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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 radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is \(120 \mathrm{~m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of \(40 \mathrm{~m}\) to the right of antenna \(B .\) The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wavelength for which there will be constructive interference at point \(Q ?\)

A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength \(790 \mathrm{nm}\) passing through a plastic substrate of refractive index \(1.8 .\) When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so the two beams interfere with each other (Fig. E35.29). What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.)

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.

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?

Two slits spaced \(0.260 \mathrm{~mm}\) apart are \(0.900 \mathrm{~m}\) from a screen and illuminated by coherent light of wavelength \(660 \mathrm{nm}\). The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0} .\) What is the distance on the screen from the center of the central maximum (a) to the first minimum; (b) to the point where the intensity has fallen to \(I_{0} / 2 ?\)
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