/*! 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 50 Searching for Starspots. The Hal... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 36: Problem 50

Searching for Starspots. The Hale Telescope on Palomar Mountain in California has a mirror 200 in. \((5.08 \mathrm{m})\) in diameter and it focuses visible light. Given that a large sunspot is about \(10,000\) mi in diameter, what is the most distant star on which this telescope could resolve a sunspot to see whether other stars have them? (Assume optimal viewing conditions, so that the resolution is diffraction limited.) Are there any stars this close to us, besides our sun?

Short Answer

Expert verified
The most distant resolvable sunspot is 1.29 light years away, with no stars that close except our Sun.

Step by step solution

01

Understanding the Diffraction Limit

The telescope's resolution is limited by diffraction, which can be calculated using the formula \( \theta = 1.22 \times \frac{\lambda}{D} \), where \( \theta \) is the angular resolution in radians, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the telescope's mirror in meters.
02

Convert the Sunspot Size

First, convert the size of the sunspot from miles to meters. Since 1 mile equals 1,609.34 meters, a sunspot 10,000 miles in diameter would be \( 10,000 \times 1,609.34 = 16,093,400 \) meters in diameter.
03

Calculate the Angular Resolution

Assume visible light has a wavelength of approximately 550 nm or \( 550 \times 10^{-9} \) meters. Substitute \( \lambda = 550 \times 10^{-9} \) meters and \( D = 5.08 \) meters into the formula to get \( \theta = 1.22 \times \frac{550 \times 10^{-9}}{5.08} \approx 1.32 \times 10^{-7} \) radians.
04

Calculate Maximum Distance for Resolution

To find the maximum distance at which the telescope can resolve a sunspot, use the small angle formula \( \theta = \frac{d}{L} \), where \( d \) is the diameter of the object (16,093,400 meters for the sunspot) and \( L \) is the distance to the object. To solve for \( L \), rearrange to \( L = \frac{d}{\theta} \). Substitute the known values to get \( L = \frac{16,093,400}{1.32 \times 10^{-7}} \approx 1.22 \times 10^{14} \) meters or approximately 12,200,000,000 km.
05

Convert Distance to Light Years

Convert the distance from kilometers to light years for astronomical context, knowing 1 light year is approximately 9.461 x 10^12 km. \( L \approx \frac{12,200,000,000,000}{9.461 \times 10^{12}} \approx 1.29 \) light years.
06

Assess the Existence of Stars at the Calculated Distance

Apart from the Sun, the closest known star is Proxima Centauri, which is about 4.24 light-years away. Since 1.29 light years is much less than 4.24 light-years, no other known stars besides our Sun are within the detectable limit for resolving a sunspot with this telescope.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Angular Resolution
Angular resolution is the ability of a telescope to distinguish between two closely spaced objects in the sky. It's like understanding how sharply an image can be seen. The smaller the angular resolution, the clearer the telescope can separate two nearby objects. Angular resolution is calculated using the formula: \[ \theta = 1.22 \times \frac{\lambda}{D} \] where \( \theta \) is the angular resolution in radians, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the telescope's mirror. This formula shows that angular resolution improves (i.e., \( \theta \) becomes smaller) either by using a larger telescope mirror, thus increasing \( D \), or by observing shorter wavelengths where \( \lambda \) is smaller. For the Hale Telescope, this means that its angular resolution in visible light can define the smallest features we might discern on distant astronomical objects. Understanding this helps astronomers determine if a telescope can observe details like sunspots on distant stars.
Telescope Resolution
Telescope resolution is a crucial factor that describes the telescope's ability to present detailed images. The resolution is inherently limited by the diffraction of light as it passes through the telescope. When discussing the resolution of the Hale Telescope, the size of the mirror (5.08 meters in diameter) becomes very important. A telescope with a larger diameter mirror can collect more light and generally achieves a better resolution. This is why telescopes with bigger mirrors are preferred for observing distant and faint objects in the universe. However, resolution is also dependent on various other factors:
  • Atmospheric conditions: Clear skies offer better viewing conditions.
  • Wavelength used: Different wavelengths can affect diffraction and thus resolution.
Even with perfect conditions, far-away tiny features such as sunspots on distant stars can only be resolved if they fall within the telescope's diffraction limit. Thus, a telescope's resolution is key for observing detailed astronomical phenomena.
Visible Light Wavelength
Wavelength is an important parameter when discussing the resolution capabilities of telescopes, because it determines how light is diffracted. Visible light, which is what human eyes can perceive, ranges in wavelength approximately from 400 nm (nanometers) to 700 nm. For the Hale Telescope, the calculations typically use an average value of about 550 nm. This wavelength range can affect how clearly specific features like stars or sunspots are resolved. Here's how wavelength impacts resolution:
  • A shorter wavelength (like blue light, close to 400 nm) means better angular resolution, allowing for finer details to be seen.
  • A longer wavelength (like red light, near 700 nm) reduces the resolution, making it harder to distinguish fine detail.
When scientists calculate the telescope's resolution, they assume optimal conditions using these known wavelengths of visible light. This helps determine the kind of astronomical detail visible through the telescope.
Astronomical Distance Calculation
Understanding how to calculate distances in space is crucial for astronomy, especially for determining whether specific celestial features are observable.For example, using the Hale Telescope to identify sunspots on distant stars involves calculating the maximum distance at which the telescope's resolution remains effective. This calculation uses the formula:\[ L = \frac{d}{\theta} \] where \( L \) is the distance to the object, \( d \) is the diameter of the object (like a sunspot), and \( \theta \) is the angular resolution. With this formula, astronomers can determine how far away a telescope can clearly distinguish sunspots or other small features. In this case, after calculations, the distance was found to be approximately 1.29 light years, indicating that no known stars other than the Sun are close enough for such detailed observations. Being able to convert these calculations into different units like kilometers or light years is also vital for contextualizing astronomical distances effectively. This understanding informs astronomers on the limitations of observing extraterrestrial features with available technology.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Observing Jupiter. You are asked to design a space telescope for earth orbit. When Jupiter is \(5.93 \times 10^{8} \mathrm{km}\) away (its closest approach to the earth), the telescope is to resolve, by Rayleigh's criterion, features on Jupiter that are 250 \(\mathrm{km}\) apart. What minimum- diameter mirror is required? Assume a wavelength of 500 \(\mathrm{nm} .\)

