Question

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

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

 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 varying$\overrightarrow{E}$ field from each slit The magnitude of the vector sum of the phasors at each point P is the amplitude $E_p$of the total $\overrightarrow{E}$filed at that point. The intensity at P is proportional to$E_p^2$

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 by$E_p$ ). From the geometry of the figure, we have

$E_p=E_0\cos \frac{\varphi }{2}\cos \varphi$ (amplitude in four-slit diffraction)

which leads to

$I=I_0\cos \frac{\varphi }{2}{\cos }^2\varphi$ (amplitude in four-slit diffraction)

where $I_0$is the intensity in the stright-ahead direction, at O, where $\varphi$= 0. We see that I = 0 (interference minimum) when $\varphi =\pi /2,\pi ,$and $\varphi =3\pi /2$.

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