W9_parallel_resonance.eps

428 Week 13: Interference and Diffraction

source in the center of the screenE 0. Then the total field at pointPis:

Etot(P) = E 0

D

r sin(kr−ωt) +E^0

D

r+ ∆rsin(kr+k∆r−ωt)

= E 0

D

r

sin(kr−ωt) +E 0

D

r

(

1 +

∆r

r

)− 1

sin(kr+k∆r−ωt)

= E 0

D

r sin(kr−ωt) +E^0

D

r

(

1 −

∆r

r +...

)

sin(kr+k∆r−ωt)

= E 0

D

r sin(kr−ωt) +E^0

D

rsin(kr+k∆r−ωt) +O

(

∆D

r

)

≈ E 0 sin(kr−ωt) +E 0 sin(kr−ωt+δ) (1020)

The last step follows becausefor a small angleθ:

r=

D

cos(θ)

≈

(

D

1 −θ 22 +...

)

≈D

(

1 +

θ^2

2

+...

)

≈D (1021)

soE 0 D/r≈E 0 forbothsources. Obviously this will not hold for largeθ(angles pointing

out at the edges of a large screen stretching to infinity on the horizon), nor will it hold

if the screen is close to the two slits (whereFresnelinterference or diffraction must be

considered, which is a lot more work and beyond the scope of this course although answers

there are certainly computable). In the last equation we also introduce thephase shift

produced by the path difference:

δ=k∆r=kdsin(θ) =^2 πd

λ

sin(θ) (1022)

To add these two waves, we could use a trigonometric identity for sinA+ sinB. Unfor-

tunately, nobody can ever remember the trig identities for things like this (supposedly

memorized back in high school), including me. For those of us who find itimpossible

to remember arbitrary things we memorized out of any context where they would be

useful to us for more than busy work, it behooves us to learn how toderivethe answer in

simple ways from things wecanremember and that make sense in context. We therefore

eschew the use of a trig identity andderivethe result from a geometric picture, aphasor

diagramjust as we did before for e.g. LRC circuits.

In figure (179) we see the requisite phasor geometry. The light from the first slit has a

field amplitude of they-component of a “vector” (phasor) of lengthE 0 at anglekr−ωt

with respect to thex-axis. The light from the second slit is they-component of a phasor

of lengthE 0 at anglekr−ωt+δ. The field amplitude of the sum is they-component

of the phasor that is the vector sum of these two phasors, addedby putting the tail of

the second at the head of the first. Since the triangle representing this sum is isoceles

it is easy to see that the two acute angles must both beδ/ 2118. The total amplitude is

thus the sum of the adjacent side lengths of the two right trianglesformed by dropping

a normal as shown:

|Etot|= 2E 0 cos(δ/2) (1023)

and the full time dependent electric field is given by:

Etot= 2E 0 cos(δ/2) sin(kr−ωt+δ/2) (1024)

(^118) The argument goes as follows: “δplus the obtuse angle at the vertex of the triangle form a straight line and hence
add up toπ. The sum of the angles in the triangle also add up toπ. Therefore the sum of the two acute angles have
to add up toδ. The triangle is isoceles, so they must be equal, hence they are eachδ/2.” This is why geometry is
betterthan algebra or trig – proving this algebraically is nearly impossible without the use of complex variables and
with trig identities it is difficult and requires knowing the relevant identity.