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\(\bullet\) 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 thin 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 thicknesses would also work? Find only the three thinnest ones.

Short Answer

Expert verified
The minimum film thickness is 48.19 nm, and other thicknesses are 144.56 nm, 240.94 nm, and 337.31 nm.

Step by step solution

01

Calculating wavelength inside the film

To start, we need to determine the wavelength of 505 nm light inside the \(\mathrm{TiO}_2\) film. The refractive index of the film is 2.62. Therefore, the wavelength of the light inside the film is given by: \[ \lambda_{film} = \frac{\lambda_{air}}{n_{film}} = \frac{505 \, \text{nm}}{2.62} = 192.75 \, \text{nm}. \]
02

Finding minimum thickness for destructive interference

To achieve cancellation (destructive interference) for the reflected waves, the path difference between the light reflecting off the top surface and the light reflecting off the film-glass interface should be half a wavelength inside the film. Thus, the minimum thickness \(t_1\) is given by: \[ t_1 = \frac{\lambda_{film}}{4} = \frac{192.75 \, \text{nm}}{4} = 48.1875 \, \text{nm}. \]
03

Finding other possible thicknesses

Additional thicknesses \(t_n\) that would also result in cancellation can be found by adding half wavelengths inside the film to \(t_1\). These thicknesses are given by: \[ t_n = \left(n + \frac{1}{2}\right) \frac{\lambda_{film}}{2}, \quad n = 0, 1, 2, \ldots \] Substituting \(n = 1, 2, 3\) gives the next three possible thicknesses:1. \(t_2 = \left(1 + \frac{1}{2}\right) \frac{192.75}{2} = 144.56 \, \text{nm}\) 2. \(t_3 = \left(2 + \frac{1}{2}\right) \frac{192.75}{2} = 240.9375 \, \text{nm}\) 3. \(t_4 = \left(3 + \frac{1}{2}\right) \frac{192.75}{2} = 337.3125 \, \text{nm}\)

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Key Concepts

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

Refractive Index
The refractive index is a crucial concept in understanding how light behaves when it passes through different materials. It describes how much light slows down when it enters a material compared to its speed in a vacuum. For example, in our exercise, the glass has a refractive index of 1.62, meaning light travels 1.62 times slower in this glass than in a vacuum. Similarly, the titanium dioxide (\(\mathrm{TiO}_2\)) coating has a refractive index of 2.62, indicating an even greater reduction in speed.

In simple terms, the larger the refractive index, the more the light bends or changes direction when entering the material. This property is vital in creating optical effects such as lenses and coatings that minimize glare, like the thin film in our exercise.
Wavelength
Wavelength is a measure of the distance between two consecutive peaks or troughs in a wave, such as light. It's usually measured in nanometers (nm) for visible light, ranging approximately from 400 nm to 700 nm.

When light enters a material with a different refractive index, its wavelength changes but its frequency remains constant. This change in wavelength inside a medium is given by \(\lambda_{film} = \frac{\lambda_{air}}{n_{film}}\). For instance, the 505 nm light in air becomes 192.75 nm inside the \(\mathrm{TiO}_2\) film in our example. Understanding how the wavelength alters in different materials is key to predicting the light's behavior, leading to phenomena such as interference.
Destructive Interference
Destructive interference occurs when two waves combine to form a reduced or zero amplitude result. For light waves, this effect can cancel reflections, reducing glare. In our problem, the coating causes the light waves reflecting from its surface and any further layers underneath to meet out of phase. This means the light waves diminish each other, resulting in less reflection.

The condition for destructive interference is that the difference in the path length traveled by the two waves should be half a wavelength (or \(\) multiples thereof). For the thin film coating, this path difference is precisely achieved via the film's thickness, calculated as \(t_1 = \frac{\lambda_{film}}{4}\), ensuring the reflected waves cancel out effectively.
Optical Coating
Optical coatings are thin layers of material applied to surfaces like glass to improve their optical properties. These coatings can minimize glare, enhance reflectivity, or provide anti-reflective properties, depending on their design.

In the exercise, an optical coating made from \(\mathrm{TiO}_2\) minimizes glare by exploiting the destructive interference principle. By carefully selecting the coating's thickness, unwanted reflections are canceled, making viewing clearer and more comfortable. Such coatings are used widely, from improving camera lenses to making eyeglasses more effective, illustrating their vital role in enhancing optical clarity.

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Most popular questions from this chapter

\(\bullet\) Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 \(\mathrm{km} .\) If they broadcast 3.6 \(\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

\(\bullet\) Electromagnetic waves of wavelength 0.173 nm fall on a crystal surface. As the angle from the plane is gradually increased, starting at \(0^{\circ},\) you find that the first strong interfer- ence maximum occurs when the beam makes an angle of \(22.4^{\circ}\) with the surface of the crystal planes in the Bragg reflection. (a) What is the distance between the crystal planes? (b) At what other angles will interference maxima occur?

\(\bullet\) Conserving energy. You want to coat the inner surfaces of your windows (which have refractive index of 1.51\()\) with a film in order to enhance the reflection of light back into the room so that you can use bulbs of lower wattage than usual. You find that \(\mathrm{MgF}_{2},\) with \(n=1.38,\) is not too expensive, so you decide to use it. Since incandescent home lightbulbs emit reddish light with a peak wavelength of approximately 650 \(\mathrm{nm}\) , you decide that this wavelength is the one to enhance in the light reflected back into the room. (a) What is the minimum thickness of film that you will need? (b) If this layer seems too thin to be able to put on accurately, what other thicknesses would also work? Give only the three thinnest ones.

\(\bullet\) Coherent light of frequency \(6.32 \times 10^{14}\) Hz passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

\(\bullet\) The lenses of a particular set of binoculars have a coating with index of refraction \(n=1.38,\) and the glass itself has \(n=1.52 .\) If the lenses reflect a wavelength of 525 \(\mathrm{nm}\) the most strongly, what is the minimum thickness of the coating?

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