Wave Interference Exercises and Problems
(for use with Wave Interference program by Mike Moloney, marketed by Physics Academic Software).
1. [50% Intensity and phase difference with two sources.]
Select 'Path Difference Effects'.
a) Move the mouse around and verify that the intensity is 1.0 at the center of each bright region. (The intensity relative to the maximum is given at the top center of the screen.)
b) Move the mouse to the center of the second bright region, just outside the grayscale area. (The beam from the blue source will have travelled about 2 wavelengths more than the beam from the red source.). Maneuver the mouse until you find a nearby spot where the intensity is nearly 0.5, meaning it is half of the maximum. Write down the values of d sin q / l at each spot, and record the picture of the two phasors, showing their phase relation. (d sin q / l is displayed in the upper righthand corner of the screen.) Work out and discuss why the intensity should be half its maximum value at these points.
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.
a) Figure out (looking at the path difference bar, and looking at d sin q / l may help) what caused exactly 8 dark bands to show up. Write down your reasoning, indicating what values of path difference cause dark bands to show up.
b) Now change the plot around so that exactly 16 dark bands show up between the sources. Report what values of d and l you used.
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.
a) Adjust the phase between them so that the maximum voltage which appears on the line is 40 volts even though both 40 volt signals are present. Approximately what phase delay is required between the two 40 volt signals? Use the 'Adding Two Waves' option, and make a reasonable estimate of the required phase, good to around 0.1 radians. Notice that there is a vertical scale with tickmarks in the upper-righthand area. Each small tick is 5 units, and each large tick is 25 units.
b) Repeat part a) if the required voltage is 60 volts, and report the phase delay used.
4. [Far field and 'not-so-far' field behavior.]
Under 'N Source Interference', 'N Sources in a Line', run the grating demo, and move the mouse out to around x=400 (where the phasors are). Note that near the x-axis the intensity nearly goes to zero (and the 6 phasors almost form a hexagon). Now change the number of sources to 10 and run the grayscale plot. Notice that the phasors now do a poorer job of forming a regular polygon than the six phasors did. Figure out why this is happening, (it has to do with path differences not being well approximated by d sin q ). Make a correction to the parameters, keeping 10 phasors, in order to get a better job of that first intensity zero near q =0.
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.)
a) Now you want the two antennas to radiate only in the direction perpendicular to the line joining them. Leave the wavelength alone and play with the spacing until you get a radiation pattern perpendicular to the line joining them. (You only want one single beam and not a bunch of beams. The intensity along the line of the sources should be zero.)
b) You would also like to steer the beam around and not have it be always pointing in one direction. Under 'More Plot Parameters', adjust the 'Src-Src f Diff' slider to create a phase difference between the sources. Put in about +1 radian of phase difference and measure the beam rotation angle q = tan-1(y/x). Write down the x and y values you used, the phase difference f and the calculated value of q.
c) Figure out how the beams are adding up constructively at the new phase difference and the new angle. (The slider gave the lower source the phase difference f, and the waves from the upper source travel farther to reach the point where they combine.) In particular, calculate the phase difference f from the angle q and show that this agrees with the phase difference you put in.
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.
a) Work out the phase of the light beam as it arrives each of the 6 sources, taking the phase at the top source to be zero. [Hint: a wavefront on the incoming light beam from above the horizontal will will reach the top source first, then reach the second source, etc.]
b) Under N sources in a line, run the single slit demo. For the spacing and wavelength in this demo, calculate the phase difference needed to obtain a beam traveling to the right and down at an angle of 25o to the horizontal. Under More Plot Parameters, add in your calculated phase difference, and run the grayscale plot. Measure the x,y values of the center of the outgoing beam, and determine whether the beam angle is indeed 25o downward from the horizontal. Write down your calculations and results.
7. [A sharper beam from several antennas.]
a) Under 'N Sources in a Line', select 6 sources, l = 0.20, and separation =0.10. (The source-source phase difference should be zero.) Run the grayscale plot and then go out with the mouse to x=400. Move the mouse down and measure the x,y coordinates where the main central beam goes to nearly zero intensity close to the x-axis. Calculate this angle (which we'll call the beam half-width) from the x,y measurements. (The 6 sources have created a 'sharper' beam than 2 sources did.)
b) Now you would like the half-width to be a lot narrower, only 10o (from the x-axis to the first minimum). Change the number of sources until you obtain the required angle in the beam. Write down the x,y and intensity values, and the calculations which show that you got the correct angle.
