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Chapter 36: Problem 27

If Superman really had \(\mathrm{x}\) -ray vision at \(0.10 \mathrm{~nm}\) wavelength and a \(4.0 \mathrm{~mm}\) pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by \(5.0 \mathrm{~cm}\) to do this?

Short Answer

Expert verified
Superman can discern heroes from villains at about 1,640 km altitude.

Step by step solution

01

Identify Given Values

We are given the wavelength of Superman's x-ray vision as \( \lambda = 0.10 \text{ nm} \) (or \( 0.10 \times 10^{-9} \text{ m} \)), the pupil diameter as \( D = 4.0 \text{ mm} \) (or \( 4.0 \times 10^{-3} \text{ m} \)), and the distance needed to resolve as \( s = 5.0 \text{ cm} \) (or \( 0.05 \text{ m} \)).
02

Use Rayleigh Criterion

The Rayleigh criterion gives the minimum resolvable angle \( \theta \) in radians as \( \theta = \frac{1.22 \cdot \lambda}{D} \). Substitute the values: \( \theta = \frac{1.22 \times 0.10 \times 10^{-9}}{4.0 \times 10^{-3}} \).
03

Calculate Minimum Resolvable Angle

Perform the calculation for \( \theta \): \[ \theta = \frac{1.22 \times 0.10 \times 10^{-9}}{4.0 \times 10^{-3}} = 3.05 \times 10^{-8} \text{ radians} \].
04

Relate Angle to Maximum Distance

Using the small angle approximation, \( \tan(\theta) \approx \theta \). We relate this to the maximum distance Superman can distinguish \( h \) using \( \theta = \frac{s}{h} \). Rearrange to find \( h \): \( h = \frac{s}{\theta} \).
05

Calculate Maximum Altitude

Substitute the known values into \( h = \frac{0.05}{3.05 \times 10^{-8}} \): \[ h = \frac{0.05}{3.05 \times 10^{-8}} \approx 1.64 \times 10^{6} \text{ m} \].

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Key Concepts

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

X-ray Vision
X-ray vision refers to the ability to see through objects, a trait popularized by superhero characters like Superman. In reality, X-rays are a form of electromagnetic radiation just like visible light, but with higher energy and shorter wavelengths. X-rays are used in medical imaging to see inside the human body. Unlike Superman's fictional ability, real-life X-rays cannot simply be controlled or focused by the human eye. Instead, specialized equipment such as X-ray machines utilize this type of radiation to produce images of internal structures.
Wavelength
Wavelength is the distance between two peaks of a wave. It is an essential concept in understanding different types of electromagnetic radiation, including X-rays. The shorter the wavelength, the higher the energy and the ability to penetrate materials. In the context of X-ray vision, we are considering very short wavelengths, such as 0.10 nm (nanometers) in the given exercise. This short wavelength allows X-rays to pass through various materials, making them useful in medical imaging and security applications, but our eyes are not able to see them.
Pupil Diameter
The pupil diameter is crucial in determining how much light enters the eye. It works like an aperture in a camera, controlling the amount of light that reaches the lens. A typical human pupil can range from about 2 to 8 mm. In the exercise, a 4.0 mm pupil diameter is used. This dimension plays a significant role in the Rayleigh Criterion, which calculates the resolving power of an optical system. The larger the diameter, the better the resolution, or the ability to distinguish between two points.
Resolution in Optics
Optical resolution is the ability to distinguish between two points that are close to each other. In optics, resolution is often determined by the Rayleigh Criterion, which relates resolution to wavelength and pupil diameter. According to the Rayleigh Criterion, the minimum resolvable angle, \( \theta \), can be calculated using the formula \( \theta = \frac{1.22 \cdot \lambda}{D} \). This formula indicates that both a shorter wavelength and a larger pupil diameter improve resolution. Thus, to achieve higher resolution, one must aim for a lower value of \( \theta \). In our exercise, this concept is critical for determining the maximum altitude at which Supermann can differentiate villains from heroes.

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Most popular questions from this chapter

In an experiment to monitor the Moon's surface with a light beam, pulsed radiation from a ruby laser \((\lambda=0.69 \mu \mathrm{m})\) was directed to the Moon through a reflecting telescope with a mirror radius of \(1.3 \mathrm{~m}\). A reflector on the Moon behaved like a circular flat mirror with radius \(10 \mathrm{~cm}\), reflecting the light directly back toward the telescope on Earth. The reflected light was then detected after being brought to a focus by this telescope. Approximately what fraction of the original light energy was picked up by the detector? Assume that for each direction of travel all the energy is in the central diffraction peak.

X rays of wavelength \(0.12 \mathrm{~nm}\) are found to undergo secondorder reflection at a Bragg angle of \(28^{\circ}\) from a lithium fluoride crystal. What is the interplanar spacing of the reflecting planes in the crystal?

(a) A circular diaphragm \(60 \mathrm{~cm}\) in diameter oscillates at a frequency of \(25 \mathrm{kHz}\) as an underwater source of sound used for submarine detection. Far from the source, the sound intensity is distributed as the diffraction pattern of a circular hole whose diameter equals that of the diaphragm. Take the speed of sound in water to be \(1450 \mathrm{~m} / \mathrm{s}\) and find the angle between the normal to the diaphragm and a line from the diaphragm to the first minimum. (b) Is there such a minimum for a source having an (audible) frequency of \(1.0 \mathrm{kHz}\) ?

A double-slit system with individual slit widths of \(0.030 \mathrm{~mm}\) and a slit separation of \(0.18 \mathrm{~mm}\) is illuminated with \(500 \mathrm{~nm}\) light directed perpendicular to the plane of the slits. What is the total number of complete bright fringes appearing between the two first-order minima of the diffraction pattern? (Do not count the fringes that coincide with the minima of the diffraction pattern.)

Babinet's principle. A monochromatic beam of parallel light is incident on a "collimating" hole of diameter \(x>\lambda\). Point \(P\) lies in the geometrical shadow region on a distant screen (Fig. \(36-39 a\) ). Two diffracting objects, shown in Fig. \(36-39 b\), are placed in turn over the collimating hole. Object \(A\) is an opaque circle with a hole in it, and \(B\) is the "photographic negative" of \(A\). Using superposition concepts, show that the intensity at \(P\) is identical for the two diffracting objects \(A\) and \(B\).
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