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Chapter 28: Problem 88

A thin soap film \((n=1.33)\) suspended in air has a uniform thickness. When white light strikes the film at normal incidence, violet light \(\left(\lambda_{\mathrm{V}}=420 \mathrm{nm}\right)\) is constructively reflected. (a) If we would like green light \(\left(\lambda_{G}=560 \mathrm{nm}\right)\) to be constructively reflected, instead, should the film's thickness be increased or decreased? (b) Find the new thickness of the film. (Assume the film has the minimum thickness that can produce these reflections.)

Short Answer

Expert verified
The thickness should be increased to 210 nm.

Step by step solution

01

Understand the Condition for Constructive Interference

For a film in air, constructive interference occurs when the path difference is equal to a multiple of the wavelength, i.e., \(2nt = m\lambda\), where \(n\) is the refractive index, \(t\) is the film thickness, and \(\lambda\) is the wavelength of light in vacuum. For our case, the condition becomes \(2nt = m\lambda_V\) for violet light and \(2nt = m'\lambda_G\) for green light.
02

Relate Thickness with Wavelengths

Given \(n = 1.33\), violet light \(\lambda_V = 420 \, \mathrm{nm}\), and green light \(\lambda_G = 560 \, \mathrm{nm}\), we first note that for constructive interference with minimum thickness, \(m\) for violet is smaller or equal to \(m'\) for green. Let's express the violet film condition as \(t = \frac{m \lambda_V}{2n}\) and the green film condition as \(t' = \frac{m' \lambda_G}{2n}\).
03

Determine the Change in Thickness

For the same film to reflect green light, \(t'\) (with green) should match \(t\) (with violet). Since \(\lambda_G > \lambda_V\), for the same order \(m = m'\), \(t'\) must be greater than \(t\). Therefore, the thickness should be increased to reflect green light.
04

Calculate the New Thickness

Assuming minimum thickness with the same order, equate \(m\lambda_V = m'\lambda_G = 2nt\). With \(t = \frac{m\lambda_V}{2n}\), and rearranging for the green light, the minimum thickness for green light becomes \(t' = \frac{\lambda_G}{2n}\). Substitute to find \(t' = \frac{560 \, \mathrm{nm}}{2 \times 1.33} = 210 \, \mathrm{nm}\).
05

Conclusion

The new thickness of the film should be increased to \(210 \, \mathrm{nm}\) if we want green light to be constructively reflected with the same order of interference.

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Key Concepts

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

Constructive Interference
Constructive interference is a fascinating concept in the realm of optics. It occurs when two or more waves overlap, and their wave amplitudes add up because they are in phase. In simple terms, constructive interference happens when the peaks of one wave coincide with the peaks of another, creating a wave with a greater amplitude. This results in brightness or color reinforcement when dealing with light waves, making certain colors much more pronounced.

For thin films, such as a soap bubble, constructive interference occurs when the optical path difference between two reflected light waves equals a whole number multiple of the light's wavelength. The equation governing this is given by:
  • Path difference = multiples of wavelength: \[2nt = m\lambda\] where:
    • \(n\) is the refractive index of the film.
    • \(t\) is the thickness of the film.
    • \(m\) is the order of interference (1, 2, 3,...).
    • \(\lambda\) is the wavelength of light in a vacuum.
With this in mind, different wavelengths (colors) can be selectively enhanced based on the film's thickness and refractive index.
Refractive Index
The refractive index is a crucial parameter in optics, particularly in topics such as thin film interference. It is a measure of how much a light wave is bent, or refracted, as it enters a new medium. The refractive index (\(n\)) indicates how fast light travels in a medium compared to the speed of light in a vacuum, where it's always 1.

Simply put, a refractive index greater than 1 means light slows down as it enters the medium. In our exercise, the refractive index of the soap film is given as 1.33, meaning light travels slower through the film compared to air. This bending of the light alters the effective wavelength within the medium, crucially affecting the color you see.

When light reflects off a thin film, the refractive index determines how the light is phase-shifted and how much of it is reflected versus transmitted, which is pivotal for creating conditions for constructive interference.
  • A higher refractive index results in more significant bending of light.
  • For the soap film in question, it's essential to consider this factor when calculating its thickness for specific interference effects.
Wavelength
Wavelength is another fundamental concept in understanding how thin film interference works. Wavelength refers to the distance between consecutive peaks (or troughs) of a wave. In optics, it is typically measured in nanometers (nm), and different wavelengths are responsible for different colors of light in the visible spectrum.

Shorter wavelengths, such as violet light at 420 nm, will appear closer together, while longer wavelengths, such as green light at 560 nm, are spaced further apart. These variations in wavelength dictate how waves interfere with one another, especially when considering complex interactions like those in a thin film.
  • Changing the wavelength changes the interference pattern.
  • In our example, the soap film was adjusted to pass from reflecting violet to reflecting green light, which required an increase in thickness.
Wavelength plays a pivotal role in adjusting the thin film's thickness to achieve the desired constructive interference. The precise tuning of the film thickness is necessary to ensure the preferred color stands out due to constructive overlap of waves.

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Most popular questions from this chapter

A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if (a) the wavelength is doubled, (b) the slit width is doubled, or (c) the distance from the slit to the screen is doubled?

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 \(1.336 .\) )

Two sources emit waves that are in phase with cach other. What is the longest wavelength that will give constructive interference at an observation point \(161 \mathrm{m}\) from one source and \(295 \mathrm{m}\) from the other source?

A spy camera is said to be able to read the numbers on a car's license plate. If the numbers on the plate are \(5.0 \mathrm{cm}\) apart, and the spy satellite is at an altitude of \(160 \mathrm{km},\) what must be the diameter of the camera's aperture? (Assume light with a wavelength of \(550 \mathrm{nm} .\) )

(a) If a thin liquid film floating on water has an index of refraction less than that of water, will the film appear bright or dark in reflected light as its thickness goes to zero? (b) Choose the best explanation from among the following: I. The film will appear bright because as the thickness of the film goes to zero the phase difference for reflected rays goes to zero. II. The film will appear dark because there is a phase change at both interfaces, and this will cause destructive interference of the reflected rays.
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