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Chapter 27: Problem 35

Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the pupils of your eyes have diameters of about \(7.0 \mathrm{mm}\). The taillights of this car are separated by a distance of \(1.2 \mathrm{m}\) and emit red light (wavelength \(=660 \mathrm{nm}\) in vacuum). How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?

Short Answer

Expert verified
The car is about 10435 meters away.

Step by step solution

01

Understand the problem

We need to find the distance at which two car taillights appear as a single light due to diffraction. Diffraction affects how light spreads through small openings, like your eyes' pupils.
02

Apply Rayleigh's Criterion

According to Rayleigh's criterion for two points of light to be just resolvable, the angular resolution limit is \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the aperture (pupil in this case).
03

Calculate the Angular Resolution

Substitute the given values: wavelength of red light \( \lambda = 660 \times 10^{-9} \) meters and pupil diameter \( D = 7.0 \times 10^{-3} \) meters into Rayleigh's criterion: \[ \theta = 1.22 \frac{660 \times 10^{-9}}{7.0 \times 10^{-3}} \approx 1.15 \times 10^{-4} \text{ radians} \].
04

Relate Angular Resolution to Distance

The two taillights form an angle \( \theta = \frac{d}{L} \), where \( d = 1.2 \) meters (separation of lights) and \( L \) is the distance we want to find. Set this equal to the angular resolution: \( 1.15 \times 10^{-4} = \frac{1.2}{L} \).
05

Solve for Distance

Solve for distance \( L \): \[ L = \frac{1.2}{1.15 \times 10^{-4}} \approx 10435 \text{ meters} \].

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Key Concepts

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

Rayleigh's Criterion
The Rayleigh's criterion is a principle used to determine the minimum angle at which two separate points of light can be just distinguished by an optical system. It is a vital concept in the field of optics and provides a specific limit, allowing us to quantify when two light sources, such as the taillights of a car, become indistinguishable from one another. When viewing distant objects, diffraction causes the light to spread out, creating a challenge for resolving distinct images.
  • The criterion is expressed mathematically by the formula: \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution limit.
  • \( \lambda \) is the wavelength of the light being used, while \( D \) is the diameter of the viewing aperture, like your eye's pupil.
By utilizing Rayleigh's criterion, one can set this theoretical resolving power limit and use it for practical purposes like observing the distance at which merging occurs.
Angular Resolution
Angular resolution pertains to an optical system's ability to distinguish fine details within the field of view. It is directly linked to Rayleigh's criterion, emphasizing how closely two light sources can be placed before they appear as one.
  • Angular resolution is inversely proportional to the distance at which the separate objects can be perceived: as the angle decreases, the distance to the point of merging increases.
  • It can be calculated using the formula for Rayleigh's criterion, where smaller angle values represent better resolution capabilities.
In the given problem, the taillights of the car are separated by a known distance, and the angular resolution helps determine the point where those lights merge into a single visual entity from your perspective. This logical approach is significant for understanding not only light behavior but also applications in diverse fields like astronomy and optical engineering.
Wavelength of Light
Wavelength describes the distance between successive peaks of a wave and is a key attribute in describing electromagnetic waves like light. Light's wavelength is central to understanding how diffraction influences visibility and differentiation of light sources.
  • Wavelengths are typically measured in nanometers (nm) or meters (m), with visible light ranging from about 400 nm (violet) to 700 nm (red).
  • In this exercise, red light at a wavelength of 660 nm is used to analyze the diffraction and merging effects of car taillights.
Since diffraction is wavelength-dependent, longer wavelengths such as red light cause more significant diffraction. This phenomenon helps illustrate why red light tends to blur into other sources more readily than shorter waves, directly impacting how we perceive overlapping or merging light sources in everyday scenarios.

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Most popular questions from this chapter

The wavelength of the laser beam used in a compact disc player is \(780 \mathrm{nm}\). Suppose that a diffraction grating produces first-order tracking beams that are \(1.2 \mathrm{mm}\) apart at a distance of \(3.0 \mathrm{mm}\) from the grating. Estimate the spacing between the slits of the grating.

A dark fringe in the diffraction pattern of a single slit is located at an angle of \(\theta_{\mathrm{A}}=34^{\circ} .\) With the same light, the same dark fringe formed with another single slit is at an angle of \(\theta_{\mathrm{B}}=56^{\circ} .\) Find the ratio \(W_{A} / W_{\mathrm{B}}\) of the widths of the two slits.

A hunter who is a bit of a braggart claims that from a distance of \(1.6 \mathrm{km}\) he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of \(498 \mathrm{nm}\) (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to \(8 \mathrm{mm},\) the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of 498 nm.

You are standing in air and are looking at a flat piece of glass \((n=1.52)\) on which there is a layer of transparent plastic \((n=1.61) .\) Light whose wavelength is \(589 \mathrm{nm}\) in vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.

A flat screen is located \(0.60 \mathrm{m}\) away from a single slit. Light with a wavelength of \(510 \mathrm{nm}\) (in vacuum) shines through the slit and produces a diffraction pattern. The width of the central bright fringe on the screen is \(0.050 \mathrm{m}\). What is the width of the slit?
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