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
length of either slit
extra path traveled
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).
a) Start at around x=350, y=0. All the arrows should pretty well add up in a line here. Move the mouse vertically down until the intensity first becomes zero. Make a large sketch of the phasors, numbering them from 1 through 8 in order. Judging from the sketch, what is the approximate phase difference between #1 and #5? #2 and #6? Are the the 1-5 and 2-6 phase differences approximately the same, or are they different? Write down an explanation which connects the 1-5 phase difference and the 2-6 phase difference.
b) Write down the value of W/l sin q and d/l sin q from the top of the screen. Which of these two values is directly related to the 1-5 and 2-6 phase differences? Which of the two is directly related to the phase difference between #1 and #2?
c) This 'clamshell' picture is for the first intensity zero near the x-axis. It could be said that this is due to 'interference'. Now continue to move the mouse downward until you come to an intensity zero which is caused by 'diffraction' from within either slit. Draw a large sketch of all 8 phasors, numbering them from 1 to 8 in order. Discuss which part of the sketch supports the idea that waves from a single slit are producing a zero of intensity (said to be due to 'diffraction' within the slit).
d) Write down the value of W/l sin q and d/l sin q from the top of the screen. Which of these two values is directly related to the effect of 'diffraction' within a slit?
12. ['Steering' a beam (as in a phased-array radar).]
[You should do 'Two antennas; single beam, rotating the beam' before this problem.]
a) Run the single slit demo under 'N Sources in A Line' . The beam here has a maximum along the x-axis (q = 0). Now go to 'More Plot Parameters' and change the source-source phase difference to around +1 radian, and rerun the grayscale plot. Move the mouse to the area of x=350 to 400 and locate the maximum of intensity. Determine the angle q by which the beam has been rotated ('steered'). Record your x and y values and the calculation of q.
b) Now run the grating demo and then add in around +1 radian of source-to-source phase difference and do another grayscale plot. Again locate the amount of 'steering' of the beam, recording x, y and giving the q calculation.
c) Why doesn't the same amount of phase shift between sources result in the same amount of beam steering in parts a) and b)? You may have to go back and re-examine the simulations in parts a) and b) to obtain the information necessary to answer this question. You may also notice that the value of d sin q / l when the phasors all line up is not the same before and after the phase shift is introduced in parts a) and b). This might help you with the explanation, or the calculation coming up in part d)
d) For the phase shift of part a) do a calculation for the angle q, showing that it agrees quite closely with what you observed.
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.
a) [First sub-maximum.] Move the mouse vertically to find the largest intensity value outside the main intensity area (the phasors will appear to be wrapped more than once around at this point, but less than twice around). Record the value of the maximum intensity you are able to find.
b) There is an easy way to estimate the intensity at this point. It assumes the phasors are distributed smoothly along a circle, and that this sub-maximum occurs when they wrap 1.5 times around. At 1 1/2 wraps, the resultant is the diameter of the circle, and the total length of the phasor arrows is the circumference, or 1.5 diameters. So the resultant amplitude is the maximum amplitude divided by 1.5 pi. Because the intensity is proportional to the square of the amplitude, the relative intensity at 1.5 wraps will be 1/(1.5 p)2 . Calculate this out and see how it compares to the value you got in part a).
c) [Second sub-maximum.] Locate the second sub-maximum and record x,y and intensity values. Calculate the intensity by the method of part b). (It should be close to the relative intensity on the screen.)
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'.
a) How many speakers will MacGyver need, at a minimum?
b) Work out where to place the speakers before doing any simulating, and write down your reasoning.
c) Decide what scaling you will use (how many pixels per meter?). Write down how you reasoned it out.
d) Carry out the simulation using the sources in a line. (Let each source represent m speakers; figuring that speakers can be stacked vertically.) Write down your results. Give x and y values, and the (relative) intensity at each location. Specifically calculate the db level at each Bad Guy to show it is above 120 db at both spots.
