homas Young, an English Physicist in 1801, reported his experiment about the interference of light - when light was made to pass through a pin-hole and then through another two pin-holes side-by-side, a variation of intensity was observed on the screen.
Later when this experiment was performed with sunlight using long parallel slits instead of pin-holes, bands of colour called "interference fringes" were produced. If monochromatic light is used instead of sunlight, dark and white fringes would appear on the screen instead. This is only observed if the two slits in the second set are placed close together, otherwise all that would appear on the screen would be two bright overlapping patches. Young's original experiments were performed with white light and he could deduce the values of wavelengths of extreme red and extreme violet lights from his observations, which were quite close to the accurate ones. The pattern of light and dark fringes is sharply defined only if light of a single wavelength is used.
The above observation can be understood using Grimaldi's concept of diffraction. When sunlight passes through the first slit, it is diffracted, and the emergent light spreads out. This emergent light reaches a screen with two more narrow slits, which act as sources of light, and diffract it once more. As the light waves form each slit spread out, they meet each other at different points in the space. At the points where these light waves are in phase, they add together to give a bright fringe - constructive interference. At those points, where these waves are completely out of phase, they cancel each other out, thus producing a dark area - destructive interference.
A more concrete explanation to the above observation can be given thus:
Let P be a point on the screen. Two light waves, one from each slit, are considered, that reach point P after covering different distances:
For a fuly destructive interference on the screen, the path difference between the two light waves should be an odd-number multiple of half-wavelengths:
Therefore, at the points where the first condition is satisied, a bright area is observed. Points where the second condition is satisfied, dark fringes appear.
In the same diagram as before, a perpendicular from S1 is dropped on S2P. Let the foot of this perpendicular be A. Let O be the point exactly between the two slits. Let d be the distance between the slits and the screen, and s the distance between the two slits, such that s<
This also explains why the fringes are only observed when the slits are near and not when they are too far from each other. If s becomes much larger than λ, the fringe-width will be very small. Hence the light and dark fringes will be so closely spaced that the screen will appear to have just one bright patch. The above equation can also be used to measure the wavelength of light in an experiment, which can be calculated by measuring d, s and the fringe-width.
Interference from thin films
Let us assume a thin transparent film of thickness = d with plane parallel faces (refer to diagram). A parallel beam of light is incident on one of these surfaces at an angle of incidence = i. One of the light rays of this incident beam is considered. A part of this light ray is reflected back into the same medium, while the major part suffers refraction at the surface of the film. As the wave that is inside the film now touches the other surface, it is again split into two parts - one that is reflected and the other, which suffers refraction. The amplitude of each of these emergent and reflected waves is different. Considering only the emergent rays first, we focus all of them at one point with the help of a convex lens. According to their phase-differences they will be producing constructive or destructive interference at the focal point. If the optical path difference between two such rays = 2μd, the phase difference:
When light is incident from an optically rarer medium, like air, to an optically denser medium, like a soap film, the reflected ray suffers a sudden phase change. But the next wave with which it interferes does not suffer such a change. Therefore, if 2μd = nλ (condition for minimum intensity in reflected rays), the second wave has to be out of phase with the first one which has suffered the phase change. Similar is the case when 2μd = (n + 1/2)λ (condition for maximum intensity in reflected rays). Using the above concepts, it can be also explained why the top of a vertically placed soap film, which is thinner, appears dark (as light rays reflected there suffer destructive interference), whereas coloured interference fringes decorate the rest of the film.