W9_parallel_resonance.eps

Week 13: Interference and Diffraction 431

Consider the general phasor diagram in figure??. We wish to add:

Etot=E 0 sin(kr−ωt) +E 0 sin(kr−ωt+δ) +E 0 sin(kr−ωt+ 2δ) (1033)

withδ=kdsin(θ) is the phase angle produced by the path difference between any two

adjacent slits^120. Examining figure (181) we see that the general result is:

E 0 cosδ

E 0 cosδ

δ

E 0

E 0

E 0

E 0

ω

ω

kr −

kr −

kr −ωt

t

δ t

δ

δ

δ

α

Figure 181: Phasor diagram for general solution for three slits. Note that the amplitude of the sum
of the three phasorsEtot=E 2 + 2E 0 cos(δ).

Etot=E 0 (1 + 2 cos(δ)) (1034)

and we rather expect that the interference pattern intensity willbe:

Itot=^1

2 μ 0 c

|Etot|^2 =I 0

(

1 + 4 cos(δ) + 4 cos^2 (δ)

)

(1035)

which equals 9I 0 whenδ= 0, 2 π, 4 π...and equalsI 0 whenδ=π, 3 π, 5 π.... Itseemsas

though it will equal zero for certain values of the phase angle as well,but how can we

determine which ones?

To answer this last question and find a more general way of determining the pattern of

maxima and minima for 3 slits (and later for more) we turn back to the phasor diagram.

Consider the four diagrams drawn in figure (182):

Clearly, we get a principle maximum whenever the three phasors line up(for simplicity

the figures are shown at a time thatkr−ωt= 0) for atotalfield amplitude of 3E 0.

This obviously occurs whenδ = 0, but it canalsocorrespond toδ = 2π, 4 π, 6 π...–

rotating any field phasor through 2πputs it back where it started. We conclude that

this arrangement leads to amaximumin intensity withIp= 9I 0 called theprinciple

maximaof the interference pattern, when the condition:

δprinciple max=

2 π

λ

dsin(θ) = 0,± 2 π,± 4 π...=± 2 π m m= 0, 1 , 2 ... (1036)

a computer to do the tedious arithmetic for us, but this course isn’t about doing hard arithmetic – seriously, stop
laughing – it is aboutideasand theideaof interference can be perfectly well understood and quantitatively analyzed
with these simplifications, idealizations, and approximations.

(^120) Note well that the angles in the corners of the symmetric trapezoid can be seen to equalδby reasoning out loud:
“δplusπ/2 plusαadd up toπbecause they make a straight line. Inside the bottom triangle,αplustπ/2 plus the
unknown angle in the corner at the origin add up toπbecause it is a triangle. Therefore the bottom angle must beδ.
And you thought high school jommetry wasn’t good for anything...