Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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
  1. The two non-adjacent pairs of slits.
  2. Each adjacent pair of slits.
  3. The two non-adjacent pairs of slits.

Step by step solution

01

 Step 1: Concept.

We can derive an expression for the intensity distribution for the four-slit diffraction pattern by the same phasor-addition method. We imagine the plane wave front at the four slits subdivided into four waves, for which we superpose their contributions at a point P on a distant screen (Fig. 1a). To do this, we use a phasor to represent the sinusoidally varyingE field from each slit The magnitude of the vector sum of the phasors at each point P is the amplitude Epof the total Efiled at that point. The intensity at P is proportional toEp2

At the point O shown in Fig. la, corresponding to the center of the pattern, the four phasors are essentially in phase (that is, have the same direction); In Fig. lb the resultant amplitude of the phasors at O is denoted by

Eg. Figure Ic shows the phasor diagram corresponding to point P; the differences in path lengths (Fig. la) lead to phase differences from adjacent slits, (the amplitude of the electric field at P is denoted byEp ). From the geometry of the figure, we have

Ep=E0cosϕ2cosϕ (amplitude in four-slit diffraction)

which leads to

I=I0cosϕ2cos2ϕ (amplitude in four-slit diffraction)

where I0is the intensity in the stright-ahead direction, at O, where ϕ= 0. We see that I = 0 (interference minimum) when ϕ=π/2,π,and ϕ=3π/2.

Referring to Fig. 1a, we observe the following. In caseϕ=π/2 , the two pairs of slits 1 and 3 and slits 2 and 4 have a phase difference of each, and thus correspond to

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A candle4.85cm tall is39.2cm to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

34.15 The thin glass shell shown in Fig. E34.15 has a spherical shape with a radius of curvature of 12cm, and both of its surfaces can act as mirrors. A seed high is placed 15.0cmfrom the center of the mirror along the optic axis, as shown in the figure. (a) Calculate the location and height of the image of this seed. (b) Suppose now that the shell is reversed. Find the location and height of the seed’s image.

Huygens’s principle also applies to sound waves. During the day, the temperature of the atmosphere decreases with increasing altitude above the ground. But at night, when the ground cools, there is a layer of air just above the surface in which the temperature increaseswith altitude. Use this to explain why sound waves from distant sources can be heard more clearly at night than in the daytime. (Hint:The speed of sound increases with increasing temperature. Use the ideas displayed in Fig. 33.36 for light.)

When a thin oil film spreads out on a puddle of water, the thinnest part of the film looks dark in the resulting interference pattern. What does this tell you about the relative magnitudes of the refractive indexes of oil and water?

34.16 A tank whose bottom is a mirror is filled with water to a depth of 20.0cm. A small fish floats motionless 7.0cm under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free