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).
a) Determine the scaling, in pixels/meter for this problem. Spell out your reasons for this choice.
b) If there is no phase shift in your detection array, at what time t will the signal from Red October first fall to zero intensity? (1 knot = 1 nautical mi/hr; 1 nautical mile = 2000 yards, but call it 2000 m for this problem).
c) Before simulating this situation, determine the amount of phase shift between detectors in order to have Red October produce a maximum signal at the time in part b).
d) Run the simulation and report how it came out with respect to your prediction. (Try to understand what went wrong if there was a problem.)
e) What amount of phase shift per second must be introduced into your detection array to keep a maximum signal showing up from Red October while she maintains 8 knots?
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.
a) Before you simulate this situation, calculate what the wavelength of the emitted sound waves would have to be. Show all your work.
b) Carry out the simulation and report your results.
c) Determine the approximate intensity from this speaker at a horizontal distance of 375 m and a height of 0.01 m above the ground.
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.
a) [1st order max.] Move the mouse to x=550 and then vertically down until you find the first order maximum. Record the x and y values and the intensity. Also write down the d sin q / l value shown at the top right of the screen. Calculate q from x and y, then calculate d sin q . Compare d sin q to l. , and to the screen ' value of d sin q / l that you wrote down from the screen.
b) [2nd order max.] Move the mouse to x=450 and then vertically down until you find the largest part of the second order maximum. Record the x and y values and the intensity. Calculate q from x and y, then compare d sin q to 2 l.
c) [Min after 1st order max.] The minimum closest to the x-axis occurs when the path difference is l/N, and the pattern is a regular polygon of N sides. (The phase difference between adjacent sources is then 2pi/N, which forms the regular polygon.) Near the first order peak, minima occur at a path difference of l + l/N and at a path difference of l - l/N between adjacent sources, each time forming the regular polygon.
Move the mouse to x=550 and then vertically down until you find the minimum beyond the first order maximum, where the path difference should be l + l/N . Record the x and y values, the intensity value, and the value at the upper right for 'd sin q / l=' . Calculate q from x and y, then compare d sin q to l + l/12 .
d) Compare your d sin q / l value to the one read off the screen. (The screen d sin q value is the actual path difference between the first and second sources, It is not calculated from d and q, so it should be a little different than what you calculate.)
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
d sin q = la + la/N = lb.
Rearranging this we get the formula for resolving power of a grating in first order:
N = la/(lb-la) = total number of sources, or
R = resolving power l/Dl = N (in first order); ( D stands for 'delta').
Under 'N Sources in a Line' select 12 sources and a separation of 0.30.
a) Move the mouse to the wavelength slider and click near the lefthand end until you get a wavelength of 0.5583. Then use the left arrow key to step the value down by 0.05 until you get a wavelength of 0.1083. Run the grayscale plot and determine the location of the first order maximum, somewhere around x=450 and y=190. Record the values of x,y and the intensity, then calculate q. If this q is greater than the angle for the first order minimum when l=0.10, then these two wavelengths can be resolved in first order. If not, then they are too close to be resolved. (You should have the q value for this first-order minimum at l=0.10 from the previous exercise; if not, change the wavelength to 0.10 and run another grayscale plot to obtain it.)
b) Again click the mouse near the lefthand end of the wavelength slider until you get a value of 1.0165 and then use the left arrow or down arrow to lower the wavelength value to 0.1165. Then run the grayscale plot and see if this wavelength can be resolved from 0.100 in first order. (Record the x,y values and intensity, then calculate q.)
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.
a) Run a simulation with 8 sources l=0.25, d=0.25 and see if the angle between the first and last phasor appears to be about 160o at half intensity. Repeat for 12 sources (the angle should not change with the number of sources).
b) What should happen if Q =2p ? What shape would the phasor diagram have?
Since Q is the phase difference between the first and last phasors in the chain, Q/N is the phase difference between adjacent phasors in the chain. This phase difference is close to (2p/l)( d sin q ). We want to know the angle q at which the chord length is 0.707 of the arc length.
c) Put all of the above ideas together, and find the angle q for 10 sources of wavelength 0.25 pixels separated by 0.5 pixels where the intensity will be half of the maximum intensity.
d) Run the simulation, record your results, and compare to the angle found in part b).
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.