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Chapter 35: Problem 24

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
The minimum thickness of the film that will cancel the reflection is approximately \(48 nm\). The next three viable thicknesses would be found using higher-order destructive interference.

Step by step solution

01

Identify the Formula for Thin Film Interference

The condition for constructive or destructive interference caused by a thin film can be given by the formula \(2nt = m\lambda\), where \(n\) is the refractive index, \(t\) is the thickness of the film, \(m\) is the order of interference, and \(\lambda\) is the wavelength of light. In our case, we are looking for destructive interference, which happens when \(m\) is a half-integer.
02

Substitute the Given Values Into the Formula

The refractive index of TiO2, \(n = 2.62\), light of a wavelength, \(\lambda = 505 nm = 505 \times 10^{-9} m\), and since we are looking for minimum thickness, the interference order, \(m = 0.5\). Substituting these into the equation, we get \(2(2.62)t = 0.5 \times 505 \times 10^{-9}\). Solving for \(t\), we get \(t \approx 48 nm\).
03

Find the Next Three Thicknesses

The thicknesses for higher-order destructive interference occur at every half-wavelength increase in thickness. So, to find the next viable thicknesses, we set \(m = 1.5\), \(m = 2.5\) and \(m = 3.5\) in the equation and solve for \(t\), respectively.

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Key Concepts

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

Destructive Interference
Destructive interference is a phenomenon where two waves combine to produce a reduction in amplitude. This is essential for reducing glare from surfaces like glass. When light strikes a thin film coating, it splits into two paths: one reflecting off the top and the other from the bottom surface of the film.
  • For destructive interference, these two reflected waves need to be out of phase by a half of the wavelength, so they cancel each other out.
  • The condition for this is achieved when the optical path difference equals an odd multiple of half wavelengths. Mathematically, this is expressed as: \[ 2nt = (m + \frac{1}{2})\lambda \] where \( n \) is the refractive index, \( t \) is the film thickness, \( m \) is the order of interference (0.5 for the smallest thickness), and \( \lambda \) is the wavelength of light.
This condition ensures that the reflected waves from the top and bottom of the coating interfere destructively, minimizing glare and allowing the artwork to be seen more clearly.
Refractive Index
The refractive index is a measure of how much the speed of light is reduced inside a medium. Different materials have different indices, and this affects how light behaves when it transitions from one medium to another.
  • A higher refractive index means that light travels more slowly in the medium, causing a greater bending of light rays.
  • In the context of thin film interference, the refractive index of both the thin film and the underlying substrate (in this case, glass) play crucial roles.
  • The refractive index of the coating material (TiO2 in this scenario) determines the optical path length that influences interference patterns.
Understanding the refractive index is vital as it determines the exact thickness needed for effective interference, as seen in setting up the correct thickness for destructive interference in the previous section.
Wavelength
Wavelength is the distance between consecutive peaks of a wave and is crucial for determining interference patterns in thin films. When light waves reflect off a thin film, their path length difference is related to the wavelength of the light.
  • For destructive interference, paths need to differ by a half wavelength to achieve cancellation.
  • The specific wavelength considered (505 nm in this case) impacts the thickness required for the film to produce destructive interference.
  • Since the wavelength changes as it enters a medium with a different refractive index, it gets shorter in a medium with a higher index like TiO2 compared to air.
Wavelength adjustments due to medium changes are captured in the formula used for calculating the minimum thickness of the film to achieve the desired interference. Thus, understanding how wavelength operates within this context helps in predicting the behavior of light to minimize unwanted reflections.

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Most popular questions from this chapter

Two thin parallel slits that are \(0.0116 \mathrm{~mm}\) apart are illuminated by a laser beam of wavelength \(585 \mathrm{nm}\). (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that \(\sin \theta\) can be? What does this tell you is the largest value of \(m ?\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is \(11.0 \mathrm{~cm}\) long and has a refractive index of \(1.55 .\) A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a manner similar to that discussed in Example 35.4 . When you view the glass plates from above with reflected white light, you observe that, at \(1.15 \mathrm{~mm}\) from the line where the sheets are in contact, the violet light of wavelength \(400.0 \mathrm{nm}\) is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength \(550.0 \mathrm{nm}\) ) and orange light (of wavelength \(600.0 \mathrm{nm}\) ) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart?

Jan first uses a Michelson interferometer with the \(606 \mathrm{nm}\) light from a krypton-86 lamp. He displaces the movable mirror away from him, counting 818 fringes moving across a line in his field of view. Then Linda replaces the krypton lamp with filtered \(502 \mathrm{nm}\) light from a helium lamp and displaces the movable mirror toward her. She also counts 818 fringes, but they move across the line in her field of view opposite to the direction they moved for Jan. Assume that both Jan and Linda counted to 818 correctly. (a) What distance did each person move the mirror? (b) What is the resultant displacement of the mirror?

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 \((380\) to \(750 \mathrm{nm})\) 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?

Two identical horizontal sheets of glass have a thin film of air of thickness \(t\) between them. The glass has refractive index \(1.40 .\) The thickness \(t\) of the air layer can be varied. Light with wavelength \(\lambda\) in air is at normal incidence onto the top of the air film. There is constructive interference between the light reflected at the top and bottom surfaces of the air film when its thickness is \(650 \mathrm{nm} .\) For the same wavelength of light the next larger thickness for which there is constructive interference is \(910 \mathrm{nm}\). (a) What is the wavelength \(\lambda\) of the light when it is traveling in air? (b) What is the smallest thickness \(t\) of the air film for which there is constructive interference for this wavelength of light?
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