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A spy satellite orbiting at \(160 \mathrm{~km}\) above Earth's surface has a lens with a focal length of \(3.6 \mathrm{~m}\) and can resolve objects on the ground as small as \(30 \mathrm{~cm}\). For example, it can easily measure the size of an aircraft's air intake port. What is the effective diameter of the lens as determined by diffraction consideration alone? Assume \(\lambda=550 \mathrm{~nm}\).

Short Answer

Expert verified
The effective diameter of the lens is approximately 0.358 meters.

Step by step solution

01

Understand the Problem

We are given the focal length of a lens, the height above the Earth's surface, the size of the objects it can resolve on the ground, and the wavelength of light. We need to find the effective diameter of the lens using the principle of diffraction.
02

Determine the Resolution Limit by Diffraction

The minimum resolvable detail by a lens is determined by the Rayleigh criterion: \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular resolution, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the lens.
03

Calculate the Angular Resolution

Convert the resolving power from linear distance to angular measure. This is done by finding the angular size using the resolution limit: \( \theta = \frac{s}{h} \), where \( s \) is the size of object on the ground (30 cm or 0.3 m), and \( h \) is the height of the satellite (160 km or 160,000 m). Therefore, \( \theta = \frac{0.3}{160000} \approx 1.875 \times 10^{-6} \) radians.
04

Rearrange the Rayleigh Criterion Equation

We have \( \theta = 1.22 \frac{\lambda}{D} \). Solve for \( D \) to find the effective diameter of the lens: \( D = 1.22 \frac{\lambda}{\theta} \).
05

Substitute Given Values into the Equation

Substitute \( \lambda = 550 \times 10^{-9} \) meters and \( \theta = 1.875 \times 10^{-6} \) radians into the equation: \( D = 1.22 \frac{550 \times 10^{-9}}{1.875 \times 10^{-6}} \).
06

Perform the Calculation

Calculate \( D = 1.22 \frac{550 \times 10^{-9}}{1.875 \times 10^{-6}} = 0.358 \text{ meters} \). Thus, the effective diameter of the lens is approximately 0.358 meters.

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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 fundamental principle used to understand how well a lens can resolve details. It helps define the limit beyond which two close point sources of light distinguishable, forming the basis for calculating angular resolution. According to this criterion, the minimum angle of separation (\( \theta \)) between two points that can still be resolved is given by the formula:
  • \( \theta = 1.22 \frac{\lambda}{D} \)
Here, \( \lambda \) represents the wavelength of the light in meters, and \( D \) is the diameter of the lens in meters. The factor 1.22 stems from the diffraction pattern of a circular aperture, known to cause the first minimum of the pattern to appear at this specific function when units are in radians.
By understanding the Rayleigh Criterion, we can effectively determine the limitations imposed by diffraction in any optical system, such as telescopes or lens systems within satellites.
Angular Resolution
Angular resolution measures how well a lens or ear can distinguish between two distinct points in the field of view. It's directly derived from the Rayleigh Criterion and is pivotal in optical instruments.
In practical applications, like the spy satellite example, angular resolution gets calculated by converting linear spatial resolution to an angular measure. This conversion can be achieved using the simple formula:
  • \( \theta = \frac{s}{h} \)
where \( s \) is the smallest object size that can be resolved (e.g., 0.3 meters on Earth), and \( h \) is the height of the observation point above the surface (e.g., 160,000 meters for a satellite).
Thus, angular resolution becomes critical when determining how detailed an image a lens can provide. It shows us exactly what size of objects can be clearly identified at a given distance, something particularly vital for surveillance or astronomical applications.
Satellite Optics
Satellite optics refers to the systems and lenses used in satellites to capture images or data from their hover above Earth. These optics must often serve in challenging conditions, including vast distances and subtle changes in environmental conditions.
One critical factor in designing these optics is understanding how diffraction affects the system. The lens's effective diameter, calculated from previously outlined principles like the Rayleigh Criterion, affects how much detail the satellite's optics can resolve. In scenarios like observing Earth from 160 km up, every centimeter in the lens size can make a significant difference in resolving fine details like aircraft structures.
Additionally, satellite optics must balance weight, cost, and practicality, often leading to advanced engineering solutions to optimize lenses to be both lightweight and duty-capable for the task. Advanced materials, precise fabrication methods, and innovative designs are crucial in ensuring that satellites achieve their resolution requirements while remaining feasible for launch.

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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.

In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui, Hawaii, and reflected back from the shuttle Discovery as it sped by \(354 \mathrm{~km}\) overhead. The diameter of the central maximum of the beam at the shuttle position was said to be \(9.1 \mathrm{~m}\). and the beam wavelength was \(500 \mathrm{~nm}\). What is the effective diameter of the laser aperture at the Maui ground station? (Hint: A laser beam spreads only because of diffraction; assume a circular exit aperture.)

In a certain two-slit interference pattern, 10 bright fringes lie within the second side peak of the diffraction envelope and diffraction minima coincide with two-slit interference maxima. What is the ratio of the slit separation to the slit width?

A diffraction grating illuminated by monochromatic light normal to the grating produces a certain line at angle \(\theta .(\mathrm{a}) \mathrm{What}\) is the product of that line's half-width and the grating's resolving power? (b) Evaluate that product for the first order of a grating of slit separation \(900 \mathrm{~nm}\) in light of wavelength \(600 \mathrm{~nm}\).

Assume that the limits of the visible spectrum are arbitrarily chosen as 430 and \(680 \mathrm{~nm}\). Calculate the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of \(20.0^{\circ} .\)

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