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Chapter 9: Problem 1

Near field and far field. How far away should you be from a double slit of slit spacing \(0.1 \mathrm{~mm}\) irradiated with visible light in order to use the far-field approximation without making use of a lens? How far should you be from two microwave antennas having spacing \(10 \mathrm{~cm}\) and emitting \(3-\mathrm{cm}\) microwaves to use the far-field approximation?

Short Answer

Expert verified
18.18 m for visible light and 0.33 m for microwaves.

Step by step solution

01

Identify the criteria for far-field approximation

The far-field approximation criteria require that the distance from the slit or antenna, denoted as \(L\), must satisfy \( L \gg \frac{a^2}{\lambda} \). Here, \(a\) represents the slit or antenna spacing, and \(\lambda\) is the wavelength.
02

Calculate minimum distance for double slit with visible light

For visible light, let's assume an average wavelength \( \lambda = 550 \text{ nm} = 550 \times 10^{-9} \text{ m}\). Given the slit spacing \( a = 0.1 \text{ mm} = 0.1 \times 10^{-3} \text{ m} \), the far-field distance is found using:\[L \gg \frac{(0.1 \times 10^{-3})^2}{550 \times 10^{-9}}\]Calculating:\[L \gg \frac{0.1^2 \times 10^{-6}}{550 \times 10^{-9}} = \frac{0.01 \times 10^{-6}}{550 \times 10^{-9}} = \frac{0.01}{550} \times 10^{3} = 18.18 \text{ m}\]
03

Calculate minimum distance for microwave antennas

For the microwave antennas: \( a = 10 \text{ cm} = 0.1 \text{ m} \) and \( \lambda = 3 \text{ cm} = 0.03 \text{ m} \). Using the criteria:\[L \gg \frac{(0.1)^2}{0.03}\]Calculating:\[L \gg \frac{0.01}{0.03} = 0.33 \text{ m} \]
04

Interpret the results

For the double slit with visible light, the observer should be well beyond 18.18 m to use the far-field approximation. For the microwave antennas, the observer should be more than 0.33 m away to ensure the far-field conditions are met.

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Key Concepts

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

Wave Theory
Wave theory is fundamental to understanding how waves behave in various scenarios. It describes how waves propagate, interact, and contribute to different phenomena. In the case of light and other electromagnetic waves, wave theory helps explain how light beams can spread out as they travel, as well as how they can bend or bounce off surfaces. Wave theory considers several key concepts:
  • Wavelength: This is the distance over which a wave's shape repeats. It's usually denoted by the symbol \( \lambda \), and different types of waves (like light, sound, or microwaves) have different characteristic wavelengths.
  • Wavefronts: These are imaginary lines or surfaces embodying the wave's phase. They help visualize how waves propagate through space.
  • Amplitude: The height of the wave, which correlates to energy or intensity.
Understanding wave theory allows us to dive deeper into how these waves create interference patterns and diffraction under different circumstances.
Interference Patterns
Interference patterns are created when two or more waves overlap and combine. This can happen with light, sound, and other types of waves. The key here is how these waves interact at various points:
  • Constructive Interference: This occurs when waves meet in such a way that their amplitudes add up, leading to a brighter or louder signal. For example, when the crests of two waves align, the resulting wave is stronger.
  • Destructive Interference: This happens when waves meet out of phase, reducing or canceling each other out. In practical terms, this might mean a darker region in a light pattern or a quieter noise. It's like two waves having peak and trough at the same point.
When we talk about double-slit experiments, as mentioned in the earlier solution, the interference pattern occurs because light waves pass through two slits, creating alternating bright and dark bands on a screen. This pattern shows where the waves physically reinforcing each other or canceling out, making interference patterns a vivid demonstration of wave behavior.
Diffraction
Diffraction is a fascinating phenomenon that occurs when waves encounter obstacles or openings, causing them to spread out. This is closely related to wave theory and is critically important in understanding light and sound behavior. Essentially, diffraction explains why waves don't simply travel in straight lines but instead bend around corners and spread when passing through small openings. The amount of diffraction depends on the size of the obstacle or opening relative to the wavelength of the wave:
  • If the opening is much larger than the wavelength, only slight diffraction occurs.
  • If the opening is comparable to or smaller than the wavelength, significant spreading or bending can be observed.
In the context of the given problem, understanding diffraction is crucial for calculating the appropriate distances to meet far-field conditions. This ensures that we can predict where and how waves will form clear interference patterns, imperative for precise scientific applications like optics and radio communications.

