**Solutions**
to 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.]**

a) The phasors should line up in the center of this beam and d sin q should be 2 wavelengths. If each phasor has amplitude A, the total amplitude will be 2A, and the intensity will be proportional to (2A)^2 =4A^2 .

b) The blue bar should line up with the 2nd tickmark. The half-intensity points on each side come at d sin q = 1.75 and 2.25 wavelengths. The phasors are 90 degrees out of phase, so that when we add them vectorially they will give A sqrt (2), and the intensity will be proportional to (A sqrt(2))^2 = 2A^2 . (Half of 4A^2.)

2. **[Dark bands and path difference.] **

a) Dark bands show up whenever the path difference is an odd number of half-wavelengths. The 4 dark bands on the top are due to path differences of 7/2, 5/2,3/2, and 1/2 wavelength. The 4 dark bands on the bottom are due to path lengths of -7/2, -5/2, -3/2, and -1/2 wavelength.

b) The maximum path difference is the separation
of the sources, so to get 16 dark bands we want more separation of the
sources. Instead of 4 wavelengths we could make it 8 wavelengths. The wavelength
could be cut in half, keeping the separation the same, or by doubling the
spacing at constant wavelength.

3. **[Combine two alternating-current voltages.]**

The phase difference between sources is exactly

a) 2 Pi/3 (about 2.1 rad; the diagram should be an equilateral triangle),

b) about 1.45 rad .

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

Being 'very far away' from the grating means all beams travel parallel to one another on their way to where they meet, and the path difference between adjacent sources is d sin q (a small fraction of a wavelength). Being closer means that path differences between adjacent sources in the array start differing from d sin q by a noticeable fraction of a wavelength. This is what causes the polygon not to close on itself.

Keep the 4:1 ratio but reduce the wavelength and d to l = 0.05 and d = 0.15. With a smaller array, the rays are more nearly parallel at say, x=400. (Holding the d/l ratio constant means the angles of the maxima stay the same.) This way we are closer to the 'far field' where the path difference can be expected to be d sin q to a small fraction of a wavelength]. The 10 phasors will add to zero much better now.

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

a) The antennas give a single beam when the spacing is half a wavelength (l=2.0, separation = 1.0) Any larger spacing and the main beam gets narrower, but you begin to get energy showing up in 'stray' beams. Any narrower spacing, and the single beam gets broader and broader.

b) The beam rotates by an angle of q =18.77o using a phase shift of 1.0088 radians.

c) The path difference of d sin q = 0.3218 results in a phase difference of 2pi/l(0.3218) = 1.01 radians. The phase from the upper source is increased by 1.01 radians when q =18.77o because its beam travelled farther. The beam from the lower source had a phase added to it of 1.0088 radians to the two add constructively when q =18.77o.

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

a) A 25o angle to the normal means the wavefront travels an extra distance d sin 25o in reaching each successive source in a slit simulation. The phase difference resulting from this extra travel is (2pi/l)(d sin 25o) = 2.655 d/l . For d/l = 0.5, the phase difference is about 1.33 radians. The phase at each of the 6 sources is then (0, 4/3, 8/3, 4, 16/3, 20/3) radians.

b) A source-source phase difference of 1.33 radians will create a beam whose maximum is found traveling 25o below the horizontal.

7. **[A sharper beam from several antennas.]**

a) Typical coordinates for the intensity minimum would be (383,135) where the intensity =0.0000. This half-angle is around 19.4o.

b) Going to 12 sources, one has a typical min at (363,61) , for a half-angle of 9.5o.

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

If the antennas were spaced one wavelength apart, there would be a strong beam along the direction of the sources. If we space the wavelengths half a wavelength apart there is no radiation at all in this direction. The antenna spacing can be taken anywhere less than a wavelength to half a wavelength to satisfy this problem. Taking a wavelength of 0.20 and a spacing of 0.10 and 40 sources, a typical half-angle min would occur at (606,30) where the intensity is 0.0005. This gives a half-angle of 2.83o. The array length would be (39)(0.10)(1.3 m /0.20) = 25.4 m using a wavelength of 1.3 m. (39 sources will give a half-angle of 2.92o, and a shorter length.)

Spacing the sources just less than a wavelength could go as follows. Set l = 0.15, and separation = 0.1376. (This is done by clicking near the left-hand end of the separation slider until 0.7876 is selected, and then using the left arrow key to step down 0.05 at a time to 0.1376.) Then 22 sources give a typical minimum at (593,30), for a half-angle of 2.90o and an overall length of (21)(0.1376)(1.3m)/(0.15) = 25.0 m. There is only a small amount of stray radiation in this configuration, as may be seen by moving the mouse along the line of the sources.

