The Double Slit

Cut a pair of narrow slits, small distance d apart (center-to-center), then shine light through (preferably laser light), toward a screen a large distance D from the slits. After going through a slit, light spreads out (diffraction) so that there is overlap between the two. At a point on the screen equidistant from the two slits, they are in phase, so there is a maximum there. Other points where a maximum occurs is where one ray travels ml farther than the other ray (m = 0, 1, 2, 3, ….). About halfway between maxima the path difference is (m+¹/₂)l , and they cancel. In the sketch below, where d<<D, the rays are extremely close to being parallel.

Hence for a maximum of intensity, dsinq = ml when d << D. (See the lower sketch above, the slits magnified.)

An alternative approach is to put a converging lens behind the slits and place the screen a distance f from the lens. Then the rays which intersect on the screen are parallel before they hit the lens, so dsinq = ml applies exactly. In practice this is not an improvement, however, because d can easily be made extremely small compared to D.

In the corresponding experiment with sound waves or microwaves, it could be inconvenient to have d<<D, so it is worth finding an exact expression for the location of the maxima. Let us call y the distance from the center on the screen to the location of the m^th maxima. Then we find that
y = (ml )[D²/(d²-m² l ²) + 1/4]^1/2. This is one your physics instructor almost certainly doesn't know about. The derivation is tedious, and you don't want to see it.

Your browser will run interference for you on your way back to the main waves and related goodies.

Other stuff:
Mechanics
Fluids, heat, electricity and magnetism
Quantum
Index

email me: fredrick.gram at tri-c.edu (remove spaces and replace at with @).