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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 Ti\(O^2\) , 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 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
Minimum thickness is 48.2 nm. Three other thin options are 144.6 nm, 241.0 nm, and 337.4 nm.

Step by step solution

01

Understand the Problem

We need to find the minimum thickness of the titanium dioxide (TiO鈧�) film such that it causes destructive interference for light of wavelength 505 nm when applied to glass with a refractive index of 1.62. This requires utilizing the concept of thin film interference. Additionally, we'll find other potential thicknesses for increased durability.
02

Set Up Destructive Interference Condition

For destructive interference, the optical path difference should lead to a half-wavelength phase shift. The condition for destructive interference is \(2t = \frac{\lambda}{n_{\text{film}}} (m + \frac{1}{2})\) where \( t \) is the thickness of the film, \( n_{\text{film}} \) is the refractive index of the film, \( \lambda \) is the wavelength of light, and \( m \) is an integer for multiple orders of thickness.
03

Calculate Minimum Thickness

To find the minimum thickness for the first order \((m = 0)\), substitute the given values in the formula: \[ t = \frac{\lambda}{4n_{\text{film}}} \]Substitute \(\lambda = 505\, \text{nm}\) and \(n_{\text{film}} = 2.62\): \[ t = \frac{505}{4 \times 2.62} \approx 48.2 \text{ nm} \]
04

Calculate Other Possible Thicknesses

To find other thicknesses that could work, use higher orders \(m = 1, 2, 3\) in the equation from Step 2:For \(m = 1\):\[ t = \frac{3 \times 505}{4 \times 2.62} \approx 144.6 \text{ nm} \]For \(m = 2\):\[ t = \frac{5 \times 505}{4 \times 2.62} \approx 241.0 \text{ nm} \]For \(m = 3\):\[ t = \frac{7 \times 505}{4 \times 2.62} \approx 337.4 \text{ nm} \]
05

Conclusion

The minimum film thickness required for destructive interference of light with a wavelength 505 nm is approximately 48.2 nm. Other possible thicknesses are approximately 144.6 nm, 241.0 nm, and 337.4 nm.

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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 fascinating phenomenon where two light waves combine in such a way that they cancel each other out. This happens when the peaks of one wave align with the troughs of another, leading to a significant reduction in light intensity. In the context of thin films, such as a layer of titanium dioxide (TiO鈧�) coating a piece of glass, this is particularly useful for minimizing reflection or glare. When you coat glass with a thin film having a different refractive index, the light reflecting off the top and bottom surfaces of the film can interfere destructively. This means they cancel each other, reducing the glare that otherwise makes it hard to see through the glass.
This process is essential for applications like anti-reflective coatings on glasses or camera lenses, where clear visibility is crucial. To achieve destructive interference, the optical path difference between the two reflecting light waves must equate to a half-wavelength shift鈥攅ffectively canceling the waves out. The key formula for this condition in thin films is given by:
  • \(2t = \frac{\lambda}{n_{\text{film}}}(m + \frac{1}{2})\)
where \(t\) is the film thickness, \(\lambda\) is the wavelength of light, \(n_{\text{film}}\) is the refractive index of the film, and \(m\) is an integer representing the order of thickness.
Optical Path Difference
Optical path difference (OPD) refers to the difference in distance traveled by two light waves, caused by variations in material thickness and refractive index. In thin film interference, knowing the OPD helps us determine how light waves will interfere鈥攅ither constructively or destructively鈥攚hen reflecting off different layers of material.
Imagine a light wave hitting the coated glass: it reflects from both the top and bottom surfaces of the TiO鈧� layer. These reflected waves potentially interfere with one another. The OPD is crucial in determining whether their interference is constructive (making light brighter) or destructive (reducing brightness).
To control reflections using coatings, we make sure the OPD equals a half-wavelength (\(\lambda/2\)) for destructive interference. This precise control of the OPD through the thickness and refractive index allows us to cancel unwanted reflections effectively.
  • For maximum effectiveness and minimal reflection, the condition \(OPD = (m + 1/2)\lambda\) is used in the formula:
  • \(2t = \frac{\lambda}{n_{\text{film}}}(m + \frac{1}{2})\)
Understanding OPD's role allows engineers to design coatings with specific thicknesses that minimize glare.
Refractive Index
The refractive index, often denoted by \(n\), is a measure of how much a material slows down light compared to vacuum. It tells us how fast light travels through a given medium. This concept is pivotal in understanding light's behavior across different media, such as glass and coatings like TiO鈧�.
Materials with a high refractive index, such as TiO鈧� (with an index of 2.62), cause light to bend more drastically than those with a lower index, like air. When you have a coating with a refractive index different from the glass (1.62 in our example), it affects how light reflects and refracts at their interface.
In thin film interference, the refractive index determines the phase change that happens when light reflects off a boundary. This phase change is integral to achieving destructive interference. By tailoring the refractive indices of coatings, manufacturers can precisely control which wavelengths of light are canceled out, reducing glare.
  • Using different refractive indices wisely leads to more efficient anti-reflective coatings.
  • Effective coatings make viewing art through glass without annoying glare possible.
Thus, understanding and selecting appropriate refractive indices for coatings is crucial for enhancing clarity in optical applications.

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Most popular questions from this chapter

Two slits spaced \(0.0720 \mathrm{~mm}\) apart are \(0.800 \mathrm{~m}\) from a screen. Coherent light of wavelength \(\lambda\) passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is \(3.00 \mathrm{~mm} .\) If the intensity at the peak of the central maximum is \(0.0600 \mathrm{~W} / \mathrm{m}^{2},\) what is the intensity at points on the screen that are (a) \(2.00 \mathrm{~mm}\) and (b) \(1.50 \mathrm{~mm}\) from the center of the central maximum?

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm\)19.0\(^\circ\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(1 \over 10\) the maximum intensity on the screen?

A uniform film of TiO\(_2\) , 1036 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 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 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 TiO\(_2\) film.

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index 1.50. At room temperature (20.0\(^\circ\)C), this film is just thick enough for light with wavelength 582.4 nm reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170\(^\circ\)C, you find that the film cancels reflected light with wavelength 588.5 nm. What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)

Eyeglass lenses can be coated on the \(inner\) surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?
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