Optical Interference is an effect that can occur when two or more light beams are superimposed. We knew that a ray of light is composed of an infinite number of waves of equal wavelength. the value of the wavelength determines the colour of light.

For the sake of simplicity, let us consider two waves, having a sinusoidal property, from two different light rays. The below figure illustrates the combined effect of the two waves of light. The two rays, A and B, are in phase at the origin O and will remain so as the rays propagate through a large distance.

Suppose the two rays have amplitudes y_{A} and y_{B},

then the resultant wave will have an amplitude y_{R} = y_{A} + y_{B}

Thus, when the two rays are in phase, the resultant amplitude is maximum and the intensity of light is also maximum. However, if the two rays are out of phase, say by an amount d, then the resultant wave will have an amplitude will be

y_{R} = (y_{A} + y_{B}) cos𝛿/2

It is clear that the combination of the two waves no longer produces maximum illumination.

Consider the case where the phase difference between the two waves is 180°. The amplitude of the resulting wave, which is shown in the above figure, is the algebraic sum of y_{A} and y_{B}. The corollary is that if y_{A} and y_{B} are equal, then y_{R} will be zero since cos(180/2) is zero. This means that complete interference between two waves having the same wavelength and amplitude produces darkness.

One of the properties of light is that light from a single source can be split into two component rays. Observing the way in which these components recombine shows us that the wavelength of light can be used for linear measurement. The linear displacement d between the wavelengths of the two light rays results in maximum interference when d = λ/2

Where λ is the wavelength of light.

Now in what way is this property going to help us in taking linear measurements? The above Figure illustrates how the property of interference of light can be used for linear measurement. Let us consider two monochromatic light rays from two-point sources, A and B, which have the same origin. The light rays are made to fall on a flat screen that is placed perpendicular to the axis OO_{1}. The axis OO_{1} is in turn perpendicular to the line joining the two-point sources, A and B. Since both rays originate from the same light source, they are of the same wavelength. Let us also assume that the distances OA and OB are equal.

Now, consider the convergence of two rays at point O_{1} on the screen. Since the distances AO_{1} and BO_{1} are equal, the two rays are in phase, resulting in maximum illumination at point O_{1}. On the other hand, at point O_{2}, the distance BO_{2} is longer than the distance AO_{2}. Therefore, by the time the two rays arrive at point O_{2}, they are out of phase. Assuming that the phase difference d = λ/2, where λ is the wavelength of light, complete interference occurs, forming a dark spot.

At point O_{3} on the screen, the distance BO_{3} is longer than AO_{3}. If the difference between the two distances, that is, BO_{3} − AO_{3}, is equal to an even number of half wavelengths, the two light rays arriving at O_{3} will be in phase, leading to the formation of a bright spot. This process repeats on either side of O_{1} on the screen, resulting in the formation of alternate dark and bright areas. This pattern of alternate bright and dark areas is popularly known as fringes. The dark areas will occur whenever the path difference of A and B amounts to an odd number of half wavelengths, and the bright areas will occur when the path difference amounts to an even number of half wavelengths.

## Conclusion

We have discussed the Optical Interference with schematic diagram. Please let us know your thought in the comment session below.

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