/*! 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 48 \(\bullet$$\bullet\) Resolution ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) Resolution of telescopes. Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcsec- onds in an arcminute).(a) Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolu- tion with \(550-\) nm light. (b) Increasing the telescope diameter beyond the value found in part (a) will increase the light- gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10 -m diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope \((5-m\) diameter) on Palomar Mountain in California? In what ways are they not superior? Explain.

Short Answer

Expert verified
(a) The diameter needed is approximately 0.45 m. (b) Keck gathers more light but has similar resolution limits to Hale.

Step by step solution

01

Understand Rayleigh's Criterion

Rayleigh's criterion provides a formula to calculate the minimum resolvable angle (resolution) for a telescope. The formula is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the resolution in radians, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the telescope's aperture.
02

Convert Resolution to Radians

Convert the given resolution from arcseconds to radians using the conversion factor \(1\text{ arcsecond} = \frac{1}{60 \times 60 \times 180/\pi} \) radians. The given resolution is 0.3 arcseconds, so \( \theta = 0.3 \times \frac{\pi}{180 \times 60 \times 60}\text{ radians}\).
03

Calculate the Diameter

Solve the Rayleigh's criterion formula for \( D \) using the converted \( \theta \) and \( \lambda = 550 \) nm (which is \( 550 \times 10^{-9} \) meters): \( D = 1.22 \times \frac{\lambda}{\theta} \). Calculate \( D \) and find its value.
04

Compare Telescope Features

Keck telescopes, each of 10 m diameter, can gather more light compared to the Hale telescope, which has a 5 m diameter. The larger aperture of Keck increases its light-gathering power, enabling observation of fainter objects. However, due to atmospheric distortion, both Keck and Hale have similar resolution limits for visible light on Earth.

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 91Ó°ÊÓ!

Key Concepts

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

Telescope Resolution
Telescope resolution refers to the ability of a telescope to distinguish between two closely spaced objects in the sky. This ability is crucial for observing the fine details in astronomical objects, such as the surface features on planets or the tight grouping of stars in a distant galaxy.

Rayleigh's Criterion is typically used to determine a telescope's resolution. It states that the minimum angle between two distinguishable sources depends on the wavelength of light (\( \lambda \)) and the diameter of the telescope's aperture (\( D \)). The formula is: \[ \theta = 1.22 \frac{\lambda}{D} \] where:
  • \( \theta \) is the smallest resolvable angle in radians.
  • \( \lambda \) is the light wavelength, generally given in meters.
  • \( D \) is the aperture diameter of the telescope.
Understanding these components helps astronomers decide how to build more effective earth-based telescopes for clear observations.
Atmospheric Distortion
Atmospheric distortion is a significant hindrance to telescope observations from Earth. As light from celestial bodies travels through the Earth's atmosphere, it encounters varying air densities that cause the light to scatter and blur. This effect is known as "seeing" and essentially limits the resolution of ground-based telescopes.

Factors influencing atmospheric distortion include:
  • Temperature variations causing air turbulence.
  • Humidity and particle content in the air.
  • Weather patterns affecting air stability.
Telescopes are often placed in high altitude locations to reduce these effects, as thinner air at higher elevations typically provides a clearer view. However, despite these measures, the turbulence still imposes a limit on the optical clarity achievable.
Light Wavelength
The wavelength of light is a measure of distance between successive peaks of a wave and it significantly affects telescope resolution. Visible light has wavelengths ranging approximately from 400 nm (violet) to 700 nm (red). In astrophotography, using specific wavelengths provides different levels of detail.

In the context of Rayleigh’s Criterion, using shorter wavelengths (e.g., blue or violet) generally yields better resolution because:
  • \( \theta \) decreases with shorter wavelengths, enhancing detail detection.
  • This is crucial when observers need to discern small details in bright celestial objects.
Choosing the optimal wavelength depends largely on the type of observation and the specific features astronomers are interested in studying.
Astronomical Observation
Astronomical observation using telescopes allows scientists to study planets, stars, galaxies, and other celestial phenomena. The choice of telescope specifications, like resolution and aperture size, can significantly affect the outcome and quality of these observations.

Key considerations include:
  • Larger telescopes, like the 10-meter Keck telescopes, boast better light-gathering ability, allowing astronomers to observe fainter objects.
  • However, beyond a certain point, increasing the aperture size does not enhance resolution due to atmospheric limitations.
  • Space-based telescopes bypass the issue of atmospheric distortion altogether, offering superior resolution for the same size due to the pristine environment of space.
Overall, careful design and location choice, along with advancements in technology, drive the continuous improvement of astronomical observations.

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

\(\bullet$$\bullet\) A glass plate \((n=1.53)\) that is 0.485\(\mu \mathrm{m}\) thick and sur- rounded by air is illuminated by a beam of white light normal to the plate. What wavelengths (in air) within the limits of the visible spectrum \((\lambda=400 \mathrm{nm}\) to 700 \(\mathrm{nm})(\mathrm{a})\) are intensified in the reflected beam, (b) are cancelled in the reflected light?

\(\bullet\) Nonglare glass. When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a thin film to cancel part of the glare. (a) If the glass has a refractive index of 1.62 and you use \(\mathrm{TiO}_{2}\) , which has an index of refraction of \(2.62,\) as the coating, what is the minimum film thickness that will cancel light of wavelength 505 \(\mathrm{nm}\) ? (b) If this coating is too thin to stand up to wear, what other thicknesses would also work? Find only the three thinnest ones.

\(\bullet\) Coherent light of frequency \(6.32 \times 10^{14}\) Hz passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

\(\bullet\) A plate of glass 9.00 \(\mathrm{cm}\) long is placed in contact with a sec- ond plate and is held at a small angle with it by a metal strip 0.0800 \(\mathrm{mm}\) thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 \(\mathrm{nm}\) . How many interfer- ence fringes are observed per centimeter in the reflected light?

\(\bullet\) A thin film of polystyrene of refractive index 1.49 is used as a nonreflecting coating for Fabulite (strontium titanate) of refractive index \(2.409 .\) What is the minimum thickness of the film required? Assume that the wavelength of the light in air is 480 \(\mathrm{nm}\) .

See all solutions

Recommended explanations on Physics Textbooks

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.