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

Trial and error: Set wavelength =0.1, and d=0.1 and try different N values, giving N=14 . Calculation: We want d sin q = l/N so that the N phasors will close back on themselves and give zero output at q = 4o. This gives (since d=l) N = 1/ sin(4o) =. 14 b) Phase difference must be such as to cancel out the phase difference of the incoming waves (see discussion under problem 6). This is the same problem as creating an output beam at an angle of 20 degrees to the normal. The phase difference should be +/- (2pi/l)(d sin 20o) = +/- 2pi sin 20o = +/- 2.15 radians. (+ for source below, - for source above)

10. **[Design a receiving antenna for the famous '21 cm line' of hydrogen.]**

This is the same problem as 8 in different clothing. It is a receiving array rather than a transmitting array and at a different wavelength, but the underlying physics is the same. The array length will again be between 19.3 and 19.5 wavelengths as in the previous exercise, for an overall length at 0.21 m of 4.1 m.

Discriminating between (resolving) sources 3o apart requires that the maximum of source A lie no closer to the maximum of the source B than the adjacent minimum of B does. [This is the so-called 'Rayleigh Criterion'.] All mins adjacent to max's have a phase difference of 2pi/N between the N phasors. Wavefronts coming in at an angle q to the normal of a line of detectors (see comments in answer to #6.) create a phase difference of (2pi/l)( d sin q), which should be (2pi/N) if the zero of intensity is to form at that angle q. Letting N=22, d=0.1376, and l = 0.15, we find that q= 2.84o, as required.

11. **[Double slit behavior.]**

a) The phase difference between #1 and #5 is pi radians, as it is between #2 and #6. The path difference between #1 and #5 is the same as the path difference between #2 and #6, so they will both have the same phase differerence.

b) At a typical value of (466,45), W sin q/l = 0.50 which says the path difference between #1 and #5 (and between #2 and #6, etc.) is half a wavelength, causing cancellation between 1 and 5, 2 and 6, etc. This causes the total intensity to vanish; it is directly related to the 'zero of intensity' due to interference between slits.

d sin q / l = 0.144 saying that the path difference between sources inside either slit is about 1/7 of a wavelength. Each group of 4 sources in a slit has a finite resultants, but the two resultants exactly cancel out.

c) The 4 sources making up a slit are 90o out of phase and add to zero (of diffraction). The sketch shows a closed square from the four phasors of one slit, and a closed square due to the same effect from the other slit.

d) At a typical value of (460,411) W sin q/l = 3.50. d sin q / l = 0.999, which applies directly to the zero of diffraction, saying as it does that the path difference between light from opposite ends of the slit is an integer number of wavelengths.

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

a) When a phase shift of 1.0376 radians is used in the single slit demo, the beam rotates by about 21.4 degrees (arctan(128/366)).

b) When a phase shift of 1.0376 radians is used in the grating demo, the beam rotates by about 3.0 degrees (arctan(20/382)).

c) The source spacings are different; in the single slit, the sources are spaced half a wavelength apart, while in the grating demo they are spaced 3 wavelengths apart. Since the phase difference is (2pi/l)( d sin q), at a given angle q, the path difference in the grating will generate 6 times as much phase difference as does the path difference in the single slit. The beam maxima occurs when the 1.0376 radians added to each source is cancelled by the 1.0376 of phase difference due to extra path.

d) The single slit extra path is d sin (arctan(128/366) = 0.330 d = 0.165 l. This gives a phase difference of (2pi)(0.165) = 1.0371 radians, corresponding to what was added.

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

a) typical values would be (394,123), Int = 0.0495.

b) 1/(1.5 pi)2 = 0.0450, which is close to 0.0495

c) For 2 1/2 wraps, 1/(1.5 pi)2 =0.0162, compared to 0.0190 from the screen; Results get closer if we use more sources. For 18, the submins have intensities of 0.0482 and 0.0175 respectively.

14. **[Intensity from a speaker.]**

(Definition: decibels = 10 log10(I/Io), where Io = 10^-12 watts/m^2.) Use the Near/Medium Field option for the circular aperture. The wavelength is 0.25 m and the diameter is 15 inches, or 0.38 m. Since we will be looking at around 4m, imagine x=400 pixels to be at 4 m, so one pixel represents 0.01 m. This makes the wavelength 25 pixels. The N phasor sources must take up 0.38 m all together, so try 10 sources at a separation of 4 pixels. The intensity drops by a factor of 5 when the db level drops by 7 db. So when the relative intensity is 0.2 we are 7 db below the maximum (where the relative intensity is 1.0). This is at about x=400, y=224, or 2.24 m to one side of the speaker axis at a distance of around 12 ft or 4 m. One may check for consistency by trying another N, spacing combination, like N=16, spacing =2.50. (It gives a relative intensity of 0.2 at x=400, y= 226 or so.)

