/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 The wings of tiger beetles (Fig.... [FREE SOLUTION] | 91Ó°ÊÓ

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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 mm as the diameter of your pupil.

Short Answer

Expert verified
The viewing distance should be approximately 27 cm.

Step by step solution

01

Understanding Rayleigh's Criterion

Rayleigh's criterion defines the minimum angular separation at which two points can be resolved. According to this criterion, the resolving angle \( \theta \) can be calculated using the formula: \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the aperture (pupil in this case).
02

Calculating the Resolving Angle

Substitute \( \lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m} \) and \( D = 3.00 \text{ mm} = 3.00 \times 10^{-3} \text{ m} \) into the Rayleigh criterion equation: \[ \theta = 1.22 \frac{550 \times 10^{-9}}{3.00 \times 10^{-3}} \]. Calculating this gives \( \theta \approx 2.233 \times 10^{-4} \text{ radians} \).
03

Estimating the Viewing Distance Using Geometry

The angular resolution \( \theta \) relates to the distance \( L \) from the wing to the observer and the resolved patch size \( s \) using simple trigonometry: \( \theta \approx \frac{s}{L} \). Here \( s = 60 \mu m = 60 \times 10^{-6} \text{ m} \). Rearrange to find \( L \): \( L = \frac{s}{\theta} \).
04

Calculating the Viewing Distance

Substitute \( s = 60 \times 10^{-6} \text{ m} \) and \( \theta = 2.233 \times 10^{-4} \text{ radians} \) into the distance formula: \[ L = \frac{60 \times 10^{-6}}{2.233 \times 10^{-4}} \]. This calculates to \( L \approx 0.269 \text{ m} \), or approximately 27 cm.

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Key Concepts

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

Resolving Angle
When we talk about the resolving angle, we're referring to how well an optical system can distinguish between two adjacent images. This concept is central to Rayleigh's criterion, which offers a quantitative way to determine if two points of light will appear distinct or blurred together. By using the formula \( \theta = 1.22 \frac{\lambda}{D} \), where \(\lambda\) is the wavelength of light and \(D\) is the diameter of the aperture (your eye’s pupil, in this context), we can calculate this resolving angle. This calculation helps us understand how close two objects can be yet still appear distinctly separate when viewed from a distance. In practice, the smaller the resolving angle, the better the system can distinguish between closely spaced features. This is crucial in applications such as observing distant stars or, in this case, viewing the intricate patterns on a wing of a tiger beetle.
Thin-Film Interference
Thin-film interference is a fascinating phenomenon that occurs when light waves reflect off the two surfaces of a thin film, creating patterns of constructive or destructive interference. When these light waves overlap, they can amplify or cancel each other out, leading to brilliant and colorful effects. This principle is what gives tiger beetle wings their varied colors. A thin layer on or within the wing causes light waves of certain wavelengths to interfere constructively, amplifying those colors to the observer. A change in viewing angle can alter the path difference between light waves reflecting from the top and bottom surfaces of the film. Hence, the colors seem to shift based on the observer’s perspective, combining elements of physics and art to produce dynamic, iridescent colors. Thin-film interference is not only fascinating to watch but is also vital in the design of anti-reflective coatings and optical instruments.
Wavelength of Light
The wavelength of light is a key factor in various optical phenomena, including diffraction and interference, playing a vital role in Rayleigh's criterion. Light exists as a wave, and the wavelength is the distance between consecutive peaks of these waves. It is typically measured in nanometers (nm), with visible light ranging approximately from 380 nm (violet) to 750 nm (red). A specific wavelength, like the 550 nm used in our problem, is within the green portion of the visible spectrum. This wavelength affects how light interacts with objects and materials, such as through reflection, refraction, and interference. When resolving features like the colorful patches on a beetle's wing, selecting a particular wavelength can influence the outcome of measurements, as shorter wavelengths generally allow for finer resolution. Knowing and applying the correct wavelength is essential for accurate calculations in optical setups.

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Most popular questions from this chapter

The pupil of a person's eye has a diameter of \(5.00 \mathrm{~mm}\). According to Rayleigh's criterion, what distance apart must two small objects be if their images are just barely resolved when they are \(250 \mathrm{~mm}\) from the eye? Assume they are illuminated with light of wavelength \(500 \mathrm{~nm}\).

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle's "shadow region." Previously, television signals had a wavelength of about \(50 \mathrm{~cm}\), but digital television signals that are transmitted from towers have a wavelength of about \(10 \mathrm{~mm}\). (a) Did this change in wavelength increase or decrease the diffraction of the signals into the shadow regions of obstacles? Assume that a signal passes through an opening of \(5.0 \mathrm{~m}\) width between two adjacent buildings. What is the angular spread of the central diffraction maximum (out to the first minima) for wavelengths of (b) 50 \(\mathrm{cm}\) and \((\mathrm{c}) 10 \mathrm{~mm} ?\)

Perhaps to confuse a predator, some tropical gyrinid beetles (whirligig beetles) are colored by optical interference that is due to scales whose alignment forms a diffraction grating (which scatters light instead of transmiting it). When the incident light rays are perpendicular to the grating, the angle between the first-order maxima (on opposite sides of the zeroth-order maximum) is about \(26^{\circ}\) in light with a wavelength of \(550 \mathrm{~nm}\). What is the grating spacing of the beetle?

A diffraction grating is made up of slits of width \(300 \mathrm{~nm}\) with separation \(900 \mathrm{~nm}\). The grating is illuminated by monochromatic plane waves of wavelength \(\lambda=600 \mathrm{~nm}\) at normal incidence. (a) How many maxima are there in the full diffraction pattern? (b) What is the angular width of a spectral line observed in the first order if the grating has 1000 slits?

Millimeter-wave radar generates a narrower beam than conventional microwave radar, making it less vulnerable to antiradar missiles than conventional radar. (a) Calculate the angular width \(2 \theta\) of the central maximum, from first minimum to first minimum, produced by a 220 GHz radar beam emitted by a \(55.0-\) cm-diameter circular antenna. (The frequency is chosen to coincide with a low-absorption atmospheric "window.") (b) What is \(2 \theta\) for a more conventional circular antenna that has a diameter of \(2.3\) \(\mathrm{m}\) and emits at wavelength \(1.6 \mathrm{~cm} ?\)

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