/*! 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 53 (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 18 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) \(8.84 \times 10^{-7}\) radians; (b) \(1.51 \times 10^{11}\) m; (c) 2.48 mm.

Step by step solution

01

Convert Given Parameters

Start by converting the given parameters to consistent units that we will use in later calculations. The telescope's lens diameter is given in centimeters and focal length in meters.Given:- Lens diameter: \(D = 76 \text{ cm} = 0.76 \text{ m}\)- Focal length: \(f = 14 \text{ m}\)- Wavelength: \(\lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m}\)
02

Calculate Angular Separation Using Rayleigh Criterion

The Rayleigh criterion gives the minimum angular separation \(\theta\) that two sources can have and still be resolved:\[\theta = 1.22 \frac{\lambda}{D}\]Substitute the values:\[\theta = 1.22 \times \frac{550 \times 10^{-9}}{0.76}\]\[\theta \approx 8.84 \times 10^{-7} \text{ radians}\]
03

Calculate Physical Separation of Stars

Use the angular separation \(\theta\) to find the physical distance \(d\) between the stars, given their distance \(L = 18 \) light-years from Earth:First, convert light-years to meters:\[L = 18 \times 9.461 \times 10^{15} \text{ m} = 1.703 \times 10^{17} \text{ m}\]The physical separation is:\[d = \theta \times L\]Substitute the values:\[d = 8.84 \times 10^{-7} \times 1.703 \times 10^{17} \]\[d \approx 1.51 \times 10^{11} \text{ m}\]
04

Calculate Diameter of First Dark Ring

For a single star, the diameter of the first dark ring in the diffraction pattern observed at the focal plane is given by:\[ r = 1.22 \frac{\lambda f}{D} \]Substitute the values:\[ r = 1.22 \times \frac{550 \times 10^{-9} \times 14}{0.76} \]\[ r \approx 1.24 \times 10^{-3} \text{ m} = 1.24 \text{ mm} \]Thus, the diameter of the first dark ring is twice this radius:\[ \text{Diameter} = 2 \times 1.24 \text{ mm} = 2.48 \text{ mm} \]

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Key Concepts

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

Rayleigh Criterion
The Rayleigh criterion is a crucial concept in understanding the resolution limits of optical systems. It helps us determine how close two point sources, like stars, can be to each other while still being individually distinguished.
The criterion is based on the principle of diffraction and is particularly useful in contexts like astronomy. When a point source, such as a star, is viewed through a telescope, its image is not a point but a diffraction pattern. The key takeaway from the Rayleigh criterion is:
  • Two sources are said to be just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other.
  • The formula to express the minimum angular separation, \( \theta \), is \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the telescope's aperture.
This formula shows that the ability to resolve two distinct light sources depends mainly on the wavelength of light and the size of the telescope's aperture. Larger apertures allow for finer resolution.
Understanding the Rayleigh criterion is essential for astronomers as it dictates the theoretical limit of resolution that can be achieved with their telescopes.
Angular Separation
Angular separation refers to the angle between two celestial bodies as observed from Earth. It is an important concept in observational astronomy to determine how far apart two objects appear in the sky.
To calculate angular separation, the Rayleigh criterion is often utilized, which explains how diffraction affects the resolution of telescopic images. Here's what you need to know:
  • The angular separation \( \theta \) in radians is the angle subtended by two objects at the observer's viewpoint.
  • Once you have \( \theta \), you can also determine the linear distance between objects if their distance from the observer is known.
In the Thaw refracting telescope example, we calculated the angular separation using the formula from Rayleigh's criterion, \( \theta = 1.22 \frac{\lambda}{D} \), to find that images of two stars are barely resolved.
This angle turns out to be approximately \(8.84 \times 10^{-7}\) radians, showcasing how minute changes in angle can be distinguishable through precision calculations. Hence, understanding angular separation allows both amateur and professional astronomers to better gauge the distances and positions of stars in the night sky.
Telescope Optics
Telescope optics is a fascinating field that explores how light is manipulated to provide ultra-clear views of celestial objects. Optical instruments like the Thaw refracting telescope rely on a series of lenses and mirrors to magnify objects in space. Here's a breakdown of the essential components and mechanisms:
  • **Objective Lens**: This is the primary lens that gathers light from distant objects. The larger the lens diameter, the more light it gathers, improving the telescope's resolving power.
  • **Focal Length**: This is the distance from the lens to the point where the image is focused. In our example, it's 14 meters, indicating a relatively long telescope, which allows for high magnification.
  • **Diffraction Patterns**: All light passing through the objective lens generates a diffraction pattern, especially noticeable in point sources like stars.
  • **Aperture**: Size matters! A larger aperture (in this case, 76 cm in diameter) enables better resolution and clarity, aligned with the Rayleigh criterion's predictions.
  • **Focal Plane**: The image is captured here, and the size of diffraction rings can be measured directly, revealing detailed features of observed objects.
Telescope optics require precise engineering, and understanding diffraction patterns helps astronomers mitigate image distortion.
This enables us to gather detailed data on stars and other celestial phenomena, enhancing our knowledge of the universe.

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Most popular questions from this chapter

Monochromatic light with wavelength \(420 \mathrm{~nm}\) is incident on a slit with width \(0.050 \mathrm{~mm}\). The distance from the slit to a screen is \(3.5 \mathrm{~m}\). Consider a point on the screen \(2.2 \mathrm{~cm}\) from the central maximum. Calculate (a) \(\theta\) for that point, (b) \(\alpha\), and (c) the ratio of the intensity at that point to the intensity at the central maximum.

a) How many bright fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if \(\lambda=550 \mathrm{~nm}, d=0.180 \mathrm{~mm}\), and \(a=30.0 \mu \mathrm{m}\) ? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at \(\theta=30^{\circ}\) ?

The telescopes on some commercial surveillance satellites can resolve objects on the ground as small as \(85 \mathrm{~cm}\) across (see Google Earth), and the telescopes on military surveillance satellites reportedly can resolve objects as small as \(10 \mathrm{~cm}\) across. Assume first that object resolution is determined entirely by Rayleigh's criterion and is not degraded by turbulence in the atmosphere. Also assume that the satellites are at a typical altitude of \(420 \mathrm{~km}\) and that the wavelength of visible light is \(550 \mathrm{~nm}\). What would be the required diameter of the telescope aperture for (a) \(85 \mathrm{~cm}\) resolution and (b) \(10 \mathrm{~cm}\) resolution? (c) Now, considering that turbulence is certain to degrade resolution and that the aperture diameter of the Hubble Space Telescope is \(2.4 \mathrm{~m}\), what can you say about the answer to (b) and about how the military surveillance resolutions are accomplished?

A grating has 600 rulings/mm and is \(6.0 \mathrm{~mm}\) wide. (a) What is the smallest wavelength interval it can resolve in the third order at \(\lambda=500 \mathrm{~nm}\) ? (b) How many higher orders of maxima can be seen?

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