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Chapter 28: Problem 32

White light is incident on a soap film \((n=1.30)\) in air. The reflected light looks bluish because the red light \((\lambda=670 \mathrm{nm})\) is absent in the reflection. What is the minimum thickness of the soap film?

Short Answer

Expert verified
The minimum thickness of the soap film is approximately 128.85 nm.

Step by step solution

01

Understanding Soap Film Reflection

When light is reflected off a thin film like soap, certain wavelengths are enhanced or cancelled depending on the film's thickness and refractive index. For destructive interference (where light is absent), the path difference must be an odd multiple of half the wavelength inside the film.
02

Using Destructive Interference Condition

For destructive interference of the reflected red light (with wavelength \(\lambda = 670 \, \text{nm}\)), the path difference must be \((m + \frac{1}{2})\lambda_n\), where \(\lambda_n = \frac{\lambda}{n}\). For the first minimum (minimum thickness), \(m = 0\).
03

Calculating the Wavelength in the Film

The wavelength of the light inside the film is given by \(\lambda_n = \frac{\lambda}{n}\). Substitute \(\lambda = 670 \, \text{nm}\) and \(n = 1.30\) into the equation: \[ \lambda_n = \frac{670 \, \text{nm}}{1.30} \approx 515.38 \, \text{nm}. \]
04

Finding Minimum Thickness

For minimum thickness with \(m=0\), the thickness \(t\) is given by \(t = \frac{(m + \frac{1}{2})\lambda_n}{2}\). Substitute \(m = 0\) and \(\lambda_n = 515.38 \, \text{nm}\) into the equation: \[ t = \frac{(0 + \frac{1}{2}) \times 515.38 \, \text{nm}}{2} = \frac{515.38 \, \text{nm}}{4} \approx 128.85 \, \text{nm}. \]

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Key Concepts

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

Destructive Interference
Destructive interference occurs when two waves combine in such a way that they cancel each other out. This can happen with light waves when they overlap with a path difference that leads to opposite phases. In thin films, like soap films, light reflecting from the top and bottom surfaces can interfere. For the reflected light to be "absent," or destructive interference to occur, the path difference must be an odd multiple of half the wavelength within the film.
  • Destructive interference conditions are crucial for determining which wavelengths will be missing in reflected light.
  • In the case of soap films, this principle helps explain why certain colors are not seen and others are intensified.
Understanding this concept is essential in solving problems related to thin film interference.
Refractive Index
The refractive index ( ) is a measure of how much light is bent, or refracted, when entering a material. It is a dimensionless number that describes how fast light travels in the material compared to the speed of light in a vacuum. In the context of a soap film, the refractive index is crucial in determining the effective wavelength of light within the film.
  • A higher refractive index means light travels slower within that medium.
  • Refractive index impacts the calculation of both wavelength in the film and the interference pattern produced.
  • The refractive index of the soap film in our example is given as 1.30, a typical value for such materials.
This concept not only changes how light waves behave but also plays a vital role in designing optical systems.
Wavelength Calculation
Wavelength calculation in a medium involves adjusting the wavelength of light in air or vacuum by its refractive index. This adjustment is necessary because light travels at different speeds in different media. For example, the red light with a wavelength of 670 nm in air has a different effective wavelength inside the soap film.
  • The formula used is \(\lambda_{n} = \frac{\lambda}{n}\).
  • Inserting our values, \(\lambda = 670 \, \text{nm}\) and \(n = 1.30\), we find that \(\lambda_{n} \approx 515.38 \, \text{nm}\).
  • This new wavelength inside the film is crucial for determining the conditions for interference.
Accurate wavelength calculations ensure that predictions of interference patterns are correct, which is essential in many areas of optics.
Optics
Optics is the study of light and its interactions with different materials. Thin film interference, like what occurs with soap films, is a fascinating optical phenomenon where light waves reflect and refract, leading to beautiful color patterns due to interference.
  • Optical phenomena like this are used in creating anti-reflective coatings and enhancing colors in materials.
  • The principles of wave reflection, refraction, and interference underpin much of optical science.
  • Optics includes a wide range of applications, from developing better lenses to understanding natural phenomena like rainbows and soap bubbles.
By mastering the fundamental concepts of optics, you can start to explore these applications and deepen your understanding of how light behaves in various settings.

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Most popular questions from this chapter

The yellow light from a helium discharge tube has a wavelength of \(587.5 \mathrm{nm}\). When this light illuminates a certain diffraction grating it produces a first-order principal maximum at an angle of \(1.250^{\circ} .\) Calculate the number of lines per centimeter on the grating.

Images produced by structures within the eye (like lens fibers or cell fragments) are referred to as entoptic images. These images can sometimes take the form of "halos" around a bright light seen against a dark background. The halo in such a case is actually the bright outer rings of a circular diffraction pattern, like Figure \(28-21\), with the central bright spot not visible because it overlaps the direct image of the light. Find the diameter of the eye structure that causes a circular diffraction pattern with the first dark ring at an angle of \(3.7^{\circ}\) when viewed with monochromatic light of wavelength \(630 \mathrm{nm} .\) (Typical eye structures of this type have diameters on the order of \(10 \mu \mathrm{m}\). Also, the index of refraction of the vitreous humor is \(1.336 .\) )

A microphone is located on the line connecting two speakers that are \(0.845 \mathrm{m}\) apart and oscillating \(180^{\circ}\) out of phase. The microphone is \(2.25 \mathrm{m}\) from the midpoint of the two speakers. What are the lowest two frequencies that produce an interference maximum at the microphone's location?

A thin layer of magnesium fluoride \((n=1.38)\) is used to (a) What thickness should the coat a flint-glass lens \((n=1.61)\). magnesium fluoride film have if the reflection of \(565-\mathrm{nm}\) light is to be suppressed? Assume that the light is incident at right angles to the film. (b) If it is desired to suppress the reflection of light with a higher frequency, should the coating of magnesium fluoride be made thinner or thicker? Explain.

A radio broadcast antenna is \(36.00 \mathrm{km}\) from your house. Suppose an airplane is flying \(2.230 \mathrm{km}\) above the line connecting the broadcast antenna and your radio, and that waves reflected from the airplane travel 88.00 wavelengths farther than waves that travel directly from the antenna to your house. (a) Do you observe constructive or destructive interference between the direct and reflected waves? (Hint: Does a phase change occur when the waves are reflected?) (b) The situation just described occurs when the plane is above a point on the ground that is two-thirds of the way from the antenna to your house. What is the wavelength of the radio waves?
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