Week 10 Lecture 3: Multiple source interference Problems F2003Q4c, 66, 67 (courseware 96-99)

Transcription

1 Week 10 Lecture 3: Multiple source interference Problems F003Q4c, 66, 67 (courseware 96-99) f we can add waves we can add more! Fortunately this is only of practical interest for waves with equal ω, A and successive δ s. Phasor Addition for N Waves see next slide (just a head to tail construction of phasors ) Convince yourself this is the same as the two vector case: (kx-ωt) is the angle to the 1 st phasor (although axis is not shown). δ is the phase difference between successive waves. R is the resultant amplitude. We must determine α - as before with waves. n the Head to Tail Construction t maps out a circle radius r (or an arc of a circle) Each phasor subtends an angle δ, for a total subtended angle of (similar triangles). Previously: for phasor case need to find R and β For N phasor case need to find R and also α to 1 get an equation representing the sum wave.

2 For waves having the same A, ω and successive δ s Resultant Amplitude = R R is the intensity along the line in the observation angle θ direction.

3 ASDE Try the phasor addition program on the website (links). You can get the diagram on the previous page ug N=10 (sources) then try δ= ο, 11 o, 30 o, 36 o (this is the phase difference between adjacent sources). So what is happening physically when you do this? You are simply increag the observation angle!! 36 deg N sources 30 deg 11 deg deg δ not θ From phasor addition Program For: δ = 0 o then = 100%, δ= o then = 99% δ= 11 o then = 73% δ= 30 o then = 3.8% δ = 36 o then = 0% Hang on thought you got a minimum when δ=π, 3π, 5π, etc. How can you get a minimum at δ=36 o? BECAUSE this is 10 sources! The earlier δ minima were for sources. Remember that δ=kdθ so this angle isn t θ but it is related to θ. Note that we don t know θ at this point because we haven t set d or λ With N=5 and δ=144 o you get a cool star!! This is a minima, because R=0! This means that at the θ angle corresponding to δ=144 o you have a minima (no intensity) 3

4 Now back to our story n order to determine resultant phasor- need α and R What is α? compare with the phasor case: phasors with δ phase difference this angle is δ/ N phasors with same δ diff this is What is R? from triangle ODP we can use the coe rule: take square root why N-1? Because N = 1 corresponds to only one phasor no phase difference. R R R R = r = r = 4r = r + ( 1 cos ) 1 r cos cos cos Now, we don t know what r is, so let s get rid of it r = r 1 α = ( N 1) δ =/ 4

5 To do this get an expression for A from triangle ODB do same coe rule analysis as we did for R. δ A = r dividing R A = r δ r R = A δ This is it!! An expression for R! So for the resultant wave the amplitude is given by R above and the angle with respect to the x-axis is (kx-ωt + α) where α is: Note deleted some words here α = ( N 1) δ 5

6 Putting everything together: So on a faraway screen at the sum of all waves at a particular point P (or angle θ) is y 1 ( ) + + = N 1 y K yn A kx ωt + δ δ Amplitude R of sum wave Angle of sum wave wrt the x axis ntensity (the square of the amplitude) = A δ where δ = kd θ = π d λ θ But A = source intensity s (for each individual wave source) so = s δ = s Nπ d θ λ π d θ λ 6

7 How does the pattern change with increag N s? Look at the case when θ=0 degrees N sources approaches When θ 0 approaches (i.e. dθ 0) then Nπ d λ θ is small so the small angle approx holds: () = ( ) Nπ d θ λ s = s N π d θ λ This tells us that the intensity increases with N. ----More slits or sources give a higher intensity of the principle maxima (see next slide) 7

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9 mportant Points to Note 1. Principal maxima shape: gets narrower and sharper with increag N according to α N. Principal maxima position for N sources is same as for N =. ie. maxima when dz = nλ l or d θ = nλ this is when π δ = d θ = 0, π, 4π, K λ 3. Minima occur? Try program N=30, δ=π/30 = 360/30=1 o 1. Secondary maxima? - roughly halfway between the adjacent minima though not exactly.. Spacing between principle maxima? (like N=) λl z = z λ spacing d or = θ = l d = s δ principle max spacing given by (ce z/l was for small angles) when π δ =, N 4π, N θ = Try phasor program N=30, δ=0 6π, N λ d K top term = 0 so minimum (only works for n N) -can also do with phasors when they sum to zero Try program δ=18 o ce next min at 4 o As d/λ gets larger you get many principle maxima. 9

10 Summary slide: MS Summary: Multiple Source nterference N sources a distance d apart all with amplitude A and the same ω for any given observation angle θ we calculate δ then overall path difference from top to bottom overall phase difference = Nkdθ = path difference between each source so δ = phase difference between each source = δ = kdθ get resultant amplitude R for this angle θ and R = 10

11 problem 11

12 Polar plot of intensity (radial) vs θ (angle): Note that you could NOT get this pattern for light waves (λ~1µm) but could for radio waves with λ from 1cm to 100 m. θ s s plotting: Nπ d θ = λ π d θ λ as a function of θ Or d/λ = ½ (source spacing is half the wavelength) Here λ/d=θ =. How does this work? This means θ 1 changes by. Why? well, θ changes by 1 going from 0 to 90 o and then again by 1 from 90 to 180 o. 1+1 =!

13 Spacing increases relative to the wavelength Plots of θ vs / for N= s Primary peak spacing is θ = λ/d = 1 Or θ = 0 ο, 90 o Peak spacing is θ = λ/d = 0.5 or θ = 0 o, 30 o and 90 o Peak spacing is θ = λ/d = 0.1 or θ = 0 o, 5.7 o,11.5 o,17.5 o All of these peaks should have the same intensity (but they didn t plot that way on my printer) 13 Note the Youngs -slit demo gave a d/λ of ~00 so lots of peaks!!

14 Plots of θ vs / s for d/λ = ½ (spacing half the wavelength) (Note change in scale of plot as N increases) 14

15 Further points to note: (following on from earlier points) Again recall: s Nπ d θ = λ π d θ λ Critical parameters are N and d/λ. Large N means more narrow and higher peak intensity at principle maxima. AND f d/λ is small get very few principle maxima. f d/λ is large get very many principle maxima. small d/λ and high N gives a very intense, directional signal. Can do this for radio waves where λ is large (1cm 100m) to get intense directional beam. (see next slide on broadside array). for visible light the wavelength λ is ~ 1µm or 10-6 m. so even the smallest spacing (say d=0.mm like in the laser demo) will give d/λ~00. This will give a huge number of primary peaks which is why we see so many peaks 15 when we do the demonstration with light!