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On December \(26,2004,\) a violent earthquake of magnitude 9.1 occurred off the coast of Sumatra. This quake triggered a huge tsunami (similar to a tidal wave) that killed more than \(150,000\) people. Scientists observing the wave on the open ocean measured the time between crests to be 1.0 \(\mathrm{h}\) and the speed of the wave to be 800 \(\mathrm{km} / \mathrm{h}\) . Computer models of the evolution of this enormous wave showed that it bent around the continents and spread to all the oceans of the earth. When the wave reached the gaps between continents, it diffracted between them as through a slit. (a) What was the wavelength of this tsunami? (b) The distance between the southern tip of Africa and northern Antarctica is about \(4500 \mathrm{km},\) while the distance between the southern end of Australia and Antarctica is about 3700 \(\mathrm{km} .\) As an approximation, we can model this wave's behavior by using Fraunhofer diffraction. Find the smallest angle away from the central maximum for which the waves would cancel after going through each of these continental gaps.

Short Answer

Expert verified
(a) 800 km, (b) African-Antarctica gap: 10.26°, Australia-Antarctica gap: 12.56°.

Step by step solution

01

Calculate the Wavelength

To find the wavelength \(\lambda\) of the tsunami, use the formula \(\lambda = v \cdot T \), where \(v\) is the speed of the wave and \(T\) is the period. The speed \(v\) is given as 800 km/h, and the period \(T\) is 1.0 h. Therefore, \(\lambda = 800 \, \text{km/h} \times 1.0 \, \text{h} = 800 \, \text{km}\).
02

Setup for Diffraction Calculation

To find the smallest angle \(\theta\) for destructive interference using Fraunhofer diffraction when the wave diffracts through a gap, use the formula for the first minimum: \(d \sin(\theta) = m\lambda\), where \(d\) is the distance (gap), \(m\) is the order of minimum (we use \(m = 1\) for first minimum), and \(\lambda\) is the wavelength calculated before.
03

Calculate Smallest Angle for African-Antarctica Gap

For the African-Antarctica gap, where \(d = 4500\, \text{km}\):\[4500 \sin(\theta) = 1 \times 800\]\[\sin(\theta) = \frac{800}{4500} \approx 0.1778\]Calculate \(\theta\) using \(\theta = \arcsin(0.1778) \approx 10.26^\circ\).
04

Calculate Smallest Angle for Australia-Antarctica Gap

For the Australia-Antarctica gap, where \(d = 3700\, \text{km}\):\[3700 \sin(\theta) = 1 \times 800\]\[\sin(\theta) = \frac{800}{3700} \approx 0.2162\]Find \(\theta\) using \(\theta = \arcsin(0.2162) \approx 12.56^\circ\).

