W9_parallel_resonance.eps

Week 13: Interference and Diffraction 441

The first of these is the central maximum. Atθ= 0, all the radiators in the slit are

basically equidistant fromPand hence all of the coherent wavelets they spawn arrive in

phase in the middle. We use this middle point of complete constructive interference of

all of the Huygens radiators to define thepeakamplitude and (time average) intensity of

the light in the diffraction pattern,E 0 andI 0 = 1/(2μ 0 C)E^20 respectively.

The second are the locations of thediffraction minima– angles at which the total am-

plitude and intensity arezero. We can find these using the following not-too-difficult

mini-argument.

13.7: Diffraction Minima, Heuristic Rule

Consider the two waves emerging from the two Huygens radiators portrayed above in

figure 184 and proceeding to the pointP. As shown, the wave from the lower slit arrives

having travelled a longer path, with a path difference of ∆r=a 2 sin(θ).

We now apply the simple heuristic concept that served us well when wewere trying to

understand the two-slit minimum. If this path difference contains exactlyλ/2 (one half

of a wavelength) then the waves from these two particular radiators willcancelatP.

Now consider the second radiator down from the top. It also has a path difference of

a

2 sin(θ) compared to the radiator second down from the middle and these two cancel. The

third down from the top cancels the third down from the middle. In fact,everyHuygens

radiator in the top half of the slit cancels the corresponding radiatora/2 beneath it in

the lower half of the slit. The field amplitude and intensity atP arezero(which is as

low as one can get), making

a

2 sin(θ) =

λ

2 , or

asin(θ) = λ (1066)

a condition for adiffraction minimum.

θ

θ

a/4

a/4

a/4

a/4

a/4 sin θ=λ/2

waves cancel at P

waves cancel at P

Figure 185: The slit, with the Huygens radiators divided into four equal segments. Light from the
two pairs indicated cancels atP when the path differencea 4 sin(θ) contains a half of a wavelength,
for all of the pairs that make up the slit.

Now imagine dividing the strip into fourths, as portrayed in figure 185. As you can see, if

the path difference between the radiator at the top (0) and the radiator ata/4 contains

λ/2 (a half a wavelength) they cancel,and so does the wave from the radiator ata/ 2

cancel the wave from the radiator at 3 a/ 4 !Every point in the first quarter cancels a point

from the second quarter and at the same time the corresponding points in the third and