/*! 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 33 Astronomers have discovered a pl... [FREE SOLUTION] | 91Ó°ÊÓ

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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 minimum aperture diameter required is approximately 2.34 mm.

Step by step solution

01

Identify Known Values

First, let's list all the known values given in the problem. The distance to the star Upsilon Andromedae from Earth is \( D = 4.2 \times 10^{17} \mathrm{~m} \). The distance of the planet from the star is \( d = 1.2 \times 10^{11} \mathrm{~m} \). The wavelength of the light used is \( \lambda = 550 \mathrm{~nm} = 550 \times 10^{-9} \mathrm{~m} \).
02

Calculate Angular Separation

The angular separation \( \theta \) between the planet and the star is calculated using the small-angle formula: \[ \theta = \frac{d}{D} \]Substituting the given values,\[ \theta = \frac{1.2 \times 10^{11} \mathrm{~m}}{4.2 \times 10^{17} \mathrm{~m}} \approx 2.857 \times 10^{-7} \mathrm{~radians} \]
03

Apply Rayleigh's Criterion

To resolve two objects, the angular resolution \( \theta \) of the telescope must be equal to or larger than the angular separation of the objects. According to Rayleigh's Criterion, the minimum angular resolution \( \theta \) is given by:\[ \theta = 1.22 \frac{\lambda}{D_{aperture}} \]Set the telescope's resolution to the calculated angular separation:\[ 2.857 \times 10^{-7} = 1.22 \frac{550 \times 10^{-9} \mathrm{~m}}{D_{aperture}} \]
04

Solve for Aperture Diameter

Rearrange the equation to solve for the aperture diameter \( D_{aperture} \):\[ D_{aperture} = 1.22 \frac{550 \times 10^{-9} \mathrm{~m}}{2.857 \times 10^{-7}} \]Calculating gives:\[ D_{aperture} \approx 0.00234 \mathrm{~m}, \text{ or about } 2.34 \mathrm{~mm} \]
05

Conclusion

The minimum necessary aperture diameter for the telescope to resolve the planet and the star is approximately 2.34 mm.

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Key Concepts

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

Angular Resolution
Angular resolution describes a telescope's ability to distinguish between two closely situated objects. It is a crucial aspect of telescope optics because it determines how detailed an image a telescope can present. The angular resolution is influenced by the wavelength of the light used and the diameter of the telescope's aperture.
The greater the aperture of the telescope, the higher its resolving power. This means it can separate two distant objects that are very close to each other in the sky.
In practical terms, if a telescope has a high angular resolution, it can identify two stars aligned nearly in the same direction from Earth as separate entities rather than a single point of light. The small-angle formula often helps provide a measure of angular separation:
  • \( \theta = \frac{d}{D} \) where \( \theta \) is angular separation, \( d \) is the distance between objects, and \( D \) is the distance to the objects
Rayleigh's Criterion
Rayleigh's Criterion is a pivotal concept in optics, particularly when discussing the resolving power of optical instruments like telescopes. It provides a formal criterion for measuring the point at which two objects can be considered distinctly separate when viewed through an optical device.
According to Rayleigh's Criterion, two light sources are resolvable if the central peak of one diffraction pattern coincides with the first minimum of another. The criterion uses the formula:
  • \( \theta = 1.22 \frac{\lambda}{D_{aperture}} \)
This states that the minimum angular resolution \( \theta \) is determined by the wavelength \( \lambda \) of the light and the aperture diameter \( D_{aperture} \).
The factor 1.22 comes from the analysis of the diffraction patterns created by a circular aperture, such as in many telescope designs. It reflects the geometry and physics governing these diffraction patterns, ensuring that the telescopic image provides accurate detail without blur.
Visible Light Wavelengths
Visible light wavelengths are critical to understanding telescope optics because they dictate how much detail a telescope can uncover about celestial bodies. The visible spectrum ranges from approximately 380 nm to 740 nm, with each wavelength corresponding to different colors we can see.
In the context of telescope optics, shorter wavelengths (such as blue light) have higher frequencies and can offer better resolution given the same aperture size. However, they are more easily scattered, which can complicate observations.
The choice of 550 nm in this particular exercise represents green light, which is in the middle of the visible spectrum, providing a balance between resolution and light gathering ability.
When designing or operating telescopes, considering the wavelength of light is crucial. The diffraction limit of a telescope increases with wavelength; hence, telescopes built to observe in the visible light spectrum must be optimized to accommodate this balancing act between detail and light availability.

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Most popular questions from this chapter

A flat observation screen is placed at a distance of \(4.5 \mathrm{~m}\) from a pair of slits. The separation on the screen between the central bright fringe and the first-order bright fringe is \(0.037 \mathrm{~m}\). The light illuminating the slits has a wavelength of \(490 \mathrm{~nm}\). Determine the slit separation.

Concept Questions (a) What, if any, phase change occurs when light, traveling in air, reflects from the interface between the air and a soap film \((n=1.33) ?\) (b) What, if any, phase change occurs when light, traveling in a soap film, reflects from the interface between the soap film and a glass plate \((n=1.52) ?(\mathrm{c})\) Is the wavelength of the light in a soap film greater than, smaller than, or equal to the wavelength in a vacuum? Problem A soap film \((n=1.33)\) is \(465 \mathrm{nm}\) thick and lies on a glass plate \((n=1.52)\) Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm}\), travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?

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.

Three, and only three, bright fringes can be seen on either side of the central maximum when a grating is illuminated with light \((\lambda=510 \mathrm{~nm}) .\) What is the maximum number of lines \(/ \mathrm{cm}\) for the grating?

Light of wavelength \(691 \mathrm{nm}\) (in vacuum) is incident perpendicularly on a soap film \((n=1.33)\) suspended in air. What are the two smallest nonzero film thicknesses (in \(\mathrm{nm})\) for which the reflected light undergoes constructive interference?

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