W9_parallel_resonance.eps

Week 13: Interference and Diffraction 447

Example 13.8.1: Diffraction Pattern of a Slit of Widtha= 4λ.

To draw the semiquantitatively correctI(θ) for a single slit, we must capture itsfeatures

• both those we can compute or discover exactly as well as those that we can only guess at
  short of plotting the exact result. We’ll find it a lot easier to plot notI(θ) butI(sin(θ)),
  so much so that I’m going to focus on this in the example. Note well that all we have to
  do to convert to or plot in terms ofθis take the inverse sines of the points we obtain.
  We have seen above that we can exactly locate the principle maximum and the minima.
  We cannotexactlylocate the secondary maxima, but we can guess their approximate
  location as roughly halfway between the minima in our drawing. Similarly,we can’t
  exactly determine the intensity of the secondary maxima, but we doknow that they
  have to getsmalleras we increase their order, quite rapidly.
  To facilitate drawing a graph with these features, we therefore begin by locating the
  minima:
  asin(θm) = mλ
  4 λsin(θm) = mλ
  sin(θm) =

m

4

θm = sin−^1

(m

4

)

(1082)

Let’s arrange these for the values ofmfor which the inverse sine exists in a table. All

angles are in radians.Don’t forget to skipm= 0, the principle maximum!

m sin(θm) θm

1 14 sin−^1

( 1

4

)

= 0. 25268

2 24 sin−^1

( 1

2

)

= 0. 52360

3 34 sin−^1

( 3

4

)

= 0. 84806

4 44 sin−^1 (1) = 1. 00000

Table 6: Diffraction minima for a single slit of widtha= 4λ.

We see that it is a lot easier to draw the plot in terms of theregularsin(θm) than it is in

terms ofθm. Of course, the latter is a lot more useful. Oh well, such is life. You should

beableto do whichever one a problem requests on the homework or a quiz orexam. One

reason I often accept results plotted in terms of sin(θm) is that one doesn’t usually need

a calculator to do a decent job.

13.9: Two Slits of Finite Width

We are now ready to consider two slits offinitewidth. The result is very simple. We

get interference maxima and minima at exactly the same angles we gotthem for very

narrow slits. However, the field strength at those angles ismodulatedby the diffraction

of the field through the individual slits. As a result, the field we observe as an angle of

θis theproductof the field expressions for interference and diffraction:

Etot(θ) = 2E 0 cos(δ/2)

(

sin(φ/2)

φ/ 2

)

(1083)

Following the usual procedure (using the time average Poynting vector and relation

betweenE 0 andB 0 ) we get the intensity

Itot(θ) = 4I 0 cos^2 (δ/2)

(

sin(φ/2)

φ/ 2

) 2

(1084)