Wave Interference Exercises and Problems

(for use with Wave Interference program by Mike Moloney, marketed by Physics Academic Software).

Due to limitations of text, q stands for the angle theta, and l stands for the wavelength.

1. [50% Intensity and phase difference with two sources.]

Select 'Path Difference Effects'.

2. [Dark bands and path difference.]

Select 'Path Difference Effects. There are exactly 8 dark bands showing on the grayscale plot. in between the sources.

3. [Combine two alternating-current voltages.]

Two sinusoidal electrical voltages having the same amplitude (40 volts) and the same frequency are sent down a transmission line.

4. [Far field and 'not-so-far' field behavior.]

5. [Two antennas; single beam, rotating the beam.]

Suppose you have two identical antennas. Each will radiate uniformly in all directions. If you put them very close compared to a wavelength, the combination of antennas will radiate as a single antenna. Try this out by going to 'N Source Interference' then 'N Sources In a Line', and set it for 2 sources at a wavelength of 2 pixels and a separation of 0.1 pixels. (When you run this, the grayscale plot is uniform, and the intensity everywhere is very close to 1.0.)

6. [A plane wave coming in at an angle to a slit.]

A beam of coherent light comes in at an angle of 25o above the normal direction of a slit. Let the slit be simulated by 6 sources with a spacing of 0.5 pixels, and l= 1 pixel.

7. [A sharper beam from several antennas.]

8. [Design a narrow-beam transmitting antenna.]

Design a transmitting antenna array at a wavelength of 1.3 m which has only a single beam with a full width of 6o and a half-width (defined in the previous problem) of 3o. It should have no significant sidelobes (there should be only one main beam, little intensity outside this beam). Give number of elements, the element spacing, and total length in meters.

9. [Detect a source which makes an angle of 20 degrees to your antenna normal .]

You are looking for a source believed to be at an angle of 70 degrees from the line of your receiving array of N equally spaced detectors ( It makes an angle of 20 degrees to the perpendicular bisector of your array.) The waves are coming from a submerged nuclear submarine at a frequency of 3000 hz. [ Speed of sound in water is 1500 m/s.] Your array has a full-width angular resolution of 4 degrees at this frequency (sources exactly 2 degrees from the perpendicular bisector of your array produce no net signal from the array).

Assuming your detectors are spaced 0.500 m apart, how many are there in the line?

10. [Design a receiving antenna for the famous '21 cm line' of hydrogen.] Design a receiving antenna system at a wavelength of 21 cm which can discriminate between distant radio sources which are only 3 degrees apart in the sky. This antenna must not have any 'aliasing'; it must not pick up strong signals from different parts of the sky at the same time. Give the number of sources, source spacing, and total antenna length in meters.

11. [Double slit behavior.]

Run the double slit simulation (under N source interference). This has 18 total sources ( l = 0.4, separation = 0.15), with the middle 10 blocked off (omitted) so that only four on each end are active. These four on each end play the role of a single slit, giving one slit on each end. Because there is a significant distance between the ends, we get interference between the two 'single' slits, as well as 'diffraction' between the four at each end. Here are the meanings of information in the top right-hand area of the screen for the double slit.



W sin q / l

d sin q / l

distance between the
top of one slit and
the top of the other

length of either slit

extra path traveled
by waves from top
of one slit over
waves from top of
other slit

extra path
traveled by
waves from
opposite ends
of a slit

In this instance, W is 2.1 (18 minus 4 spacings), the top-to-top spacing ( or center-to-center spacing) of the slits. The slit width d is 0.6 (4 spacings, taking each source to be a strip 0.15 wide).

12. ['Steering' a beam (as in a phased-array radar).]

[You should do 'Two antennas; single beam, rotating the beam' before this problem.]

13. [Intensity 'sub-maximum' values from an array]

Go to 'N Sources in a Line', and select a wavelength of 0.25, a spacing of 0.10, and 12 sources. Then run the grayscale plot and run the mouse out to x=350 to 400.

14. [Intensity from a speaker.]

A circular speaker mounted in a flat surface acts approximately like a circular aperture. A 15-inch speaker mounted in a wall produces a maximum sound intensity of 95 db at a point 12 feet directly in front of it at a frequency of 1360 hz. Determine the area around this point where the sound level will be greater than 88 db. (Note: decibels = 10 log10(I/Io), where Io = 10-12 watts/m2.)