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

a) To get the extra 10 db, one needs 10 times as much intensity. The maximum intensity depends on the square of the amplitude, and the maximum amplitude of N sources is N times an individual amplitude. This means N^2>10, so N must be at least 4. (And MacGyver will need some leeway.)

b) The speakers must produce maxima 2 m apart at a distance of 100 m. So d sin q is one wavelength when q is about 2/100. This gives us d=12.5 m as a separation for two sources. (But we need 4 sources to do the job, so we will use two sources, each representing 2 speakers on top of one another).

c) The maximum comfortable screen x value is around 400, so let's try 400 pixels equals 100m. This means that a wavelength of 1 pixel corresponds to 0.25 m. Separation of sources by 12.5 m corresponds to 50 pixels separation. The 2 m separation of the bad guys will become 8 pixels.

d) Using 2 sources in 'sources in a line' (each representing 2 speakers), with the parameters above we find a precise but slim maximum at each bad guy. The maxima are separated by 8 pixels (2m) as they should. There is 122 db at each max, but one pixel (25 cm) away the intensity is down to 121 db.

16. **[Tracking 'Red October'.]**

a) Scaling of 0.1 pixel per meter permits x=400 to model 4000 m. The wavelength is then 0.1 as is the source separation d.

b) Zero of intensity happens when the phase shift between detectors is 2p/N, or when the path difference from the source to adjacent detectors is l/N. l = d, so this occurs when sin q = 1/25 = 0.04. Red October will move 160 m to make this angle happen, which will take (160 m)/[8 x 2000 m / 3600 s] = 36 sec.

c) Phase shift should be 2pi/25 = 0.2513; will put this in the simulation for source-source phase difference.

d) The max signal occurs at y= 16, corresponding to 160 m. d sin q / l = 0.039, quite close to the theoretical value of 0.04.

e) Phase shift/sec = .2513/36 or 0.00698 rad/s.

17. **[Audio version of Lloyd's mirror.]**

a) Because of the phase shift on reflection from the ground, we use 2 sources, the true source and an 'image' which is pi radians out of phase. The source separation is 2.0 m. tan q equals 32/362, and d sin q = 0.348 m, which should be the wavelength.

b) Simulation gives a max at (362,32) with l = 0.35, 2 sources, d = 2.0 and a source-source phase difference of pi.

c) Intensity is zero, since path difference is nearly zero, & reflected waves are shifted by pi.

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

a) typical screen values at the first max: (531,187), Int= 0.9962, d sin q / l = 0.997. Calculated q is 19.40 degrees, and calculated d sin q / l = 0.9965.

b) typical screen values at the 2nd max: (420,375), Int=0.9987, d sin q / l = 1.998 Calculated q is 41.76 degrees, and calculated d sin q / l = 1.998.

c) typical screen values at min after 1st max: (610,236), Int=0.0008, d sin q / l = 1.083 Calculated q is 21.15 degrees, and calculated d sin q / l = 1.08246.

d) The calculated d / l sin q value is 1.0825, compared to 1.083 on the screen.

19. **[Resolving power of a grating.]**

a) typical screen values at the first max: (582,225), Int= 0.9984, d sin q / l = 0.999. Calculated q is 21.14 degrees, and calculated d sin q / l = 0.9990. These two wavelengths are right on the edge of being resolved, the max of one on top of the min (21.15 degrees) of the other.

b) This wavelength is clearly resolvable from l = 0.10.

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

a) Angle looks about right at both 8 and 12 sources. (out at around x=400)

b) We would get a regular polyhedron whose resultant is zero. That's predicted by sin (Q/2) /(Q/2) when Q = 2pi

c) Q/N = 0.2784 = (2pi/l)( d sin q ). For N=10 and d/l =2, 4p sin q = 0.2784, or q = 0.02215 radians. At a distance x=400, this gives y about 9.

d) The simulation measures about 8 (it's hard to get intensity right at 0.5)

21. **[Far Field Figure of Merit.]**

The FOM is given by ((x2+(L/2)2)1/2 - x )/l . For an array of 6 elements separated by 0.5 pixels, L=2.5 and L/2 = 1.25. With x=400 and l = 1, the FOM is 0.002. For an array of 10 elements at the same x, spacing and wavelength, the FOM is 0.025, or about 12 times higher than for 6 elements.