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Tsunamit On December \(26,2004,\) a violent magnitude 9.1 earthquake 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) Africa gap: 10.25°, Australia gap: 12.29°.

Step by step solution

01

Calculate the Wavelength

The formula to find the wavelength \( \lambda \) is \( \lambda = v \cdot T \), where \( v \) is the speed of the wave and \( T \) is the period of the wave. The speed is given as 800 km/h and the period is 1.0 h. Plug these values into the equation to find the wavelength \( \lambda = 800 \times 1 = 800 \) km.
02

Diffraction through the African Gap

For Fraunhofer diffraction, the condition for the first minimum angle \( \theta_1 \) is given by \( \sin(\theta_1) = \frac{m\lambda}{a} \), where \( m = 1 \) for the first minimum, \( \lambda = 800 \) km (from Step 1), and \( a = 4500 \) km for the African gap. Calculate \( \theta_1 \) using this formula: \( \sin(\theta_1) = \frac{1 \times 800}{4500} \), leading to \( \theta_1 = \arcsin(0.1778) \approx 10.25^{\circ} \).
03

Diffraction through the Australian Gap

Use the same formula for the first minimum angle \( \theta_2 \). Here, \( a = 3700 \) km for the Australian gap. Calculate \( \theta_2 \) using the formula: \( \sin(\theta_2) = \frac{1 \times 800}{3700} \), leading to \( \theta_2 = \arcsin(0.2162) \approx 12.29^{\circ} \).

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Key Concepts

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

Wavelength Calculation
To understand how tsunami waves behave, we first need to calculate their wavelength. The wavelength of a wave is the distance over which the wave's shape repeats. For a tsunami, which is essentially a really long wave, this distance can be enormous.
To calculate the wavelength, we use the formula \( \lambda = v \cdot T \), where \(v\) is the speed of the wave, and \(T\) is the period, or time between crests. For the 2004 tsunami, the wave speed \(v\) was measured to be 800 km/h, and the period \(T\) was 1 hour. Plug these into the equation to get:
  • \(\lambda = 800 \times 1 = 800\) km.
This simple calculation reveals that a tsunami can have a wavelength comparable to the distance between two large cities!
Fraunhofer Diffraction
When a wave encounters an obstacle, such as continents, it bends around or passes through gaps like a slit, a phenomenon known as diffraction. Fraunhofer diffraction deals with wave behavior at considerable distances from the openings.
Mathematically, the angle \(\theta\) where diffraction results in a minimum (cancellation) is given by \(\sin(\theta) = \frac{m\lambda}{a}\). Here, \(m\) is the order of the minimum (usually 1 for the first minimum), \(\lambda\) is the wavelength, and \(a\) is the width of the gap.
In the exercise, the wave travels through gaps between continents:
  • The gap between Africa and Antarctica is 4500 km.
  • Between Australia and Antarctica, it is 3700 km.
These large distances lead to significant bending effects as the wave spreads through these gaps.
Wave Speed
Understanding wave speed is crucial for predicting how fast and far a tsunami can travel. Wave speed depends on both the medium (e.g., deep ocean) and the wave properties (e.g., wavelength and period).
For the 2004 tsunami, the wave speed was 800 km/h in the open ocean. This speed is typical for tsunamis formed by large seismic events such as continental earthquakes. The fast-moving waves can cross entire oceans in hours, leading to widespread impacts across distant shores.
Wave speed affects:
  • The time between wave crests, known as the wave period.
  • How quickly a tsunami can reach shorelines, impacting warning times.
  • The energy distribution as the wave travels.
Hence, understanding wave speed is essential for tsunami forecasting and mitigating damage.
Continental Earthquake Impact
Continental earthquakes, especially those occurring underwater, can have devastating effects by generating tsunamis. When tectonic plates shift abruptly, they displace huge amounts of water, forming waves that propagate across entire ocean basins.
In the case of a major event like the 2004 Sumatra earthquake:
  • Enormous energy is released, leading to powerful waves.
  • The interaction between seismic waves and ocean water drags the latter into motion.
  • Consequential tsunamis can affect coastlines thousands of miles away.
These earthquakes are pivotal in tsunami generation, making it crucial to monitor seismic activities in vulnerable regions. Early warning systems are vital for saving lives by alerting populations of imminent wave threats.

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Most popular questions from this chapter

X rays of wavelength 0.0850 \(\mathrm{nm}\) are scattered from the atoms of a crystal. The second-order maximum in the Bragg reflection occurs when the angle \(\theta\) in Fig. 36.23 is \(21.5^{\circ} .\) What is the spacing between adjacent atomic planes in the crystal?

A converging lens 7.20 \(\mathrm{cm}\) in diameter has a focal length of 300 \(\mathrm{mm}\) . If the resolution is diffraction limited, how far away can an object be if points on it 4.00 \(\mathrm{mm}\) apart are to be resolved (according to Rayleigh's criterion)? Use \(\lambda=550 \mathrm{nm} .\)

Parallel rays of light with wavelength 620 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of 40.0 \(\mathrm{cm}\) . The diffraction pattern is observed in the focal plane of the lens, and the distance from the center of the central maximum to the first minimum is \(36.5 \mathrm{cm} .\) What is the width of the slit? (Note: The angle that locates the first minimum is not small.)

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 \(\mathrm{mm}\) wide. At the point in the pattern \(3.25^{\circ}\) from the center of the central maximum, the total phase difference between wavelets from the tup and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is \(I_{0} ?\)

Parallel rays of green mercury light with a wavelength of 546 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of \(60.0 \mathrm{cm} .\) In the focal plane of the lens the distance from the central maximum to the first minimum is 10.2 \(\mathrm{mm}\) . What is the width of the slit?

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