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What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 \(\mathrm{nm} ?\) The index of refraction of the film is \(1.33,\) and there is air on both sides of the film.

Short Answer

Expert verified
The thinnest soap film that appears black is approximately 90.226 nm thick.

Step by step solution

01

Understand the Problem

We need to find the minimum thickness of a soap film that causes destructive interference for light with a wavelength of 480 nm, making it appear black. This occurs when the reflected light waves cancel each other out.
02

Identify Conditions for Destructive Interference

For destructive interference in a thin film with an index of refraction greater than the surroundings (air), a half-wavelength shift occurs upon reflection. We use the formula for destructive interference: two times the thickness of the film equals an odd multiple of half-wavelengths in the film.
03

Calculate Wavelength in the Film

Convert the wavelength of light in air (480 nm) to the wavelength in the film. This is done using the formula: \( \lambda_{film} = \frac{\lambda_{air}}{n} \) where \( n \) is the index of refraction. Here, \( \lambda_{film} = \frac{480 \text{ nm}}{1.33} = 360.902 \text{ nm} \).
04

Determine the Minimum Film Thickness

The minimum thickness \( t \) for destructive interference (causing black appearance) is when it equals \( \frac{(2m+1)\lambda_{film}}{4} \) where \( m = 0 \) to minimize thickness. Substitute \( \lambda_{film} = 360.902 \text{ nm} \) and \( m = 0 \), to find \( t = \frac{360.902 \text{ nm}}{4} \approx 90.226 \text{ nm} \).
05

Conclusion

Thus, the thinnest soap film that appears black has a thickness of approximately 90.226 nm when illuminated with light of wavelength 480 nm.

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Key Concepts

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

Destructive Interference
In optics, destructive interference is a phenomenon where two or more light waves superimpose to form a resultant wave of lower amplitude. It occurs when the crest of one wave aligns with the trough of another, leading to cancellation. When analyzing thin film interference, destructive interference is vital in determining what causes a soap film to appear black.

For constructive interference, two waves combine to enhance their amplitude, amplifying their brightness. But for a thin film to exhibit destructive interference, specific conditions must be met. In particular, the thickness of the film plays a crucial role. The rules of interference tell us that for a film, the path difference between waves undergoing reflection must be an odd multiple of half-wavelengths.

When these waves meet, they are directly out of phase, leading to darkness or minimized reflection in specific colors, like the black appearance in the soap film example.
Wavelength Conversion
Understanding wavelength conversion is essential to correctly calculate interference patterns in thin films. Since a wave's speed changes when entering a medium with a different index of refraction, its wavelength also changes, while the frequency remains constant.

To convert the wavelength of light in air to that within a material like a soap film, we use the formula:
  • \( \lambda_{film} = \frac{\lambda_{air}}{n} \)
where \( \lambda_{air} \) is the wavelength in air, and \( n \) is the index of refraction of the film.

This conversion allows us to understand how the light's behavior will change when it travels through different media. For the soap film with a 480 nm wavelength in air and an index of refraction of 1.33, converting it gives us a wavelength of approximately 360.902 nm in the soap film. This adjusted wavelength is crucial for applying interference formulas to determine film characteristics.
Index of Refraction
The index of refraction is a fundamental concept in optics, describing how light propagates through different materials. It is symbolized by \( n \) and defined as:
  • \( n = \frac{c}{v} \)
where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium. A higher index of refraction means that light travels more slowly through the material.

In the context of thin film interference, the index of refraction affects both the speed and wavelength of light within the film. It's critical to know this value in order to convert wavelengths and accurately calculate interference patterns and resultant film thicknesses.

For instance, in a soap film surrounded by air, with an index of refraction of 1.33, light moves more slowly inside the film than in the air. This results in a shorter wavelength within the film, which then impacts how interference occurs at different thicknesses.
Optics
Optics is the branch of physics that deals with the behavior and properties of light. It encompasses the study of light reflection, refraction, and diffraction. These principles help us understand interesting phenomena like rainbow-colored films or soap bubbles.

In the realm of thin films, optics explains how seemingly transparent materials can display vibrant colors through interference. White light, when interacting with a thin film, undergoes reflection and refraction at various interfaces, causing certain wavelengths to be enhanced or canceled out depending on conditions like the thickness or the refractive index.

Thus, optical principles provide the groundwork for understanding complex phenomena such as thin film interference. By studying light's behavior when interacting with materials, we can predict and explain the visual effects we often encounter, whether it’s in everyday life or technical applications in fields like lens design or telecommunications.

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Most popular questions from this chapter

A source \(S\) of monochromatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass and are separated by a horizontal distance \(x .\) Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

Two thin parallel slits that are 0.0116 \(\mathrm{mm}\) apart are illuminated by a laser beam of wavelength 585 \(\mathrm{nm}\) . (a) On a very large distant screen, what is the total number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest value of \(m ?\) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

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 -nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index \(1.50 .\) At room temperature \(\left(20.0^{\circ} \mathrm{C}\right),\) this film is just thick enough for light with wavelength 582.4 \(\mathrm{nm}\) 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 \(170^{\circ} \mathrm{C},\) you find that the film cancels reflected light with wavelength 588.5 \(\mathrm{nm}\) . 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.)

Young's experiment is performed with light from excited helium atoms \((\lambda=502 \mathrm{nm}) .\) Fringes are measured carefully on a screen 1.20 \(\mathrm{m}\) away from the double slit, and the center of the 20 \(\mathrm{th}\) fringe (not counting the central bright fringe) is found to be 10.6 \(\mathrm{mm}\) from the center of the central bright fringe. What is the separation of the two slits?

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