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Chapter 18: Problem 107

Find the diameter of the aperture (opening) of a camera that can resolve detail on the ground the size of a person \((2.0 \mathrm{~m})\) from an SR-71 Blackbird flying at an altitude of \(27 \mathrm{~km}\). Assume that the light used to form the image has a wavelength of \(450 \mathrm{~nm}\). (Hint: Set the angle for the first-order dark fringe of the diffraction pattern equal to the angle subtended by a person at \(27 \mathrm{~km}\).)

Short Answer

Expert verified
The diameter of the aperture is approximately 7.4 mm.

Step by step solution

01

Understand the Concept

The problem is asking to resolve a detail, which implies using the concept of resolving power of a camera lens given by the Rayleigh criterion. According to the criterion, two objects can be resolved if the central maximum of one image coincides with the first minimum of the other, leading to an equation involving the aperture diameter.
02

Identify the Formula

We use the formula from the Rayleigh criterion for the angular resolution: \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angle, \( \lambda \) is the wavelength of the light, and \( D \) is the diameter of the aperture. This formula will allow us to find \( D \).
03

Calculate the Angle

Convert the resolving distance and altitude into an angle. Use the formula \( \theta \approx \frac{h}{d} \), where \( h = 2 \) meters (height of person) and \( d = 27,000 \) meters (altitude):\[ \theta \approx \frac{2}{27,000} \approx 7.41 \times 10^{-5} \text{ radians} \]
04

Rearrange the Rayleigh Criterion

Rearrange the Rayleigh formula to solve for \( D \):\[ D = 1.22 \frac{\lambda}{\theta} \]
05

Substitute the Values

Plug in the known values: \( \lambda = 450 \times 10^{-9} \) meters, and \( \theta = 7.41 \times 10^{-5} \) radians, into the rearranged formula:\[ D = 1.22 \frac{450 \times 10^{-9}}{7.41 \times 10^{-5}} \approx 7.4 \times 10^{-3} \text{ meters} \]
06

Conclusion

Convert the diameter from meters to millimeters for practicality: \( 7.4 \times 10^{-3} \text{ meters} = 7.4 \text{ mm} \). Thus, the diameter of the aperture needed is approximately \(7.4 \text{ mm}\).

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Key Concepts

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

Angular Resolution
Angular resolution refers to the ability of an optical instrument to distinguish between two closely spaced objects. It's essentially the instrument's capability to see small details. The smaller the angle between the two objects that can be just resolved, the better the angular resolution.
  • The angle \( \theta \), often measured in radians, indicates how far apart the images of two close objects can be detected as separate.
  • The Rayleigh criterion provides a way to calculate this angle, prescribing when two images are resolvable based on diffraction effects.
To enhance angular resolution, an optical instrument generally needs to:
  • Have a larger aperture. This reduces the diffraction effects, which broadens the light from each point object, limiting resolution.
  • Utilize shorter wavelengths of light, as shorter wavelengths can help produce sharper images.
Understanding angular resolution is crucial when dealing with cameras, telescopes, and microscopes. In the context of the exercise, the angular resolution determined the smallest details, like a person, that the camera could resolve from a high altitude.
Diffraction Pattern
A diffraction pattern is the spread of light waves as they pass through a small aperture or around an obstacle. Instead of traveling in straight lines, the light waves bend around the edges, causing a pattern of light and dark bands.
  • The central bright fringe, called the central maximum, is the most intense part of the pattern.
  • Dark fringes occur at points where destructive interference cancels out the light waves.
The appearance of diffraction patterns is an inevitable consequence of wave nature, which affects the clarity of the images formed by optical instruments.
  • By controlling the diffraction pattern, the resolution of an image can be improved.
  • In most practical scenarios, minimizing diffraction effects can allow for clearer and sharper images.
In the SR-71 Blackbird exercise, the diffraction pattern limits how much detail can be resolved on the ground from high up. The first dark fringe aligns with where the Rayleigh criterion predicts, helping to calculate the required aperture size.
Resolving Power
Resolving power is an extension of angular resolution. It's how well an optical system can distinguish between two points that are very close together in the object being viewed. A system with high resolving power can differentiate two closely spaced objects as separate entities.
  • Improved resolving power means better detail and clarity in the resulting image.
  • The resolving power is directly influenced by both the aperture of the system and the wavelength of light used.
In the exercise, determining the resolving power of the camera (mounted on the SR-71) allowed us to calculate the aperture size needed to distinguish a person on the ground. Innovations in optical technology often focus on enhancing resolving power to see finer details over greater distances.
Wavelength of Light
The wavelength of light is a fundamental property that affects many aspects of optics. It's the distance between consecutive crests of a wave and is usually measured in nanometers (nm).
  • Shorter wavelengths correspond to higher frequencies and can provide better resolution because they cause less diffraction.
  • Visible light ranges from approximately 400 nm (violet) to 700 nm (red).
For most optical systems, using a shorter wavelength can enhance the overall image sharpness and detail resolution.
  • In our exercise, the light used has a wavelength of 450 nm, placing it in the blue-violet part of the visible spectrum.
  • This shorter wavelength assists in achieving a finer image resolution than longer wavelengths would allow.
The choice of wavelength is essential in optical design, particularly when precision and detail are paramount, such as in high-altitude photography or microscopy.

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Most popular questions from this chapter

Describe how the pixels on a TV screen give the sensation of all possible colors. Also, take a look at a comic strip with a magnifying glass. Discuss what you see there in the context of the diffraction of light.

Two point sources of light with a wavelength of \(690 \mathrm{~nm}\) are viewed through the \(5.1\)-mm-diameter pupil of an eye. What is the minimum angle of separation between the two sources if they are to be seen as being separate? (Hint: The desired minimum angle is equal to the angle to the first dark fringe of the diffraction pattern.)

What is iridescence?

Light with a wavelength of \(520 \mathrm{~nm}\) is used in a two-slit experiment. If the angle to the first bright fringe above the central fringe is \(0.067^{\circ}\), what is the separation of the slits?

A thin film of oil ( \(n=1.2\) ) coats the surface of a glass lens ( \(n=1.3\) ). What is the phase change of light that reflects from (a) the air-to-oil boundary and (b) the oilto-glass boundary? Explain in each case.
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