the distance traveled by two waves differs byλ/2, so thewaves are 180 degrees out of phase
and the fields from the two slits cancel.
We can compute this location by looking at the above diagram. We assume that the distance to the
screen is much greater thand. For light detected at an angleθ, the extra distance traveled from
slit 1 is justdsinθ. So the angle of thefirst minimum(or null) can be found from the equation
dsinθ=λ 2.
More generally we will get amaximum if the paths from the slits differ by an integer
number of wavelengthsdsinθ=nλand we will get anull when the paths differ by a half
integer number wavelengths.dsinθnull=(n+1) 2 λ.
Although it is very difficult because electrons are charged,2 slit electron diffractionhas also
been observed.
So, all kinds of particles seem to diffract indicating there is some kind of wave involved. We will
now continue with some thought experiments on diffraction to illustrate the physics that Quantum
Mechanics needed to match.
- See Example 3.6.1: Derive the location of the nodes in the diffraction pattern from two narrow
slits a distancedapart. Now try to compute the intensity distribution.*