/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A uniform film of \(\mathrm{TiO}... [FREE SOLUTION] | 91影视

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Chapter 35: Problem 25

A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength \(520.0 \mathrm{nm}\) falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of \(\mathrm{TiO}_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wavelengths of the light in the \(\mathrm{TiO}_{2}\) film.

Short Answer

Expert verified
The minimum additional thickness that should be added to the TiO2 film is approximately 154.85 nm. The path difference between the light reflected off the top of the film and the light that cancels it is 3120 nm, or approximately 15.72 wavelengths of the light in the TiO2 film.

Step by step solution

01

Calculate Initial Path Difference

The initial path difference can be calculated by multiplying the thickness of the TiO2 film by the refractive index. The path difference will be \(1036nm * 2.62 = 2714.32 nm\). This is the distance that the reflected wave traverses in the film.
02

Calculate the Thickness Required for Destructive Interference

For destructive interference to occur, the total path difference needs to be an odd multiple of half the wavelength of light used. Given that the wavelength is 520.0 nm, the closest odd multiple of 260 nm (half of 520 nm) to our initial path difference is 2600 nm. Therefore, the additional path difference required to reach 2600 nm is \(2600nm - 2714.32nm = -114.32nm\). Because this value is negative, it appears that the initial thickness of the film was too large. However, keep in mind that light could traverse the film several times before destructively interfering. Therefore, we need to add an entire wavelength (520 nm) to the additional path to make it positive. So, the adjusted additional path is \(-114.32nm + 520nm = 405.68nm\).
03

Convert the Additional Path to Thickness

To convert this additional path into an equivalent thickness to be added to the existing film, it needs to be divided by the refractive index of TiO2. Therefore, the additional thickness required is \(405.68nm / 2.62 = 154.85 nm\).
04

Calculate the Total Path Difference

After adding the extra thickness, the total path difference achieved is equal to the initial path difference plus the extra distance the wave travels due to the extra thickness. This equals \(2714.32nm + 405.68nm = 3120 nm\). Therefore, the path difference is 3120 nm.
05

Express Path Difference in Wavelengths

Finally, express this path difference in terms of the wavelength of light in the TiO2 film. The wavelength of light in the film is the wavelength in air divided by the refractive index of TiO2, or \(520nm/2.62 = 198.47 nm\). So, the path difference in terms of wavelengths is \(3120nm / 198.47nm = 15.72\) wavelengths.

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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 occurs when light waves reflected from the top and bottom surfaces of a thin film, like oil on water, interfere with one another. This interference can be either constructive or destructive.
When a wave reflects from a medium with a higher refractive index than the medium it came from, it undergoes a phase shift of half a wavelength. This phase shift is crucial in determining whether the interference is constructive or destructive. The thickness of the film and the wavelength of the light also play significant roles in the interference pattern.
  • If the path difference between the two waves is an integer multiple of the wavelength, constructive interference increases the intensity of the light.
  • If the path difference is an odd multiple of half the wavelength, destructive interference occurs and cancels out the light.
This principle is used in various applications such as anti-reflective coatings on lenses and mirrors.
Destructive Interference
Destructive interference, specifically within thin film interference, results in the cancellation of light waves.
This phenomenon occurs when the combined effect of the light waves is a reduction in intensity, leading to darkness or dimness in certain areas. For it to happen, the path difference between the light waves after passing through the film should be an odd multiple of half the wavelength, expressed mathematically as \[ ext{(2n + 1)} \frac{\lambda}{2}\]where \(n\) is an integer and \(\lambda\) is the wavelength in the film.
This property is used to create anti-glare screens and in paint materials that need to counteract light reflection.
Refractive Index
The refractive index of a material quantifies how much light slows down when passing through it.For instance, a refractive index of 2.62 for titanium dioxide (TiO2) means light travels more than twice as slow compared to in a vacuum.
Refractive index affects the wavelength of light inside the medium and causes the bending of light at interfaces. This bending, or refraction, is crucial for accomplishing desired interference patterns, such as adjusting thickness in thin film interference to achieve destructive interference.
Formulaically:\[v = \frac{c}{n}\]where \(v\) is the speed of light in the medium, \(c\) is the speed of light in a vacuum, and \(n\) is the refractive index.
Understanding refractive index is essential for designing lenses or coatings where light's path must be carefully controlled.
Wavelength in Medium
Wavelength in a medium, like a film of TiO2, differs from its wavelength in air.When light enters a different medium, its speed and wavelength change while the frequency remains constant. The wavelength in the medium can be calculated using:\[\lambda_{medium} = \frac{\lambda_{air}}{n}\]where \(\lambda_{medium}\) is the wavelength in the medium, \(\lambda_{air}\) is the wavelength in air, and \(n\) is the refractive index of the medium.
This change in wavelength is fundamental to thin film interference and plays a crucial role in designing optical coatings. In our example, the wavelength of 520 nm air light alters when passing into the TiO2 film, affecting the interference conditions and helping to determine the appropriate thickness for interference applications.

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Most popular questions from this chapter

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{~Hz}\) 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?

The professor returns the apparatus to the original setting. She then adjusts the speakers again. All of the students who had heard nothing originally now hear a loud tone, while you and the others who had originally heard the loud tone hear nothing. What did the professor do? (a) She turned off the oscillator. (b) She turned down the volume of the speakers. (c) She changed the phase relationship of the speakers. (d) She disconnected one speaker.

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength \(\lambda_{0}\) in vacuum at normal incidence onto the surface of the glass. When \(\lambda_{0}=496 \mathrm{nm},\) constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is \(386 \mathrm{nm}\). (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

Red light with wavelength \(700 \mathrm{nm}\) is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe \((m=3)\) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

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.
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