/*! 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 39 Astronomers have discovered a pl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 39

Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance of \(4.2 \times 10^{17} \mathrm{m}\) from the earth. One planet is believed to be located at a distance of \(1.2 \times 10^{11} \mathrm{m}\) from the star. Using visible light with a vacuum wavelength of \(550 \mathrm{nm}\), what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?

Short Answer

Expert verified
The telescope must have an aperture diameter of at least 2.34 mm.

Step by step solution

01

Understand the Problem

We need to find the minimum aperture diameter of the telescope to resolve the planet from the star using visible light. This is a diffraction-limited problem, best approached using the Rayleigh criterion formula.
02

Apply the Rayleigh Criterion Formula

The Rayleigh criterion for resolving two objects is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the aperture diameter. \( \theta \) is the angular separation.
03

Calculate Angular Separation

Calculate the angular separation \( \theta \) between the planet and the star using \( \theta = \frac{d}{L} \), where \( d = 1.2 \times 10^{11} \mathrm{m} \) is the distance between the planet and the star, and \( L = 4.2 \times 10^{17} \mathrm{m} \) is the distance to the star.\[ \theta = \frac{1.2 \times 10^{11}}{4.2 \times 10^{17}} = 2.857 \times 10^{-7} \text{ radians} \]
04

Solve for Telescope Aperture Diameter

Using the Rayleigh criterion \( \theta = 1.22 \frac{\lambda}{D} \), convert our wavelength from nm to meters: \( \lambda = 550 \times 10^{-9} \mathrm{m} \). Solve for \( D \):\[ D = 1.22 \frac{\lambda}{\theta} = 1.22 \frac{550 \times 10^{-9}}{2.857 \times 10^{-7}} \approx 0.00234 \text{ meters} \]
05

Convert Diameter to Millimeters

Convert the diameter from meters to millimeters:\[ D = 0.00234 \text{ m} \times 1000 = 2.34 \text{ mm} \]

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.

Rayleigh Criterion
The Rayleigh Criterion is a key principle for understanding how telescopes resolve two closely spaced point sources, like stars or planets. This criterion states that two objects are considered resolved when the center of one object's diffraction pattern coincides with the first minimum of the other's diffraction pattern.
The formula for Rayleigh Criterion is:\[ \theta = 1.22 \frac{\lambda}{D} \]
where:
  • \(\theta\) is the angular separation between the objects.
  • \(\lambda\) is the wavelength of the light used.
  • \(D\) is the diameter of the telescope's aperture.

In simple terms, the larger the aperture \(D\), the smaller the angle \(\theta\), and the finer the details that can be resolved. This is crucial in astronomy as it helps determine the capability of telescopes to distinguish fine details of celestial objects.
Angular Separation
Angular Separation is a measure of how far apart two objects appear in the sky. It鈥檚 like measuring the angle between two spots as seen from Earth. This separation is vital for astronomers to determine how close objects like stars and planets are to each other from our point of view.
The formula for calculating angular separation \(\theta\) is:
\[ \theta = \frac{d}{L} \]
where:
  • \(d\) is the physical distance between the two objects.
  • \(L\) is the distance from the observer (Earth) to the objects.

Angular separation allows us to understand the spatial relationship between celestial bodies. A smaller \(\theta\) implies the objects appear closer together in the sky, requiring better telescope resolution to distinguish them separately.
Diffraction Limit
The Diffraction Limit refers to the fundamental limit on the resolving power of any optical system, governed by the wave nature of light. When light waves pass through the aperture of a telescope, they spread out, creating a diffraction pattern. This phenomenon is unavoidable and sets a limit on the resolution of telescopic images.
The formula to express this limit is tightly connected to the Rayleigh Criterion:
\[ D = 1.22 \frac{\lambda}{\theta} \]
where:
  • \(D\) is the aperture diameter.
  • \(\lambda\) is the wavelength of light.
  • \(\theta\) is the angular separation.

In essence, the diffraction limit defines the smallest detail size that can be resolved. The smaller the wavelength and larger the aperture, the finer the details that can be discerned. This is critical information for designing and using telescopic equipment in astronomy to obtain sharper, more detailed images of the cosmos.

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

Two stars are \(3.7 \times 10^{11} \mathrm{m}\) apart and are equally distant from the earth. A telescope has an objective lens with a diameter of \(1.02 \mathrm{m}\) and just detects these stars as separate objects. Assume that light of wavelength \(550 \mathrm{nm}\) is being observed. Also assume that diffraction effects, rather than atmospheric turbulence, limit the resolving power of the telescope. Find the maximum distance that these stars could be from the earth.

In Young's experiment a mixture of orange light \((611 \mathrm{nm})\) and blue light \((471 \mathrm{nm})\) shines on the double slit. The centers of the first- order bright blue fringes lie at the outer edges of a screen that is located \(0.500 \mathrm{m}\) away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in which direction (toward or away from the slits) should the screen be moved so that the centers of the first-order bright orange fringes will just appear on the screen? It may be assumed that \(\theta\) is small, so that \(\sin \theta \approx \tan \theta\).

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.

Light waves with two different wavelengths, \(632 \mathrm{nm}\) and \(474 \mathrm{nm}\), pass simultaneously through a single slit whose width is \(7.15 \times 10^{-5} \mathrm{m}\) and strike a screen \(1.20 \mathrm{m}\) from the slit. Two diffraction patterns are formed on the screen. What is the distance (in \(\mathrm{cm}\) ) between the common center of the diffraction patterns and the first occurrence of a dark fringe from one pattern falling on top of a dark fringe from the other pattern?

In a Young's double-slit experiment, two rays of monochromatic light emerge from the slits and meet at a point on a distant screen, as in Figure \(27.6 a .\) The point on the screen where these two rays meet is the eighth-order bright fringe. The difference in the distances that the two rays travel is \(4.57 \times 10^{-6} \mathrm{m} .\) What is the wavelength (in \(\mathrm{nm}\) ) of the monochromatic light?
See all solutions

Recommended explanations on Physics Textbooks

Solid State Physics

Read Explanation

Translational Dynamics

Read Explanation

Scientific Method Physics

Read Explanation

Nuclear Physics

Read Explanation

Energy Physics

Read Explanation

Rotational Dynamics

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.