/*! 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 74 An astronaut in a space shuttle ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An astronaut in a space shuttle claims she can just barely resolve two point sources on Earth's surface, \(160 \mathrm{~km}\) below. Calculate their (a) angular and (b) linear separation, assuming ideal conditions. Take \(\lambda=540 \mathrm{~nm}\) and the pupil diameter of the astronaut's eye to be \(5.0 \mathrm{~mm}\).

Short Answer

Expert verified
Angular separation: \(1.32 \times 10^{-4}\) radians; Linear separation: 21.1 meters.

Step by step solution

01

Understand the Rayleigh Criterion

To resolve two point sources, we use the Rayleigh Criterion, which states that the smallest angular separation \( \theta \) that can be resolved is given by \( \theta = 1.22 \frac{\lambda}{D} \), where \( \lambda \) is the wavelength and \( D \) is the diameter of the aperture.
02

Calculate the Angular Separation

Insert the given values into the formula: \( \lambda = 540 \mathrm{~nm} = 540 \times 10^{-9} \mathrm{~m} \) and \( D = 5.0 \mathrm{~mm} = 5.0 \times 10^{-3} \mathrm{~m} \).\[\theta = 1.22 \frac{540 \times 10^{-9}}{5.0 \times 10^{-3}}.\]Calculate \( \theta \) to find the angular separation.
03

Solve for Angular Separation

Calculate \( \theta \):\[\theta = 1.22 \times \frac{540 \times 10^{-9}}{5.0 \times 10^{-3}} = 1.3188 \times 10^{-4} \mathrm{~radians}.\]Thus, the angular separation is approximately \( 1.32 \times 10^{-4} \mathrm{~radians} \).
04

Calculate Linear Separation

To determine the linear separation \( s \) between the point sources, use the relationship \( s = L \theta \), where \( L \) is the distance from the observer to the object (160 km).\[s = 160 \times 10^{3} \times 1.3188 \times 10^{-4} \mathrm{~radians}.\]Calculate \( s \) to find the linear separation.
05

Solve for Linear Separation

Calculate \( s \):\[s = 160 \times 10^3 \times 1.3188 \times 10^{-4} = 21.101 \mathrm{~meters}.\]Thus, the linear separation is approximately 21.1 meters.

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 91Ó°ÊÓ!

Key Concepts

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

Angular Separation
Angular separation is a measure of how far apart two objects appear in the field of view of an observer. It is an angle, usually small and measured in radians, which the observer perceives between two distinct objects.
In the context of observing from a space shuttle, angular separation helps determine if two point sources, such as lights on Earth, can be visually distinguished by the observer.
The Rayleigh Criterion is essential for understanding this concept, stating that the smallest angle at which two points can be resolved is given by the equation:
  • \( \theta = 1.22 \frac{\lambda}{D} \)
Where \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the observer's aperture, such as the pupil of the eye.
These factors dictate the precision of visual resolution from a distance.
In our exercise, the angular separation calculated was approximately \( 1.32 \times 10^{-4} \) radians, which signifies the minimal angle for the astronaut to just discern the two points on Earth.
Linear Separation
Linear separation refers to the actual distance separating two points or objects, as seen from a particular vantage point. This concept is critical when translating the angular separation into a tangible distance using the formula:
  • \( s = L \theta \)
Here, \( s \) is the linear separation, \( L \) is the distance from the observer to the objects (160 km for the astronaut), and \( \theta \) is the angular separation.
This calculation allows us to understand how far apart the points appear on the ground.
In the given problem, this turns out to be about 21.1 meters.
This means that from the astronaut's perspective in the shuttle, the two point sources on Earth must be 21.1 meters apart to be distinguished as separate entities.
Wavelength
Wavelength is a key factor in the ability to resolve two objects, as it represents the distance between successive peaks of a wave—such as light. Shorter wavelengths allow for better resolution, which is why light's wavelength is crucial in the Rayleigh Criterion.
In this exercise, the wavelength \( \lambda \) is given as 540 nm (nanometers), a common wavelength for green light.
The calculation for angular separation directly incorporates the wavelength, showing that light's characteristics influence how well we can differentiate between points.
A shorter wavelength would reduce the calculated angular separation, potentially allowing us to resolve closer objects, while a longer wavelength would increase it.
Pupil Diameter
The pupil diameter, represented by \( D \), is another critical component in visual resolution described in the Rayleigh Criterion.
The pupil acts as the aperture through which light enters the eye, and its size affects the resolving power.
A larger diameter enables the eye to collect more light, thus enhancing the ability to distinguish between closely spaced points.
For the astronaut, the pupil diameter is 5.0 mm, which, when combined with the wavelength and distance, determines the angular resolution.
If the eye's pupil were smaller, the angular separation would increase, making it harder to resolve separate points. Conversely, a larger pupil would facilitate resolving finer details.

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

A beam of light consisting of wavelengths from \(460.0 \mathrm{~nm}\) to \(640.0 \mathrm{~nm}\) is directed perpendicularly onto a diffraction grating with 160 lines/mm. (a) What is the lowest order that is overlapped by another order? (b) What is the highest order for which the complete wavelength range of the beam is present? In that highest order, at what angle does the light at wavelength (c) \(460.0 \mathrm{~nm}\) and (d) \(640.0 \mathrm{~nm}\) appear? (e) What is the greatest angle at which the light at wavelength \(460.0 \mathrm{~nm}\) appears?

Light of wavelength \(600 \mathrm{~nm}\) is incident normally on a diffraction grating. Two adjacent maxima occur at angles given by \(\sin \theta=0.2\) and \(\sin \theta=0.3 .\) The fourth-order maxima are missing. (a) What is the separation between adjacent slits? (b) What is the smallest slit width this grating can have? For that slit width, what are the (c) largest, (d) second largest, and (e) third largest values of the order number \(m\) of the maxima produced by the grating?

A beam of light with a narrow wavelength range centered on \(450 \mathrm{~nm}\) is incident perpendicularly on a diffraction grating with a width of \(1.80 \mathrm{~cm}\) and a line density of 1400 lines \(/ \mathrm{cm}\) across that width. For this light, what is the smallest wavelength difference this grating can resolve in the third order?

Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(2000 \mathrm{~km}\). One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of \(1.40 \mathrm{~nm} .\) The element that emits light is the end of a wire with diameter \(0.200 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) By what factor is the beam intensity reduced in transit to the target? (The laser is fired from space, so that atmospheric absorption can be ignored.)

The wall of a large room is covered with acoustic tile in which small holes are drilled \(5.0 \mathrm{~mm}\) from center to center. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer's eye to be \(4.0 \mathrm{~mm}\), and the wavelength of the room light to be \(550 \mathrm{~nm}\) ?

See all solutions

Recommended explanations on Physics Textbooks

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.