Explanation 1) Concept: We have the equation for Rayleigh’s criterion. We also have the relation between distance between two points, the length from which we are observing those points, and small angle. We rearrange the equation to find the required answer. 2) Formula: i) θ = D L ii) θ R = 1.22 λ d 3) Given: i) L = 400 k m = 400 × 10 3 m ii) d = 5 m m = 0.005 m iii) λ = 550 n m = 550 × 10 - 9 m 4) Calculation: a) We have θ = D L We can write this as D = θ L But according to Rayleigh’s criterion, θ R = 1.22 λ d D = 1

10th Edition

ISBN: 9781118230725

Author: David Halliday, Robert Resnick, Jearl Walker

Publisher: Wiley, John & Sons, Incorporated

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Question

Chapter 36, Problem 22P

To determine

To find:

a) Smaller liner width on Earth’s surface that the astronaut can resolve

b) Can the astronaut resolve the Great Wall of China?

c) Would the astronaut be able to resolve any unmistakable sign of intelligent life on Earth’s surface?

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Chapter 36, Problem 21P

Chapter 36, Problem 23P

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Assume that Rayleigh's criterion gives the limit of resolution of an astronaut's eye looking down on Earth's surface from a typical space shuttle altitude of 428 km. (a) Under that idealized assumption, estimate the smallest linear width on Earth's surface that the astronaut can resolve. Take the astronaut's pupil diameter to be 6 mm and the wavelength of visible light to be 540 nm. (b) Can the astronaut resolve the Great Wall of China (the figure below), which is more than 3000 km long, 5 to 10 m thick at its base, 4 m thick at its top, and 8 m in height? (c) Would the astronaut be able to resolve any unmistakable sign of intelligent life on Earth's surface?

Assume that Rayleigh’s criterion gives the limit of resolution of an astronaut’s eye looking down on Earth’s surface from a typical space shuttle altitude of 400 km. (a) Under that idealized assumption, estimate the smallest linear width on Earth’s surface that the astronaut can resolve.Take the astronaut’s pupil diameter to be 5 mm and the wavelength of visible light to be 550 nm. (b) Can the astronaut resolve the Great Wall of China , which is more than 3000 km long, 5 to 10 m thick at its base, 4 m thick at its top, and 8 m in height? (c) Would the astronaut be able to resolve any unmistakable sign of intelligent life on Earth’s surface?

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Chapter 36 Solutions

Fundamentals of Physics Extended

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a) Smaller liner width on Earth’s surface that the astronaut can resolve b) Can the astronaut resolve the Great Wall of China? c) Would the astronaut be able to resolve any unmistakable sign of intelligent life on Earth’s surface?

Solution Summary: The author explains the equation for Rayleigh's criterion and the relation between distance between two points, length from which we are observing those points and small angle.

Concept explainers

To find: a) Smaller liner width on Earth’s surface that the astronaut can resolve b) Can the astronaut resolve the Great Wall of China? c) Would the astronaut be able to resolve any unmistakable sign of intelligent life on Earth’s surface?

Want to see the full answer?

Chapter 36 Solutions

To find:

a) Smaller liner width on Earth’s surface that the astronaut can resolve

b) Can the astronaut resolve the Great Wall of China?

c) Would the astronaut be able to resolve any unmistakable sign of intelligent life on Earth’s surface?