An astronaut in the space shuttle can just resolve two point sources on earth that are 65.0 \(\mathrm{m}\) apart. Assume that the resolution is diffraction limited and use Rayleigh's criterion. What is the astronaut's altitude above the earth? Treat his eye as a circular aperture with a diameter of 4.00 \(\mathrm{mm}\) (the diameter of his pupil), and take the wavelength of the light to be 550 \(\mathrm{nm} .\) Ignore the effect of fluid in the eye.

Resolution of the Eye. The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0\(\mu \mathrm{m}\) in diameter) limits the size of an object at the near point \((25 \mathrm{cm})\) of the eye to a height of about 50\(\mu \mathrm{m} .\) To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50 -\mum-tall object at 25 \(\mathrm{cm}\) from the eye with light of wavelength 550 \(\mathrm{nm}\) ? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the \(25-\mathrm{cm}\) near point with light of wavelength 550 \(\mathrm{nm} ?(\mathrm{c})\) What angle would the object in part (b) subtend at the eye? Express your answer in minutes \(\left(60 \min =1^{\circ}\right),\) and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

A slit 0.360 \(\mathrm{mm}\) wide is illuminated by parallel rays of light that have a wavelength of 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 1.20 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\) . (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?\)

NASA is considering a project called Planet Imager that would give astronomers the ability to see details on planets orbiting other stars. Using the same principle as the Very Large Array (see Section 36.7 ), Planet Imager will use an array of infrared telescopes spread over thousands of kilometers of space. (Visible light would give even better resolution. Unfortunately, at visible wavelengths, stars are so bright that a planet would be lost in the glare. This is less of a problem at infrared wavelengths.) (a) If Planet Imager has an effective diameter of \(6000 \mathrm{~km}\) and observes infrared radiation at a wavelength of \(10 \mu \mathrm{m},\) what is the greatest distance at which it would be able to observe details as small as \(250 \mathrm{~km}\) across (about the size of the greater Los Angeles area) on a planet? Give your answer in light-years (see Appendix E). (Hint: Use Rayleigh's criterion.) (b) For comparison, consider the resolution of a single infrared telescope in space that has a diameter of \(1.0 \mathrm{~m}\) and that observes \(10-\mu \mathrm{m}\) radiation. What is the size of the smallest details that such a telescope could resolve at the distance of the nearest star to the sun, Proxima Centauri, which is 4.22 light-years distant? How does this compare to the diameter of the earth \(\left(1.27 \times 10^{4} \mathrm{~km}\right) ?\) To the average distance from the earth to the sun \(\left(1.50 \times 10^{8} \mathrm{~km}\right) ?\) Would a single telescope of this kind be able to detect the presence of a planet like the earth, in an orbit the size of the earth's orbit, around any other star? Explain. (c) Suppose Planet Imager is used to observe a planet orbiting the star 70 Virginis, which is 59 lightyears from our solar system. A planet (though not an earthlike one) has in fact been detected orbiting this star, not by imaging it directly but by observing the slight "wobble" of the star as both it and the planet orbit their common center of mass. What is the size of the smallest details that Planet Imager could hope to resolve on the planet of 70 Virginis? How does this compare to the diameter of the planet, assumed to be comparable to that of Jupiter \(\left(1.38 \times 10^{5} \mathrm{~km}\right) ?\) (Although the planet of 70 Virginis is thought to be at least 6.6 times more massive than Jupiter, its radius is probably not too different from that of Jupiter. The reason is that such large planets are thought to be composed primarily of gases, not rocky material, and hence can be greatly compressed by the mutual gravitational attraction of different parts of the planet.)
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Wave Optics

Read Explanation

Solid State Physics

Read Explanation

Energy Physics

Read Explanation

Fields in Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.