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An astronaut in orbit can just resolve two point sources on the earth that are 75.0 \(\mathrm{m}\) apart. Assume that the resolution is diffraction limited, and use Rayleigh's criterion. What is the astronaut's altitude above the earth? Treat her eye as a circular aperture with a diameter of 4.00 \(\mathrm{mm}\) (the diameter of her pupil), and take the wavelength of the light to be 500 \(\mathrm{nm}\) .

Short Answer

Expert verified
The astronaut's altitude is approximately 491,803 meters above the Earth.

Step by step solution

01

Understanding Rayleigh's Criterion

Rayleigh's criterion for the resolution states that two point sources are just resolvable when the central maximum of the diffraction pattern of one image coincides with the first minimum of the other. The formula for Rayleigh's criterion is \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the aperture.
02

Calculate Angular Resolution

Insert the given values for \( \lambda = 500 \times 10^{-9} \) meters and \( D = 4.00 \times 10^{-3} \) meters into the formula. Thus, the angular resolution \( \theta = 1.22 \times \frac{500 \times 10^{-9}}{4.00 \times 10^{-3}} = 1.525 \times 10^{-4} \) radians.
03

Find the Astronaut's Altitude

Assume the altitude \( h \) and use the small angle approximation \( \theta \approx \frac{s}{h} \), where \( s = 75.0 \) meters is the separation between the point sources. Thus, \( h = \frac{s}{\theta} = \frac{75.0}{1.525 \times 10^{-4}} = 491,803.28 \) meters.

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Key Concepts

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

Diffraction Limited Resolution
Diffraction limited resolution is an essential concept in optics, describing the ability of an optical system to resolve detail within an image. When light passes through an aperture, it forms a diffraction pattern, and the resolution is said to be diffraction limited if these patterns influence how distinctly two closely spaced objects appear. In practical terms, this means that two point sources of light are only distinguishable from each other if their diffraction patterns do not overlap too much.
Given a circular aperture, Rayleigh's criterion is commonly used to define the resolution limit, mathematically expressed as \( \theta = 1.22 \frac{\lambda}{D} \). This gives the minimum angular separation (\( \theta \)) that is resolvable, based on the wavelength of light (\( \lambda \)) and the diameter of the aperture (\( D \)). The 1.22 factor is specific to circular apertures and originates from the first zero of the Bessel function, which governs the intensity pattern of diffraction in such cases.
Understanding this concept helps to grasp why in the given exercise, resolving two points on Earth from space depends heavily on the aperture size and the wavelength of light.
Angular Resolution
Angular resolution refers to the smallest angle over which a system, such as a telescope or camera, can distinguish two point sources separately. This is crucial for any observations needing precise detail, such as astronomical observations or photographing distant objects.
The formula for angular resolution, \( \theta = 1.22 \frac{\lambda}{D} \), directly ties into Rayleigh's criterion. The parameters—wavelength \( \lambda \) and aperture diameter \( D \)—determine how finely an image can be resolved. Essentially, the smaller the angular resolution, the better the system is at distinguishing fine details.
To put the theory into practice, consider the astronaut in orbit. She aims to resolve two distinct points on Earth with her eye acting as the observing system. Using the values given for this exercise: a light wavelength of 500 nm and a 4mm diameter pupil, she achieves an angular resolution of \( 1.525 \times 10^{-4} \) radians. This means that, within this angle, she can differentiate two distinct sources of light from her unique vantage point.
Astronaut Observations
Astronauts utilize their understanding of optics and angular resolution to make valuable observations from space. In this scenario, the astronaut's eye behaves like an optical instrument, and resolving two points 75 meters apart on Earth leverages diffraction-limited resolution principles.
The small angle approximation comes into play here. When the distance \( s \) between two objects and the altitude (or distance to the observer) is very large, the angle \( \theta \), determining their separation from the observer's perspective, can be approximated as \( \frac{s}{h} \). Plugging in the values we see: \( \theta \approx 1.525 \times 10^{-4} \) and \( s = 75 \) meters allows calculating the height \( h \) at approximately 491,803 meters.
This example shows the practical application of these optical principles, demonstrating how astronauts can determine their altitude by understanding the resolution and characteristics of the optics they use, whether in their eyes or equipment, to observe the Earth and beyond.

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Most popular questions from this chapter

Monochromatic light with wavelength 620 \(\mathrm{nm}\) passes through a circular aperture with diameter 7.4\(\mu \mathrm{m}\) . The resulting diffraction pattern is observed on a screen that is 4.5 \(\mathrm{m}\) from the aperture. What is the diameter of the Airy disk on the screen?

Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 \(\mathrm{km}\) . If they broadcast \(3.6-\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

If a diffraction grating produces its third-order bright band at an angle of \(78.4^{\circ}\) for light of wavelength 681 \(\mathrm{nm}\) , find (a) the number of slits per centimeter for the grating and \((b)\) the angular location of the first-order and second-order bright bands. (c) Will there be a fourth-order bright band? Explain.

Monochromatic light illuminates a pair of thin parallel slits at normal incidence, producing an interference pattern on a distant screen. The width of each slit is \(\frac{1}{7}\) the center-to-center distance between the slits. (a) Which interference maxima are missing in the pattem on the screen? (b) Does the answer to part (a) depend on the wavelength of the light used? Does the location of the missing maxima depend on the wavelength?

Parallel rays of light with wavelength 620 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of 40.0 \(\mathrm{cm}\) . The diffraction pattern is observed in the focal plane of the lens, and the distance from the center of the central maximum to the first minimum is \(36.5 \mathrm{cm} .\) What is the width of the slit? (Note: The angle that locates the first minimum is not small.)

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