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Chapter 27: Problem 31

Multiple-Concept Example 7 illustrates how the concepts needed in this problem are applied. The largest refracting telescope in the world is at the Yerkes Observatory in Williams Bay, Wiscon sin. The objective of the telescope has a diameter of \(1.02 \mathrm{~m}\). Two objects are \(3.75 \times 10^{4} \mathrm{~m}\) from the telescope. With light of wavelength \(565 \mathrm{~nm}\), how close can the objects be to each other so that they are just resolved by the telescope?

Short Answer

Expert verified
The minimum separation resolved is approximately 0.02531 meters.

Step by step solution

01

Understanding the Problem

We need to calculate the minimum angular separation that can be resolved by the telescope, known as the resolving power. The telescope's lens diameter, distance between objects and the telescope, and the light's wavelength will be used to determine the minimum separation between two objects that can be resolved.
02

Relevant Formula

The resolving power of a telescope is given by the formula: \[ \theta = \frac{1.22 \lambda}{D} \]where \( \lambda \) is the wavelength of light (565 nm or \(565 \times 10^{-9} \text{ m} \)) and \( D \) is the diameter of the objective lens (1.02 m). \( \theta \) is the smallest angular separation the telescope can resolve, given in radians.
03

Calculate Angular Resolution

Substituting the known values into the formula, we calculate:\[ \theta = \frac{1.22 \times 565 \times 10^{-9}}{1.02} \]This will give us the angular resolution in radians.
04

Convert Wavelength and Calculate

Converting the wavelength:\( 565 \times 10^{-9} = 565 \text{ nm} \to 565 \times 10^{-9} \text{ m} \).Now compute:\[ \theta = \frac{1.22 \times 565 \times 10^{-9}}{1.02} \approx 6.75 \times 10^{-7} \text{ radians} \]
05

Determine Minimum Distance Between Objects

The smallest distance \( s \) between the objects that the telescope can resolve is given by \[ s = L \times \theta \]where \( L \) is the distance to the objects (\( 3.75 \times 10^4 \text{ m} \)). Calculate:\[ s = 3.75 \times 10^4 \times 6.75 \times 10^{-7} \approx 0.02531 \text{ m} \]
06

Final Result

The calculation shows that the minimum distance between the objects so that they are just resolved by the telescope is approximately 0.02531 meters.

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Key Concepts

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

Refracting Telescope
A refracting telescope is a type of optical telescope that uses lenses to gather and focus light. This is quite distinct from reflecting telescopes, which use mirrors instead. The Yerkes Observatory's telescope is a perfect example of a large refractor. It utilizes an objective lens to collect and concentrate light from distant objects.
This objective lens is crucial in determining how much light the telescope can gather and is therefore essential in enhancing the image's brightness and clarity.
  • Light enters through the lens.
  • It bends and focuses to form an image.
  • An eyepiece magnifies this image for better visualization.
In simple terms, the refracting telescope's ability to focus light through its lenses allows it to provide clear and detailed views of astronomical objects, making it a valuable tool for astronomers.
Angular Resolution
Angular resolution is the ability of a telescope to distinguish between two closely spaced objects. In simpler words, it's a measure of how detailed the telescope's image can be. Higher resolution means more detail and the ability to see two points distinctly instead of a single blurred dot.
The resolving power is determined by the formula:\[ \theta = \frac{1.22 \lambda}{D} \]
In this equation, \(\theta\) represents the smallest angle between two points that the telescope can still resolve. A smaller \(\theta\) signifies that the telescope can distinguish objects that are closer together. The values of \(\lambda\) (wavelength of light) and \(D\) (diameter of the objective lens) play vital roles in this precision.
Wavelength of Light
Wavelength of light, denoted by \(\lambda\), is the distance between successive crests of a wave. In the context of telescopes, it's the wavelength of the light being observed. Visible light has wavelengths ranging approximately from 400 nm (violet) to 700 nm (red).
The wavelength of light affects how well a telescope can distinguish details. Shorter wavelengths can resolve smaller details compared to longer wavelengths. For the Yerkes telescope, we commonly utilize a wavelength of 565 nm (nanometers), typical of green light.
Using light with a shorter wavelength enhances the telescope's resolving power, allowing it to distinguish closely spaced objects better.
Objective Lens Diameter
The objective lens diameter, often simply referred to as the aperture, plays a crucial role in a telescope's capabilities. It defines the amount of light the telescope can capture; a larger diameter means more light and brighter images.
For the Yerkes telescope, the diameter is 1.02 meters, making it one of the largest refracting telescope apertures in the world. This size enables it to gather substantial light, which is essential for observing faint and distant objects.
  • Larger diameter = more light gathering capability.
  • Improves brightness and clarity of images.
  • Directly influences the telescope's resolving power.
Sufficient aperture is fundamental for detailed and accurate astronomical observations, making the diameter of the objective lens a key aspect of a refracting telescope's design.

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Most popular questions from this chapter

In a Young's double-slit experiment, the wavelength of the light used is \(520 \mathrm{nm}\) (in vacuum), and the separation between the slits is \(1.4 \times 10^{-6} \mathrm{~m}\). Determine the angle that locates (a) the dark fringe for which \(m=0,\) (b) the bright fringe for which \(m=1,\) (c) the dark fringe for which \(m=1,\) and \((\mathrm{d})\) the bright fringe for which \(m=2\)

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is \(675 \mathrm{~nm}\). Determine the angle that locates the first dark fringe when the width of the slit is (a) \(1.8 \times 10^{-4} \mathrm{~m}\) and (b) \(1.8 \times 10^{-6} \mathrm{~m}\).

Orange light \(\left(\lambda_{\text {vacuum }}=611 \mathrm{nm}\right.\) ) shines on a soap film \((n=1.33)\) that has air on either side of it. The light strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes it to look bright in reflected light?

(a) As Section \(17.3\) discusses, high-frequency sound waves exhibit less diffraction than low-frequency sound waves do. However, even high-frequency sound waves exhibit much more diffraction under normal circumstances than do light waves that pass through the same opening. The highest frequency that a healthy ear can typically hear is \(2.0 \times 10^{4} \mathrm{~Hz}\). Assume that a sound wave with this frequency travels at \(343 \mathrm{~m} / \mathrm{s}\) and passes through a doorway that has a width of \(0.91 \mathrm{~m}\). Determine the angle that locates the first minimum to either side of the central maximum in the diffraction pattern for the sound. This minimum is equivalent to the first dark fringe in a single-slit diffraction pattern for light. (b) Suppose that yellow light (wave length \(=580 \mathrm{~nm}\), in vacuum) passes through a doorway and that the first dark fringe in its diffraction pattern is located at the angle determined in part (a). How wide would this hypothetical doorway have to be?

(a) In a single-slit diffraction pattern the width of the central bright fringe is defined by the location of the first dark fringe that lies on either side of it. For a given slit width, does the width of the central bright fringe increase, decrease, or remain the same as the wavelength of the light increases? (b) For a given wavelength, does the width of the central bright fringe increase, decrease, or remain the same as the slit width increases? (c) When both the wavelength and the slit width change, it is possible for the width of the central bright fringe to remain the same. What condition must be satisfied for this to happen? In each case, give your reasoning. A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{~m}\). When light with a wavelength of \(\lambda_{1}=510 \mathrm{~nm}\) passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2}\) ) and a wavelength of \(\lambda_{2}=740 \mathrm{~nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2}\).
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