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Most popular questions from this chapter

Burn a piece of toilet paper and look at it through your diffraction grating (held, as always, close in front of one eye). Notice the beautifully clear "first-order flame." This shows that the soft yellow light is almost monochromatic, with very little "white light" color spectrum due to hot carbon. The yellow that you see is the by now familiar (we hope) sodium doublet of wavelengths 5890 and \(5896 \AA\). Now that you recognize "sodium yellow," light an ordinary paper match and look at it with your grating. Most of the light is "hot carbon yellow," which is not really yellow but a complete "white" color spectrum. But look closely! In the yellow part of the hot carbon spectrum, down low next to the cardboard, where the flame is "blue" looking-below the blindingly bright hot carbon spectrum-do you see a crisp, clear little monochromatic match flame? If you don't, try again! Now burn other things and look. You may well conclude that everything is made of salt or is at least contaminated by it.

Eye-pupil size and mental activity. If someone shows you a picture of a good- looking individual of the opposite sex, your eye-pupil diameter may increase by as much as \(30 \%\), according to Eckhard \(\mathrm{H}\). Hess, Scientific American p. 46 (April, 1965 ). This large a change is very easy to detect in your own pupil by using a pinhole in a piece of aluminum foil that covers one eye, with a bright source illuminating the pinhole, as discussed in Sec. 9.7. Perhaps by just thinking, you can vary your pupil size, depending on what you think about. Have someone read to you. (Concentrate on listening, not on the pupil size.)

A plane slab of glass of thickness \(t\) and index \(n\) is inserted between an observer's eye and a point source. Show that the point source appears to be displaced to a point closer to the observer by approximately \([(n-1) / n] t . \quad\) Use small-angle approximations.

When you put your face under water and try looking without a face mask, everything looks blurred, because the change of index of refraction in going from water to eye is not very great. As a simplification, assume there is no change in index. Also assume your eye lens has very little effect, as if all the focusing were done at the first air-to-eye interface. (This is a crude approximation. Actually, you can see underwater to same extent.) Assume the focal length of that first surface is \(3 \mathrm{~cm}\), and that a parallel beam of light in air is brought to a focus at the retina. When you look underwater, you lose that focusing action. Design glasses that can be worn underwater so as to enable you to see clearly. Use glass with index of refraction 1.5. Show that if the focal length when used underwater is \(3 \mathrm{~cm}\), then the focal length when used in air is about \(1 \mathrm{~cm} .\) If one of these glass lenses is used as an ordinary magnifying glass what is its magnification? Suppose you use an ordinary glass marble for the lens. You want to form an image (of a parallel beam in water) \(3 \mathrm{~cm}\) behind the rear surface of the marble. What should be the diameter of the marble?

Light is emitted from an unpolarized point source. First it passes through a linear polarizer with easy transmission axis at 45 deg to the \(x\) and \(y\) axes. Then it is incident on a double slit. Each slit is covered by a linear polarizer, one slit having the polarization axis along \(\hat{x}\), the other having it along \(\hat{y}\). (a) Suppose you look at the interference pattern with the unaided eye. Do you expect to have the usual two-slit interference pattern? What do you expect? (b) Next suppose you look at the interference pattern while holding a polaroid linear polarizer in front of one eye. What do you expect to see? What happens as you rotate the polaroid in front of your eye? (c) Now suppose you look at the pattern through a circular polarizer run backward as an analyzer. What pattern do you expect to see? There are many nice variations you can make on this problem: (i) Put a righthanded circular polarizer over one slit and a left-banded circular polarizer over the other and repeat the above observations. (ii) Add a quarter- or half-wave plate just behind the slits, etc.
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