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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 h and the speed of the wave to be 800 km/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 km, while the distance between the southern end of Australia and Antarctica is about 3700 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) 10.23° for Africa-Antarctica, 12.52° for Australia-Antarctica.

Step by step solution

01

Understand the Given Problem

We need to calculate the wavelength of the tsunami wave using its speed and the time interval between wave crests. Then we will use Fraunhofer diffraction to find the smallest angle for wave cancellation between two continental gaps.
02

Calculate the Wavelength of the Tsunami

To find the wavelength, use the formula: \( \lambda = \frac{v}{f} \). Given the speed \( v = 800 \, \text{km/h} \) and the period of the wave \( T = 1.0 \, \text{h} \), and knowing that frequency \( f = \frac{1}{T} \), we have:\[ \lambda = \frac{800 \, \text{km/h}}{1.0/h} = 800 \, \text{km} \].Thus, the wavelength is \( 800 \, \text{km} \).
03

Use Fraunhofer Diffraction for Continental Gaps

For Fraunhofer diffraction, the formula for the angle \( \theta \) where destructive interference occurs is given by \( a \sin \theta = m\lambda \), where \( a \) is the width of the gap, \( \lambda \) is the wavelength, and \( m \) is the order of the minimum (for the first minimum, \( m = 1 \)).
04

Calculate Minimum Angle for the First Gap

For the gap between Africa and Antarctica, \( a = 4500 \, \text{km} \).Using \( \lambda = 800 \, \text{km} \) in the diffraction formula:\[ 4500 \sin \theta = 1 \times 800 \]\[ \sin \theta = \frac{800}{4500} \approx 0.1778 \]Thus, \( \theta = \arcsin(0.1778) \approx 10.23^\circ \).
05

Calculate Minimum Angle for the Second Gap

For the gap between Australia and Antarctica, \( a = 3700 \, \text{km} \).Using \( \lambda = 800 \, \text{km} \) in the diffraction formula:\[ 3700 \sin \theta = 1 \times 800 \]\[ \sin \theta = \frac{800}{3700} \approx 0.2162 \]Thus, \( \theta = \arcsin(0.2162) \approx 12.52^\circ \).

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Key Concepts

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

Wavelength Calculation
To determine the wavelength of a tsunami, we apply basic wave physics. The wavelength (\( \lambda \)) of a wave is calculated using the formula:
  • \( \lambda = \frac{v}{f} \)
where \( v \) is the speed of the wave, and \( f \) is its frequency. For our tsunami, the speed \( v \) is given as 800 km/h. The time between wave crests is 1.0 hour, representing the period \( T \). Frequency is the reciprocal of the period:
  • \( f = \frac{1}{T} \)
For this tsunami, \( f \) equals 1/h, substituting into the formula, the wavelength is:
  • \( \lambda = \frac{800 \, \text{km/h}}{1/h} = 800 \, \text{km} \)
So, this wave travels with crests 800 km apart in open ocean, highlighting its enormous nature.
Fraunhofer Diffraction
Fraunhofer diffraction describes how waves behave when they encounter an obstacle, such as the continental gaps here. It's crucial for understanding how the tsunami spread through narrow passages without losing much energy. This model is typically applied when wave sources and observation points are far from the diffracting object, which is the case with vast oceanic distances. In such diffraction scenarios, destructive interference – when waves cancel each other out – occurs at specific angles. These angles (\( \theta \)) can be calculated using the formula:
  • \( a \sin \theta = m \lambda \)
Here, \( a \) is the width of the gap, \( \lambda \) is the wavelength (800 km here), and \( m \) is the order of the minimum (starting at \( m = 1 \) for the first minimum). By solving this, you can find angles where the wave impact is least felt.
Wave Speed Calculation
The speed of a wave tells us how fast it moves across a medium, and it's an integral part of understanding tsunamis. For this problem, we calculated the speed of the wave (\( v \)) and were given as 800 km/h. This value helps determine other crucial wave properties, such as wavelength and frequency.Wave speed (\( v \)) is related to wavelength (\( \lambda \)) and frequency (\( f \)) by the fundamental wave equation:
  • \( v = \lambda \cdot f \)
Given the massive scale of a tsunami, such a high speed plays a critical role in how far and fast a wave can travel, impacting vast regions and posing serious challenges to coastal safety.
Destructive Interference
Destructive interference is a fascinating phenomenon where two waves of similar amplitude meet, and effectively "cancel" each other out. In our context, as the tsunami wave passes through continental gaps, it diffuses and can interfere with itself destructively. For destructive interference to occur, differences in path length between two waves must be an odd multiple of half-wavelengths. In simpler terms, this means waves need to be out of phase, causing the heights of crests and troughs to match oppositely. Using the Fraunhofer diffraction formula, angles where this cancellation occurs can be found, explaining why some areas might experience reduced wave impact. Thus, even a massive, energetic tsunami can occasionally be less destructive, thanks to the principles of interference.

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Most popular questions from this chapter

An interference pattern is produced by light of wavelength 580 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 mm. (a) If the slits are very narrow, what would be the angular positions of the first-order and second-order, two-slit interference maxima? (b) Let the slits have width 0.320 mm. In terms of the intensity \(I_0\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

You are asked to design a space telescope for earth orbit. When Jupiter is 5.93 \(\times\) 10\(^8\) km away (its closest approach to the earth), the telescope is to resolve, by Rayleigh's criterion, features on Jupiter that are 250 km apart. What minimum-diameter mirror is required? Assume a wavelength of 500 nm.

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 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 top 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 \(I_0\)?

Light of wavelength 585 nm falls on a slit 0.0666 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?

If the planes of a crystal are 3.50 \(\AA\) (1 \(\AA\) = 10\(^{-10}\) m = 1 \(\AA\)ngstrom unit) apart, (a) what wavelength of electromagnetic waves is needed so that the first strong interference maximum in the Bragg reflection occurs when the waves strike the planes at an angle of 22.0\(^\circ\), and in what part of the electromagnetic spectrum do these waves lie? (See Fig. 32.4.) (b) At what other angles will strong interference maxima occur?

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