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Chapter 26: Problem 18

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
Minimum thickness: 96.37 nm; other thicknesses: 289.11 nm, 481.84 nm.

Step by step solution

01

Identify Problem Type

We are dealing with thin film interference, where the film's thickness affects destructive interference conditions to reduce glare.
02

Destructive Interference Condition

For destructive interference, the condition is that the path difference between the reflected waves should be equal to a half wavelength in the film. Mathematically, this condition is given by \(2t = (m + \frac{1}{2}) \frac{\lambda}{n_f}\). Here, \(t\) is the film thickness, \(m\) is the interference order (0 for minimum thickness), \(\lambda\) is the wavelength in the air, and \(n_f\) is the refractive index of the film material.
03

Calculate Minimum Thickness

Substitute the known values: \(n_f = 2.62\), \(\lambda = 505 \mathrm{nm}\), and \(m = 0\) for minimum thickness. Plug these into the formula: \(2t = (0.5) \frac{505}{2.62}\). Solve for \(t\): \(t = \frac{505}{2 \times 2.62} \times 0.5 = 96.37 \mathrm{nm}\). The minimum thickness that will work is approximately 96.37 nm.
04

Calculate Second Thickness

Use \(m = 1\) in the formula: \(2t = (1.5) \frac{505}{2.62}\). Solve for \(t\): \(t = \frac{1.5 \times 505}{2 \times 2.62} = 289.11 \mathrm{nm}\). The second possible thickness is approximately 289.11 nm.
05

Calculate Third Thickness

Use \(m = 2\) in the formula: \(2t = (2.5) \frac{505}{2.62}\). Solve for \(t\): \(t = \frac{2.5 \times 505}{2 \times 2.62} = 481.84 \mathrm{nm}\). The third possible thickness is approximately 481.84 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, commonly symbolized by 'n', is a measure of how much the speed of light is reduced inside a medium compared to its speed in a vacuum. This property is crucial in thin film interference, as it determines the effective wavelength of light within the film.

When light moves from one medium to another, its speed changes, causing it to bend, or refract. The refractive index quantifies this bending, and is calculated using the ratio:
  • \[ n = \frac{c}{v} \]
  • where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium. In the case of using titanium dioxide (\( \text{TiO}_2 \)), the refractive index is 2.62, meaning light travels much slower in \( \text{TiO}_2 \) compared to a vacuum.
Understanding the refractive index helps predict how light will behave when passing through various materials, which is crucial for designing coatings to manage interference based on specific needs.
Destructive Interference
Destructive interference occurs when two or more waves overlap in such a way that they cancel each other out. This phenomenon is used in thin film interference to reduce glare by canceling specific wavelengths of light.When light strikes a thin film like the \( \text{TiO}_2 \) coating on glass, it reflects off both the top and bottom surfaces of the film. These reflections can interfere with each other. For destructive interference to occur, the path difference between the two reflecting waves must be equal to an odd multiple of half the wavelength inside the film:
  • \[ 2t = (m + \frac{1}{2}) \frac{\lambda}{n_f} \] where \( t \) is the film thickness, \( \lambda \) is the wavelength in air, \( n_f \) is the refractive index of the film, and \( m \) is an integer representing the interference order.

  • This ensures the crest of one wave aligns with the trough of another, effectively canceling each other out.
This technique is widely used in optics to create anti-reflective coatings, improving clarity by minimizing unwanted reflections.
Wavelength of Light
Wavelength is the distance between successive crests of a wave, particularly in the context of light. It is crucial in determining colors we perceive and plays a pivotal role in thin film interference by dictating how interference conditions will manifest.Light behaves both as a particle and a wave, and its wavelength is usually measured in nanometers (nm). When considering thin film interference:
  • Air's wavelength \( \lambda \) is adjusted by the refractive index when light enters a medium. This becomes critical when calculating interference conditions as the effective wavelength inside the film is \( \frac{\lambda}{n_f} \).
  • For destructive interference, determining the right thickness of a thin film to cancel out parts of the wavelength is essential. In this scenario, to reduce glare for light of wavelength 505 nm, this calculation allows the design of coatings that effectively diminish reflections.
Being proficient with these calculations means you can skillfully analyze and apply optical phenomena to practical problems, like reducing glare on glass displays.

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Most popular questions from this chapter

Red light of wavelength \(633 \mathrm{nm}\) from a helium-neon laser passes through a slit \(0.350 \mathrm{~mm}\) wide. The diffraction pattern is observed on a screen \(3.00 \mathrm{~m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

A laser beam of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disk. (a) The tracks of tiny pits in which information is coded onto the \(\mathrm{CD}\) are \(1.60 \mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only \(0.740 \mu \mathrm{m}\) apart. Repeat the calculation of part (a) for the DVD.

Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge, so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by \(546 \mathrm{nm}\) light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by \(0.460 \mathrm{~mm}\). In the resulting interference pattern on a screen \(2.20 \mathrm{~m}\) away, adjacent bright fringes are separated by \(2.82 \mathrm{~mm}\). What is the wavelength of the light that falls on the slits?

When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?
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