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Chapter 36: Problem 28

The wings of tiger beetles (Fig. \(36-41\) ) are colored by interference due to thin cuticle-like layers. In addition, these layers are arranged in patches that are \(60 \mu \mathrm{m}\) across and produce different colors. The color you see is a pointillistic mixture of thin-film interference colors that varies with perspective. Approximately what viewing distance from a wing puts you at the limit of resolving the different colored patches according to Rayleigh's criterion? Use \(550 \mathrm{~nm}\) as the wavelength of light and \(3.00 \mathrm{~mm}\) as the diameter of your pupil.

Short Answer

Expert verified
The limit of resolving the different colored patches is approximately 26.8 cm away from the wing.

Step by step solution

01

Understand Rayleigh's criterion

Rayleigh's criterion states that two images are just resolvable when the center of the diffraction pattern of one image coincides with the first minimum of the diffraction pattern of the other. Mathematically, the minimum resolvable angle \( \theta \) is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the aperture (pupil in this case).
02

Convert units

We need to ensure consistent units in our calculations. \( \lambda = 550 \mathrm{~nm} = 550 \times 10^{-9} \mathrm{~m} \) and \( D = 3.00 \mathrm{~mm} = 3.00 \times 10^{-3} \mathrm{~m} \).
03

Calculate the minimum resolvable angle \( \theta \)

Substitute the values of \( \lambda \) and \( D \) into Rayleigh's equation. This gives:\[ \theta = 1.22 \times \frac{550 \times 10^{-9}}{3.00 \times 10^{-3}} \approx 2.237 \times 10^{-4} \text{ radians} \]
04

Relate angle to viewing distance

Using the small angle approximation, \( \theta \approx \frac{s}{L} \), where \( s \) is the size of the patch (\( 60 \mu \mathrm{m} = 60 \times 10^{-6} \mathrm{~m} \)) and \( L \) is the viewing distance. Rearrange this to find \( L \): \[ L = \frac{s}{\theta} = \frac{60 \times 10^{-6}}{2.237 \times 10^{-4}} \approx 0.268 \text{ m} \]
05

Convert the viewing distance to practical units

Convert \( L \) into centimeters for a more commonly used unit: \[ L \approx 26.8 \mathrm{~cm} \]

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Key Concepts

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

Thin-Film Interference
Thin-film interference occurs when light waves reflect off multiple surfaces to create colorful patterns. Imagine the thin layer on a soap bubble or an oily puddle. These beautiful colors arise because light waves bouncing off the top and bottom surfaces interfere with each other.
  • When waves meet, they may enhance each other, creating vivid colors (constructive interference).
  • Alternatively, they may cancel each other out, resulting in no light or a dark spot (destructive interference).
The light's wavelength, the thin film's thickness, and the angle at which light hits influence these interactions.

In the case of tiger beetle wings, thin layers act like these films. They create patterns of colors based on how they interfere with specific wavelengths of light. Each patch of color you see depends on the light's wavelength used to illuminate it.
Diffraction Pattern
A diffraction pattern results from light bending around the edges of an object. You can visualize this by imagining waves of water that curl and spread when they hit a barrier.

When light passes through an aperture, like the pupil of your eye, it scatters into a pattern.
  • This pattern consists of bright central areas flanked by decreasing bright and dark regions, much like ripples in water.
  • The core concept here is that light doesn't just move in straight lines but also spreads to occupy space around obstacles.
In viewing beetle wings, diffraction can alter how we perceive their colors.

Understanding the diffraction pattern is crucial when trying to distinguish fine details in a colorful array. By interpreting these patterns, one can resolve different colored patches on beetles according to Rayleigh's criterion.
Resolvable Angle
The resolvable angle differentiates between two points in a field of view. Rayleigh's criterion helps determine this, and it tells us at what angle two distinct points become distinguishable.

The formula is \( \theta = 1.22 \frac{\lambda}{D} \), where
  • \( \lambda \) is the wavelength of light, which dictates the color we see.
  • \( D \) is the aperture's diameter, here measured as the eye's pupil.
When you want to resolve light patches on beetle wings:
  • The smaller the angle \( \theta \), the closer two points can be for them to be recognized as distinct features.
Considering the beetle's wing, resolution varies with light's wavelength, demanding precision to separate each colorful patch visually.
Wavelength of Light
The wavelength of light is a critical parameter affecting many optical phenomena.

Wavelength, defined as the distance between consecutive peaks of light waves, largely determines what color we perceive.
  • Shorter wavelengths correspond to blue and violet, while longer wavelengths relate to red and orange areas.
  • In many optical instruments and phenomena like thin-film interference and diffraction, we use the specific value of wavelength for precise calculations.
In our beetle example, using 550 nm as the wavelength becomes a standard measure for calculations due to its representation of green light, often central to human perception.

Selecting a specific wavelength allows us to accurately model how light will interact with beetle wing structures. Thus, wavelengths aren't just abstract numbers but key elements guiding our understanding and predictions of light's behavior.

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Most popular questions from this chapter

The telescopes on some commercial surveillance satellites can resolve objects on the ground as small as \(85 \mathrm{~cm}\) across (see Google Earth), and the telescopes on military surveillance satellites reportedly can resolve objects as small as \(10 \mathrm{~cm}\) across. Assume first that object resolution is determined entirely by Rayleigh's criterion and is not degraded by turbulence in the atmosphere. Also assume that the satellites are at a typical altitude of \(400 \mathrm{~km}\) and that the wavelength of visible light is \(550 \mathrm{~nm}\). What would be the required diameter of the telescope aperture for (a) \(85 \mathrm{~cm}\) resolution and (b) \(10 \mathrm{~cm}\) resolution? (c) Now, considering that turbulence is certain to degrade resolution and that the aperture diameter of the Hubble Space Telescope is \(2.4 \mathrm{~m}\), what can you say about the answer to (b) and about how the military surveillance resolutions are accomplished?

Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(2000 \mathrm{~km}\). One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of \(1.40 \mathrm{~nm}\). The element that emits light is the end of a wire with diameter \(0.200 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) What is the ratio of the beam intensity at the target to that at the end of the wire? (The laser is fired from space, so neglect any atmospheric absorption.)

Find the separation of two points on the Moon's surface that can just be resolved by the 200 in. \((=5.1 \mathrm{~m})\) telescope at Mount Palomar, assuming that this separation is determined by diffraction effects. The distance from Earth to the Moon is \(3.8 \times\) \(10^{5} \mathrm{~km}\). Assume a wavelength of \(550 \mathrm{~nm}\) for the light.

With light from a gaseous discharge tube incident normally on a grating with slit separation \(1.73 \mu \mathrm{m}\), sharp maxima of green light are experimentally found at angles \(\theta=\pm 17.6^{\circ}, 37.3^{\circ},-37.1^{\circ}\), \(65.2^{\circ}\), and \(-65.0^{\circ}\). Compute the wavelength of the green light that best fits these data.

An x-ray beam of a certain wavelength is incident on an \(\mathrm{NaCl}\) crystal, at \(30.0^{\circ}\) to a certain family of reflecting planes of spacing \(39.8 \mathrm{pm}\). If the reflection from those planes is of the first order. what is the wavelength of the \(\mathrm{x}\) rays?
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