W9_parallel_resonance.eps

446 Week 13: Interference and Diffraction

This occurs when:

φ

2

=

πa

λ

sin(θ) =π, 2 π, 3 π... (1077)

or when:

asin(θ) =mλ m=⊗, 1 , 2 , 3 , ... (1078)

as before, so our heuristic rule is precisely derived.

We can now also (at least in principle) tackle themaxima. We will get a maximum in

intensity at the values ofφfor which:

dItot

dφ

= 0 (1079)

and which aren’t the minima (which will also occur, recall, at the zeros inthe slope of

the intensity). Physics majors and advanced students will enjoy this exercise in calculus,

which leads one to the relatively simple result that maxima occur when the transcendental

equation^128

φ

2

= tan

(

φ

2

)

(1080)

is satisfied. If one plotsφ/2 and tan(φ/2) simultaneously on a single set of axes, the

intersections of the two lines are the relevant zeros. As one can see (once one does this)

the maxima occur at anglesclose to(and just before) the condition(s):

φ/2 = 0 (exact), 3 π/ 2 , 5 π/ 2 , 7 π/ 2 ... (1081)

(note well the skipping ofπ/2).

Secondary Maxima

Minima

E (^0) Principle Maximum
Figure 189: Phasor diagrams representing successive minima and maxima for single slit diffraction.
In figure 189 the principle maximum (of lengthE 0 is illustrated for angleφ= 0. The
next two phasors show the (exact) conditions for minima, whereE 0 is wrapped first one
time aroundφ= 2πor twice aroundφ= 4π. Note that the diameter of the circle has to
get smaller as one wraps more than once! The secondary maxima arenow easy enough
to understand. We don’t get one atφ=πbecause we are still between the principle
maximum and the first minimum, there is no maximum here. Atφ= 3π/2 (dashed
circle and arrow) we can gain a tiny bit of length by rolling the circle backto a slightly
larger diameter, ditto atφ= 5π/2, although both of these figures are probably a bit
exaggerated.
It is now time to put it all together with a few examples.
(^128) Wikipedia: http://www.wikipedia.org/wiki/Transcendental Equation.