/*! 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 29 (A) What is the angular separati... [FREE SOLUTION] | 91Ó°ÊÓ

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(A) What is the angular separation of two stars if their images are barely resolved by the Thaw refracting telescope at the Allegheny Observatory in Pittsburgh? The lens diameter is \(76 \mathrm{~cm}\) and its focal length is \(14 \mathrm{~m}\). Assume \(\lambda=550 \mathrm{nm}\). (b) Find the distance between these barely resolved stars if each of them is 10 light-years distant from Earth. (c) For the image of a single star in this telescope, find the diameter of the first dark ring in the diffraction pattern, as measured on a photographic plate placed at the focal plane of the telescope lens. Assume that the structure of the image is associated entirely with diffraction at the lens aperture and not with lens "errors."

Short Answer

Expert verified
(a) 8.81 x 10^-7 radians; (b) 8.34 x 10^10 meters; (c) 0.247 mm.

Step by step solution

01

Understanding Angular Resolution using Rayleigh Criterion

For the angular resolution, we use the Rayleigh criterion: \( \theta = 1.22 \times \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the lens. Here, \( \lambda = 550 \times 10^{-9} \text{ meters} \) and \( D = 0.76 \text{ meters} \). Substitute these values: \( \theta = 1.22 \times \frac{550 \times 10^{-9}}{0.76} \). This will provide the angular separation in radians.
02

Calculating the Angular Separation

Substituting the values, we get \( \theta = \frac{1.22 \times 550 \times 10^{-9}}{0.76} = 8.8145 \times 10^{-7} \text{ radians} \). Thus, the angular separation of the two stars is approximately \( 8.8145 \times 10^{-7} \text{ radians} \).
03

Calculating Distance Between Stars

Now, use the small angle approximation: \( s = d \times \theta \), where \( s \) is the linear separation, \( d = 10 \times 9.461 \times 10^{15} \text{ meters} \) (since 10 light-years is the distance to the stars from Earth). Thus, \( s = 10 \times 9.461 \times 10^{15} \times 8.8145 \times 10^{-7} = 8.341 \times 10^{10} \text{ meters} \).
04

Diameter of First Dark Ring in Diffraction Pattern

The first dark ring diameter formula is \( D' = 2.44 \times \frac{\lambda f}{D} \), where \( f = 14 \text{ m} \), \( \lambda = 550 \times 10^{-9} \text{ meters} \). Substitute: \( D' = 2.44 \times \frac{550 \times 10^{-9} \times 14}{0.76} \). Simplifying gives \( D' \approx 2.44 \times 1.0129 \times 10^{-4} = 2.47 \times 10^{-4} \text{ meters} \). Thus, the diameter is approximately \( 0.247 \text{ mm} \).

