Explanation:
Interference
The double slit
If light is incident onto an obstacle which contains two very small slits a distance d apart, then the wavelets emanating from each slit will interfere behind the obstacle. Waves passing through each slit are diffracted and spread out. At angles where the single slit diffraction pattern produces nonzero intensity, the waves from the two slits can now constructively or destructively interfere.
If we let the light fall onto a screen behind the obstacle, we will observe a pattern of bright and dark stripes on the screen, in the region where with a single slit we only observe a diffraction maximum. This pattern of bright and dark lines is known as an interference fringe pattern. The bright lines indicate constructive interference and the dark lines indicate destructive interference.
The bright fringe in the middle of the diagram on the right is caused by constructive interference of the light from the two slits traveling the same distance to the screen. It is known as the zero-order fringe. Crest meets crest and trough meets trough. The dark fringes on either side of the zero-order fringe are caused by destructive interference. Light from one slit travels a distance that is ½ wavelength longer than the distance traveled by light from the other slit. Crests meet troughs at these locations. The dark fringes are followed by the first-order fringes, one on each side of the zero-order fringe. Light from one slit travels a distance that is one wavelength longer than the distance traveled by light from the other slit to reach these positions. Crest again meets crest.
Note: We need single-slit diffraction to observe double-slit interference. Without the spreading, waves light waves passing through different slits would not meet and therefore could not interfere.
The diagram on the right shows the geometry for the fringe pattern. If light with wavelength λ passes through two slits separated by a distance d, we will observe constructively interference at certain angles. These angles are found by applying the condition for constructive interference, which is
d sinθ = mλ, m = 0, 1, 2, ... .
The distances from the two slits to the screen differ by an integer number of wavelengths. Crest meets crest.
The angles at which dark fringes occur can be found be applying the condition for destructive interference, which is
d sinθ = (m+½)λ, m = 0, 1, 2, ... .
The distances from the two slits to the screen differ by an integer number of wavelengths + ½ wavelength. Crest meets trough.
If the interference pattern is viewed on a screen a distance L from the slits, then the wavelength can be found from the spacing of the fringes.
We have sinθ = z/(L2 + z2)½ and λ = zd/(m(L2 + z2)½), where z is the distance from the center of the interference pattern to the mth bright line in the pattern.
If L >> z then (L2 + z2)½ ~ L and we can write
λ = zd/(mL).
