of refraction is greater than 1. This is because light slows down, leading to compression of the wave crests.
For a film with an index of refraction \(n\), the wavelength \(\lambda\) inside the film is related to the wavelength in air \(\lambda_0\) by:
For a film with an index of refraction \(n\), the wavelength \(\lambda\) inside the film is related to the wavelength in air \(\lambda_0\) by:
- \(\lambda = \frac{\lambda_0}{n}\)
Other exercises in this chapter
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