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Chapter 27: Problem 12

You are standing in air and are looking at a flat piece of glass \((n=1.52)\) on which there is a layer of transparent plastic \((n=1.61) .\) Light whose wavelength is \(589 \mathrm{nm}\) in vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.

Short Answer

Expert verified
Thicknesses are approximately 56.77 nm and 170.32 nm.

Step by step solution

01

Understand the Problem

We need to find the smallest two nonzero thicknesses of a plastic layer (with refractive index \(n_2 = 1.61\)) coated on glass (with \(n_3 = 1.52\)) such that the reflection of light (with wavelength \(\lambda_0 = 589\, \text{nm}\)) looks dark, indicating destructive interference.
02

Identify Conditions for Destructive Interference

For destructive interference in a thin film, the condition \( 2 n_2 t = (m + \frac{1}{2}) \lambda \) holds, where \( t \) is the thickness, \( m \) is an integer representing the order of interference, and \( \lambda \) is the wavelength of light within the medium.
03

Find Wavelength in Plastic Layer

Determine the wavelength of light in the plastic by considering the refractive index. Using \( \lambda = \frac{\lambda_0}{n_2} \), we find \( \lambda = \frac{589\, \text{nm}}{1.61} \approx 365.22\, \text{nm} \).
04

Solve for Thickness

Using the destructive interference condition, solve for thickness \( t \): \( 2 \times 1.61 \times t = \left( m + \frac{1}{2} \right) \times 365.22 \). For the smallest two nonzero thicknesses, use \( m = 0 \) and \( m = 1 \).
05

Step 4.1: Calculate Thickness for m=0

Substitute \( m = 0 \) into the equation: \( 2 \times 1.61 \times t = \left( 0 + \frac{1}{2} \right) \times 365.22 \). Solve this to find \( t = \frac{1}{2} \cdot \frac{365.22}{2 \times 1.61} \approx 56.77\, \text{nm} \).
06

Step 4.2: Calculate Thickness for m=1

Substitute \( m = 1 \) into the equation: \( 2 \times 1.61 \times t = \left( 1 + \frac{1}{2} \right) \times 365.22 \). Solve this to find \( t = \frac{3}{2} \cdot \frac{365.22}{2 \times 1.61} \approx 170.32\, \text{nm} \).
07

Conclusion

The two smallest possible nonzero thickness values of the plastic layer for destructive interference are approximately \( 56.77 \text{ nm} \) and \( 170.32 \text{ nm} \).

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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 occurrence in thin film optics, where reflections of light lead to a cancelation effect, resulting in darkness or reduced brightness. This happens when two light waves overlap and are out of phase, causing their amplitudes to effectively subtract.

For a thin film, such as a coating on glass, destructive interference occurs when:
  • Light reflects from both the top and bottom surfaces of the film.
  • The light waves from these reflections interfere destructively.
  • The path difference between these two waves is such that they cancel out.

This is described mathematically by the condition: \[ 2n_2 t = \left(m + \frac{1}{2}\right) \lambda \] where:
  • \(n_2\) is the refractive index of the medium (here, plastic),
  • \(t\) is the film thickness,
  • \(\lambda\) is the wavelength of light in the medium,
  • \(m\) is an integer (known as the interference order).
This formula helps to find the film's thicknesses where constructive or destructive interference occurs.
Refractive Index
The refractive index is a measure of how much the speed of light is reduced in a medium compared to its speed in a vacuum. Different materials bend and slow down the light to various extents, which affects how interference takes place in thin films.

Key points about the refractive index include:
  • A higher refractive index means that light travels more slowly through the material.
  • In our example, the plastic layer has a refractive index \(n_2 = 1.61\), which is higher than the surrounding glass \(n_3 = 1.52\).
  • This means that when light enters the plastic, its speed decreases, and its wavelength also changes as a result of the medium's properties.

Understanding the refractive index is crucial for predicting how light behaves at the interfaces of different materials, impacting phenomena like interference, refraction, and reflection.
Wavelength in Medium
When light travels through a material with a refractive index different from a vacuum, its wavelength shortens proportionally to the index of refraction. This altered wavelength is critical to understanding and predicting interference patterns in thin films.

Here's how it works:
  • The wavelength of light in a medium \(\lambda\) is given by \(\lambda = \frac{\lambda_0}{n}\), where \(\lambda_0\) is the original wavelength in vacuum and \(n\) is the refractive index of the medium.
  • In our specific case, with \(\lambda_0 = 589\ ext{nm}\) in vacuum and \(n_2 = 1.61\) for plastic, the wavelength in the plastic becomes \(\lambda = \frac{589\ ext{nm}}{1.61} \approx 365.22\ ext{nm}\).
  • This adjusted wavelength is used in the condition for destructive interference to find the optimal film thickness.
By calculating the wavelength within materials, scientists can design coatings and layers with precise optical properties, leading to applications in lenses, displays, and other optical technologies.

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Most popular questions from this chapter

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00) .\) A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and \(1.40 .\) Light that has a wavelength of \(625 \mathrm{nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is \(242 \mathrm{nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

A soap film \((n=1.33)\) is \(375 \mathrm{nm}\) thick and coats a flat piece of glass \((n=1.52) .\) Thus, air is on one side of the film and glass is on the other side, as the figure illustrates. Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through air and strikes the film nearly perpendicularly. Concepts: (i) What, if any, phase change occurs when light, traveling in air, reflects from the air-film interface? (ii) What, if any, phase change occurs when light, traveling in the film, reflects from the film-glass interface? (iii) Is the wavelength of the light in the film greater than, smaller than, or equal to the wavelength in a vacuum? Calculations: For what wavelengths in the range of 380 to 750 nm does constructive interference cause the film to look bright in reflected light?

The wavelength of the laser beam used in a compact disc player is \(780 \mathrm{nm}\). Suppose that a diffraction grating produces first-order tracking beams that are \(1.2 \mathrm{mm}\) apart at a distance of \(3.0 \mathrm{mm}\) from the grating. Estimate the spacing between the slits of the grating.

When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is \(296 \mathrm{nm} .\) What is the vacuum wavelength of the light in \(\mathrm{nm} ?\)

A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{m} .\) When light with a wavelength of \(\lambda_{1}=510 \mathrm{nm}\) passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2}\) ), and a wavelength of \(\lambda_{2}=740 \mathrm{nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2}\)
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