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Key Concepts

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

Angular Resolution
Angular resolution is a critical parameter in the design and operation of optical telescopes, such as the Thaw refracting telescope at the Allegheny Observatory. It defines a telescope's ability to distinguish between two closely spaced objects, like stars.
In the context of optical astronomy, angular resolution refers to the smallest angle between two objects which can be distinguished as separate. The smaller this angle, the better the resolution.
A commonly used criterion for determining whether two points can be resolved is the Rayleigh criterion. It is calculated using the formula \( \theta = 1.22 \times \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light (in meters), and \( D \) is the lens diameter (in meters).
  • The factor 1.22 arises from the central peak of the Airy disk, which describes the diffraction pattern of a circular aperture.
  • The value obtained gives you the angular separation in radians.
  • For visible light (with a typical wavelength of about 550 nm), the angular resolution of a 76 cm diameter telescope is approximately \( 8.8145 \times 10^{-7} \) radians.
Understanding angular resolution helps astronomers determine how "sharp" an image is and how much detail can be observed.
Diffraction Pattern
When light passes through an aperture, such as a telescope's objective lens, it does not simply converge to a point; it spreads out and creates a diffraction pattern. This happens due to the wave nature of light.
At the heart of this pattern is the Airy disk, which is a bright central spot surrounded by dark and bright concentric rings. This occurs regardless of the telescope's lens or mirror quality and involves no lens errors.
In the context of the exercise, the diameter of the first dark ring in the diffraction pattern can be determined using the formula \( D' = 2.44 \times \frac{\lambda f}{D} \), where \( f \) is the focal length of the telescope.
  • The first dark ring surrounding the Airy disk signifies where the intensity of light first drops to zero due to destructive interference.
  • The calculation yields a first dark ring diameter of approximately 0.247 mm when observing with a 76 cm diameter telescope with a focal length of 14 meters and a wavelength of 550 nm.
Understanding these diffraction patterns is vital because they define the limitations of a telescope’s imaging capabilities.
Rayleigh Criterion
The Rayleigh criterion is an essential concept in understanding how optical instruments like telescopes achieve resolution. It provides a practical measure for the limit of angular resolution and is particularly relevant when discussing diffraction's impact on image clarity.
According to the Rayleigh criterion, two point sources are considered barely resolved when the peak of one image's Airy disk coincides with the first minimum of another. This is the most applicable test of resolution in practice. This calculation is critical:
  • It helps determine how close two objects can be, angularly, before they blur together into one.
  • For the given telescope, using a wavelength \( \lambda \) of 550 nm and an aperture diameter \( D \) of 76 cm, the Rayleigh criterion gives an angular resolution of \( \theta = 8.8145 \times 10^{-7} \) radians.
This criterion is utilized by astronomers to quantify a telescope's resolving power and determine its efficacy in making clear observations of closely spaced celestial objects.

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Most popular questions from this chapter

The two headlights of an approaching automobile are \(1.4 \mathrm{~m}\) apart. At what (a) angular separation and (b) maximum distance will the eye resolve them? Assume that the pupil diameter is \(5.0 \mathrm{~mm},\) and use a wavelength of \(550 \mathrm{nm}\) for the light. Also assume that diffraction effects alone limit the resolution so that Rayleigh's criterion can be applied.

Sound waves with frequency \(3000 \mathrm{~Hz}\) and speed \(343 \mathrm{~m} / \mathrm{s}\) diffract through the rectangular opening of a speaker cabinet and into a large auditorium of length \(d=100 \mathrm{~m}\). The opening, which has a horizontal width of \(30.0 \mathrm{~cm},\) faces a wall \(100 \mathrm{~m}\) away (Fig. \(36-36\) ). Along that wall, how far from the central axis will a listener be at the first diffraction minimum and thus have difficulty hearing the sound? (Neglect reflections.)

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle's "shadow region." Previously, television signals had a wavelength of about \(50 \mathrm{~cm}\), but digital television signals that are transmitted from towers have a wavelength of about \(10 \mathrm{~mm}\). (a) Did this change in wavelength increase or decrease the diffraction of the signals into the shadow regions of obstacles? Assume that a signal passes through an opening of \(5.0 \mathrm{~m}\) width between two adjacent buildings. What is the angular spread of the central diffraction maximum (out to the first minima) for wavelengths of (b) \(50 \mathrm{~cm}\) and (c) \(10 \mathrm{~mm} ?\)

In a single-slit diffraction experiment, there is a minimum of intensity for orange light \((\lambda=600 \mathrm{nm})\) and a minimum of intensity for blue- green light \((\lambda=500 \mathrm{nm})\) at the same angle of \(1.00 \mathrm{mrad} .\) For what minimum slit width is this possible?

The distance between the first and fifth minima of a single-slit diffraction pattern is \(0.35 \mathrm{~mm}\) with the screen \(40 \mathrm{~cm}\) away from the slit, when light of wavelength \(550 \mathrm{nm}\) is used. (a) Find the slit width. (b) Calculate the angle \(\theta\) of the first diffraction minimum.

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