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We now consider the simplest case which exhibits interference, two equivalent, narrow slits (Figure 3). Of course, you can get the two-slit pattern by substituting N=2 into the formula we shall eventually derive for N slits. But, deriving the two slit result as a special case first allows us to illustrate several useful results and techniques for complex arithmetic that you will have to master, in particular computing the amplitude of the complex number tex2html_wrap_inline1078 with the formula tex2html_wrap_inline1080 . This section appears very mathematical because we show all of the steps. As you learn to do these steps, they will become simpler and simpler for you.

Figure 3: Double-slit interference experiment

According to our main result, Eq. 8, the intensity at a point tex2html_wrap_inline974 on the screen for just two slits is


where tex2html_wrap_inline1084 and tex2html_wrap_inline1086 , tex2html_wrap_inline1088 and tex2html_wrap_inline1090 , and tex2html_wrap_inline1092 and tex2html_wrap_inline1094 , respectively, are (a) the intensities which each slit (if alone) would make at point tex2html_wrap_inline974 , (b) the phases of the waves emerging from each slit, and (c) the distance from each slit to the point tex2html_wrap_inline974 .

Now, because we assume the two slits to be equivalent, they must let the same intensity of waves through, and so the intensities which each separately would create on the screen are very nearly the same, so that tex2html_wrap_inline1100 . (There will be a slit difference in the distance from each slit to the point tex2html_wrap_inline974 on the screen, but so long as the screen is much further away than the separation of the slits, R>>a, this difference is negligible.) For simplicity, let us call this intensity which either slit alone creates on the screen tex2html_wrap_inline1106 .

Next, because the waves coming from behind the diffraction grating arrive at both slits in phase (Figure 3), the phases of the waves coming out of the slits will be equal, tex2html_wrap_inline1108 . For simplicity, let us call this phase tex2html_wrap_inline1110 . As we shall see in a moment, the fact that these phases are equal means that they cancel from the final result. Note that this is only true if the waves coming from behind the grating arrive at the slits with equal phases!

Putting this all together we have,


where, again, we have used the facts that tex2html_wrap_inline1034 and tex2html_wrap_inline1114 for any complex phase factor. Note that, as we had mentioned, in the end, the phases indeed cancel out. Also note that the entire result now depends only on the difference in lengths between the paths from the observation point to each slit, tex2html_wrap_inline1116 .

The entire computation of I now boils down to finding tex2html_wrap_inline1120 . A very useful way to find the square-magnitude of a complex number tex2html_wrap_inline1078 is to use the following fact


where, for any complex number tex2html_wrap_inline1078 , we define the complex conjugate as tex2html_wrap_inline1126 . Mathematically, taking the complex conjugate simply amounts to finding all occurrences of i and substituting tex2html_wrap_inline1130 .

Using the procedure of Eq. 12, we now find


where on the last line we find a somewhat simpler form by using the trig identity tex2html_wrap_inline1132 . Combining this with Eq. 10, we have the final result for two-slits,


Figure 4: Double-slit interference experiment: enlarged view

All that remains is for us to learn to compute the path difference tex2html_wrap_inline1134 . Figure 4 shows that the path difference tex2html_wrap_inline1136 corresponds very closely to the opposite side of a right triangle of angle tex2html_wrap_inline1040 and hypotenuse d. Accordingly, our final result is Eq. 13 with the definition


next up previous contents
Next: Interpretation Up: Two narrow slits Previous: Two narrow slits

Tomas Arias
Thu Sep 13 15:26:14 EDT 2001