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Chapter 35: Problem 28

35.28. Nonglare Glass. When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. (a) If the glass has a refractive index of 1.62 and you use \(\mathrm{TiO}_{2}\) , which has an index of refraction of \(2.62,\) as the coating, what is the minimum film thickness that will cancel light of wavelength 505 \(\mathrm{nm}\) ? (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.

Short Answer

Expert verified
Minimum thickness is about 96.6 nm. Other thinness: ko;n br

Step by step solution

01

Understanding the Problem

We need to find the thickness of a thin film of TIO2 that coats the glass to minimize glare. This involves optics, specifically understanding how light interferes when it reflects off surfaces.
02

Refractive Indices Identification

Given that the refractive index of glass $value is, _{glass} = 1.62$ and the refractive index of TiO2 is __2.62. We work with the reference wavelength ($n_{film} =2.62$). The wavelength ($lambda = 505 nm$) to determine the film's thickness for constructive interference.
03

Calculate Minimum Film Thickness

For maximum light intensity cancellation, the equation notice needs refraction registration: simples 2nt = (m+1/2)_lambda_). Simplifying: t = (m+1/2)lambda_/) for minimal thickness cancellation. i isn_fil_/) For \(minimonee m= fihs^mo\)ing,\(t = [505 nm/(2x2.62)]\) Finally, solving adjustsments round.
04

Calculate Alternative Thicknesses

The thickness calculations adjust the range as more intensities lose lockreflection, not improving incident f indic confirmation ray results (film change patternopard). m values one greater and two greater set. Stand = 1^{st2} (1+1/2x505/{2x2.62})=0.8993piersn to:#346 thicks terms not infinity him specific finish fix x eaiserite,

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thin Film Interference
Thin film interference is a fascinating optical phenomenon that occurs when light waves reflect off the top and bottom surfaces of a film, like an oil surface on water or a coated piece of glass. When these reflected light waves interact with each other, they can either amplify or cancel each other out, creating colorful patterns.
In this context, using thin film interference to reduce glare involves coating a surface with a thin layer that causes destructive interference. This means the reflected light waves cancel each other out, reducing the light that's reflected back to the viewer's eyes and minimizing glare.
To achieve this destructive interference, the film needs to be of a specific thickness relative to the light's wavelength. Experimenting with the correct thickness ensures the reflected light waves are perfectly out of phase, effectively reducing the glare.
Refractive Index
The refractive index is a measure of how much a material slows down light passing through it compared to light passing through a vacuum. It's a fundamental property of optics and crucial for understanding how light behaves in different materials.
When dealing with thin film interference, the refractive index ( _{film} ) of both the film and the underlying material are important. In the given problem, the refractive index of the glass is 1.62, while the refractive index of the TiO2 film is 2.62. This implies that light travels slower in the film than in the glass.
The difference in refractive indices is what causes the reflected light waves to interfere with each other. By carefully selecting materials with the correct refractive indices, we can optimize the coating for minimal glare. The contrast between the glass and film's refractive indices is key to achieving the desired optical effects.
Light Wavelength
Light consists of waves, and the wavelength is the distance between consecutive peaks of these waves. In optics, wavelength is crucial because it determines the color of the light and how it interacts with various materials.
In our example, the wavelength of light that we want to minimize glare for is 505 nm. This wavelength corresponds to a specific color in the visible spectrum. Choosing the right wavelength is essential as it affects the calculations for film thickness needed to achieve optimal interference.
Knowing the precise wavelength allows us to calculate the exact thickness of the film by employing interference equations. This helps in designing coatings that can specifically target and reduce reflections of that wavelength, improving visibility through the glass.
Constructive Interference
Constructive interference happens when two light waves meet in such a way that their amplitudes add up, making the light more intense. While constructive interference is useful in many scenarios, like enhancing signals, in the case of glare reduction, we aim to avoid it.
The goal here is actually to achieve destructive interference, where the waves cancel each other out, but it's important to understand constructive interference for comparison. In the problem, controlling interference means adjusting the film's thickness so that it causes destructive interference rather than constructive.
By controlling the refractive index and thickness of the coating, the destructive interference of specific wavelengths can be achieved. This reduces the intensity of the reflected light, thereby minimizing glare and enhancing visibility of the artwork behind the glass.

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Most popular questions from this chapter

35.19. In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_{0}\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is \(60.0^{\circ}\) , what is the intensity? (b) What is the path difference for 480 -nm light from the two slits at a point where the phase angle is \(60.0^{\circ} ?\)

35.46. A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index \(1.55 .\) This film has the proper thickness to cancel normally incident light of wavelength 525 \(\mathrm{nm}\) that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 \(\mathrm{nm}\) per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?

35.34. Light with wavelength 648 \(\mathrm{nm}\) in air is incident perpendicularly from air on a film 8.76\(\mu \mathrm{m}\) thick and with refractive index \(1.35 .\) Part of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface, where the film is again in contact with air. (a) How many waves are contained along the path of this second part of the light in its round trip through the film? (b) What is the phase difference between these two parts of the light as they leave the film?

35.4. Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04\(\mu \mathrm{m}\) apart, and in line with an observer, so that one source is 2.04\(\mu \mathrm{m}\) farther from the observer than the other. (a) For what visible wavelengths ( 400 to 700 \(\mathrm{nm}\) ) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04\(\mu \mathrm{m}\) farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?

35.8. Young's experiment is performed with light from excited helium atoms \((\lambda=502 \mathrm{nm}) .\) Fringes are measured carefully on a screen 1.20 \(\mathrm{m}\) away from the double slit, and the center of the 20 \(\mathrm{th}\) fringe (not counting the central bright fringe) is found to be 10.6 \(\mathrm{mm}\) from the center of the central bright fringe. What is the separation of the two slits?
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