15. [MacGyver and the 'Audio' Hostage Rescue]

The television character MacGyver must rescue a hostage from two Bad Guys. The two Bad Guys are 2 m apart and the hostage halfway between them. MacGyver can set up some super speakers along a line parallel to the line of the Bad Guys, where the perpendicular distance between the line of speakers and the line of the Bad Guys is 100 m. At that distance, each super speaker can create a sound level of 110 db at a frequency of 1360 Hz (a wavelength of 0.25 m) where the Bad Guys are. MacGyver is going to use several speakers strategically placed to create a sound level in excess of 120 db (the threshold of pain) at each of the Bad Guys, while the hostage is in a quiet zone, and will not have her ears 'blown off'.

16. [Tracking 'Red October'.]

You are aboard a hypothetical U. S. attack boat quietly tracking the Russian 'Red October' submarine. She has a problem and is creating sound signals at a frequency of 1500 Hz (l= 1.0 m). Your tracking array consists of 25 elements 1.0 m apart along your fore-aft axis (the y-axis). At t=0, Red October is exactly abeam of you at 4000 m, making 8 knots perpendicular to the line between the two boats (8 knots in the y-direction).

17. [Audio version of Lloyd's mirror.] A single speaker is located 1.0 m above the surface of a hard flat area. Sound waves travel directly to a spot above the ground, but also bounce off the surface before arriving at that spot. The waves which bounce off the ground suffer a phase change of 180o . At a horizontal distance of 362 m from this source at a height of 32 m above the ground there is a maximum of intensity.

18. ['Orders' of a grating, and minima near the peaks.]

Under 'N Sources in a Line', select 12 sources, wavelength = 0.1, spacing = 0.3, and run the grayscale plot. In a grating, the large constructive interference peaks are called 'orders'. You should see 'zeroth' order along the x-axis (y=0) and bright bands for both first and second order. (There is a third order right along the line of the sources, too.) Each order occurs when the path difference (very nearly d sin q) between adjacent sources is an integer number of wavelengths. For the zeroth order the path difference is zero, while for first and second order the path difference is one or two wavelengths.

19. [Resolving power of a grating.]

[This should be done after the previous exercise, 'Orders of a grating, and minima near the peaks'.]

Background. When two wavelengths la and lb simultaneously enter a grating, two sets of peaks will be generated. (The intensities add incoherently because different frequencies do not stay in phase with one another.) We wish to check for N slits (sources) how close these two wavelengths can be and still be 'resolved'. Let la >lb. The first order maximum of lb will occur when d sin q = lb. This should be at or beyond the minimum just beyond the first order maximum of la, which occurs at d sin q = la + la/N. When the max of lb is at the same q as the min of lb, we have

Rearranging this we get the formula for resolving power of a grating in first order:

Under 'N Sources in a Line' select 12 sources and a separation of 0.30.

20. [Half-intensity values for N sources.]

With quite a few phasors added together, they look like the arc of a circle, and the resultant is the chord of that arc. When the arc of N phasors is 'straightened out', we have maximum intensity. This tells us that the length of the arc is about equal to the maximum value of the resultant.

The half-intensity points will occur when the length of the chord (the resultant) is around 0.707 the length of the arc. (Remember, the intensity is proportional to the square of the resultant amplitude.) The arc has some hypothethcal radius R and subtends some hypothetical angle Q. (Notice that the angle Q is the phase difference between the first and last phasor in the chain.) The arc length is just RQ. With a bit of trig you can see that the chord length is 2 R sin Q/2 . So the ratio of chord to arc length is sin (Q/2) /(Q/2), and we want this to be 0.707 for a half-intensity point. Solving sin x = 0.707 x on a calculator gives x = 1.392, or Q=2.784 radians = 159o.

Note: solving sin x / x = (I / Imax)1/2 will let us find the angle q for any relative intensity.

21. [Far Field Figure of Merit.]

In the far field, rays that travel from the array at q=0 should all arrive in phase at a 'field point' on the line bisecting the array. One figure of merit (FOM) for the far field is that the path difference between a ray from the array center and a ray from the array end should differ by a small fraction of a wavelength. The path difference per unit wavelength should be as small as possible, and tend toward zero farther from the array. Go back to exercise 3, and work out this FOM using 6 elements and using 10 elements. Report the FOM for both 6 and 10 elements at the same spacing.