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Key Concepts

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

Understanding Tsunamis
A tsunami is often described as a series of massive ocean waves caused primarily by underwater disturbances such as earthquakes, volcanic eruptions, or landslides. The 2004 Indian Ocean tsunami, triggered by a major earthquake off the coast of Sumatra with a magnitude of 9.1, serves as a striking example. Tsunamis can be incredibly destructive, as they transfer immense amounts of energy over large distances across oceans, affecting coastal regions thousands of kilometers away.
Unlike regular ocean waves, which are caused by winds and typically have shorter wavelengths and periods, tsunamis have very long wavelengths, often exceeding 100 kilometers. Their periods can range from several minutes to a few hours. This long wavelength allows them to travel across deep ocean waters at high speeds almost undetected and build up to ferocious heights as they approach shorelines.
  • Key Causes: Underwater earthquakes are the primary cause of tsunamis, though volcanic eruptions and landslides can also trigger them.
  • Impact: They can lead to massive loss of life and property, particularly in coastal regions.
Wave Speed Essentials
Wave speed is a fundamental characteristic of waves, calculated by the equation \( v = \lambda \cdot f \) where \( v \) is the wave speed, \( \lambda \) is the wavelength, and \( f \) is the frequency. In the case of the tsunami mentioned, the wave speed was 800 km/h.
Wave speed is crucial in understanding how quickly a wave front can travel across a medium. For tsunamis, this rapid speed ensures they can cover large oceanic distances swiftly, which contributes to their potentially devastating impact.
Several factors govern wave speed:
  • Medium: Different mediums affect wave speeds differently—waves tend to travel faster in deeper water.
  • Wavelength and Frequency relation: Longer wavelengths typically result in faster speeds if other factors are constant.
Understanding wave speed helps in modeling and predicting the behavior of tsunamis as they travel vast oceanic expanses.
Exploring Wavelength
Wavelength, denoted as \( \lambda \), is the spatial period of a wave—the distance over which the wave's shape repeats. It is measured in the direction of the wave's propagation.
For the tsunami in this exercise, the wavelength was found to be 800 km. This large wavelength is characteristic of tsunamis, distinguishing them from the shorter wavelengths of tidal or wind-driven waves.
  • Characteristics: Longer wavelengths enable these waves to travel across entire ocean basins with little energy loss.
  • Calculation: Wavelength is calculated via \( \lambda = v \cdot T \), where \( v \) is wave speed and \( T \) the period.
  • Impact on Shore: As waves approach shallow waters, their speed decreases, and they grow in height.
A profound understanding of wavelength is vital when predicting and analyzing wave behavior, especially in potential disaster scenarios like tsunamis.
Fraunhofer Diffraction Explained
Fraunhofer diffraction is a principle that describes wave behavior when it encounters an obstacle or slit that is comparable in size to its wavelength. It's based on the idea of interference and diffraction patterns.
In the context of tsunamis, as these massive waves pass through gaps between continents, their behavior can be analyzed using Fraunhofer diffraction principles. Although normally applied in optics for light, this concept helps in modeling how a wave "bends" around obstacles.
This exercise used the formula \( d \sin(\theta) = m\lambda \) to calculate the smallest angle \( \theta \) for destructive interference:
  • First Minimum Formula: \( m = 1 \) for the first minimum (where waves cancel).
  • Destructive Interference: When waves from two paths meet out of phase, they cancel each other out, leading to a reduction in wave height.
  • Applications: Helps scientists understand how tsunami waves interact with geographical features.
Understanding Fraunhofer diffraction can be crucial in the study of complex wave interactions in various environments, providing valuable insight into wave behavior across different scenarios.

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Most popular questions from this chapter

In a large vacuum chamber, monochromatic laser light passes through a narrow slit in a thin aluminum plate and forms a diffraction pattern on a screen that is 0.620 \(\mathrm{m}\) from the slit. When the aluminum plate has a temperature of \(20.0^{\circ} \mathrm{C},\) the width of the central maximum in the diffraction pattern is 2.75 \(\mathrm{mm}\) . What is the change in the width of the central maximum when the temperature of the plate is raised to \(520.0^{\circ} \mathrm{C}\) ? Does the width of the central diffraction maximum increase or decrease when the temperature is increased?

If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 \(\mathrm{nm}\) ) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 \(\mathrm{nm} ) ?\)

Light of wavelength 585 nm falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be? (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

The intensity of light in the Fraunhofer diffraction pattern of a single slit is $$I=I_{0}\left(\frac{\sin \gamma}{\gamma}\right)^{2}$$ where $$\gamma=\frac{\pi a \sin \theta}{\lambda}$$ (a) Show that the equation for the values of \(\gamma\) at which \(I\) is a maximum is \(\tan \gamma=\gamma .(\) b) Determine the three smallest positive values of \(\gamma\) that are solutions of this equation. (Hint: You can use a trial-and-error procedure. Guess a value of \(\gamma\) and adjust your guess to bring \(\tan \gamma\) closer to \(\gamma .\) A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.)

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

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