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Chapter 26: Problem 17

The lead architect on the design team of a new skyscraper decides that the outer surface of the building's windows needs to be coated with a reflective layer in order to lower airconditioning costs. \(\mathrm{MgF}_{2}\) is chosen for the reflective layer, which has a refractive index \(n=1.38 .\) Assume that the refractive index of the window glass is 1.5 and that the reflective coating should work optimally at the peak of the sunlight spectrum, \(500 \mathrm{nm}\). (a) What is the minimum thickness of film that is needed? (b) Suppose the layer needs to be thicker than \(200 \mathrm{nm}\) in order to hold up to the weather. What other thicknesses would also work? Give only the three thinnest ones.

Short Answer

Expert verified
(a) Minimum thickness is about 45.29 nm. (b) Three smallest thicknesses above 200 nm are approximately 271.74 nm, 452.90 nm, and 634.06 nm.

Step by step solution

01

Understanding the Problem

We need to determine the minimum thickness of a thin film coating (made of MgF2 with a refractive index, n = 1.38) on a glass window (refractive index = 1.5) for optimal reflection at 500 nm. The film thickness will result in constructive interference of reflected light.
02

Condition for Constructive Interference

For constructive interference in a film sandwiched between two different refractive indices, we use the condition for a film of thickness \(t\) and wavelength \(\lambda\): \(2nt = m\lambda/2\), where \(m\) is an integer (order of interference). For the minimum thickness, use \(m = 1\).
03

Calculate Minimum Film Thickness

Using \(m = 1\) for the minimum thickness, set up the equation: \[2 \cdot 1.38 \cdot t = \frac{500 \, nm}{2}\]. Solve for \(t\) to find the minimum thickness: \[t = \frac{500 \, nm}{2 \cdot 1.38 \cdot 2} = \frac{500}{5.52} \, nm \approx 45.29 \, nm\].
04

Condition for other Thicknesses

Other thicknesses will be multiples of the half-wavelength in MgF2. Use the formula \(2nt = m\lambda/2\), allowing \(m = 3, 5, 7, \ldots\) to find solutions greater than 200 nm.
05

Find the Next Three Thicknesses

For \(m = 3\), \(t = \frac{3 \cdot 500}{2 \cdot 2 \cdot 1.38} \, nm = \frac{1500}{5.52} \, nm \approx 271.74 \, nm\). For \(m = 5\), \(t = \frac{5 \cdot 500}{2 \cdot 2 \cdot 1.38} \, nm = \frac{2500}{5.52} \, nm \approx 452.90 \, nm\). For \(m = 7\), \(t = \frac{7 \cdot 500}{2 \cdot 2 \cdot 1.38} \, nm = \frac{3500}{5.52} \, nm \approx 634.06 \, nm\). The three smallest thicknesses above 200 nm are approximately 271.74 nm, 452.90 nm, and 634.06 nm.

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Key Concepts

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

Refractive Index
The concept of refractive index is crucial in understanding how light behaves as it moves through different materials. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. This property affects how much the path of light is bent, or refracted, when entering a material. For instance, in our problem involving window glass and a MgF2 coating, the refractive indices are 1.5 and 1.38, respectively. Such differences in refractive index are critical in the design of optical coatings because they determine how light is transmitted or reflected. Understanding the refractive index helps predict how efficiently a thin film can cause light to undergo constructive interference, optimizing reflective properties for energy efficiency. This principle is applied in designing window films to control heat and light transmission through the glass.
Constructive Interference
Constructive interference occurs when waves overlap and their amplitudes add together, creating a larger wave. This is an important concept when dealing with thin film interference like that seen in optical coatings. For constructive interference to take place within a thin film, the path difference between the reflected light waves must be an integer multiple of the wavelength (\(m\lambda/2\)). In the given problem, the optimal reflection is achieved at 500 nm. The film causes light waves reflecting off the top and bottom surfaces to constructively interfere, enhancing reflection. To achieve the required constructive interference with the MgF2 coating (having a refractive index of 1.38), its thickness must be carefully calculated using the formula \(2nt = m\lambda/2\). Using this condition, we find minimum thicknesses and alternative thicknesses that meet design requirements.
Optical Coatings
Optical coatings are thin layers added to surfaces to modify the way light interacts with that surface鈥攅ither enhancing reflection or transmission. They are used extensively in lenses, mirrors, and windows. The focus of the exercise on using MgF2 coatings highlights a real-world application. This coating is designed to increase reflection, aiming to improve building energy efficiency for the skyscraper by reducing cooling costs. When designing such coatings, one must consider both refractive indices involved鈥攈ere of MgF2 and the glass鈥攁s well as the desired wavelength (500 nm). The thickness of the coating layer plays a pivotal role in determining its reflective qualities, governed by thin film interference principles. This manipulation of light through physics allows for practical solutions to modern engineering problems such as energy conservation.

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Most popular questions from this chapter

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute). (a) Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with \(550 \mathrm{nm}\) light. (b) Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of \(10 \mathrm{~m}\) diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope ( 5 m diameter) on Palomar Mountain in California? In what ways are they not superior? Explain.

A laser beam of wavelength \(600.0 \mathrm{nm}\) is incident normally on a transmission grating having 400.0 lines \(/ \mathrm{mm} .\) Find the angles of deviation in the first, second, and third orders of bright spots.

Coherent light of wavelength \(525 \mathrm{nm}\) passes through two thin slits that are \(0.0415 \mathrm{~mm}\) apart and then falls on a screen \(75.0 \mathrm{~cm}\) away. How far away from the central bright fringe on the screen is (a) the fifth bright fringe (not counting the central bright fringe); (b) the eighth dark fringe?

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.

Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single frequency that can be varied between 300 and \(600 \mathrm{~Hz}\). The speed of sound is \(340 \mathrm{~m} / \mathrm{s}\). You find that where you are standing, you hear minimum-intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved \(39.8 \mathrm{~cm}\) toward you, the sound you hear has maximum intensity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?
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