IN THIS SECTION YOU WILL UNDERSTAND WHY BRIGHT AND DARK FRINGES ARE FORMED
Michelson interferometer principle
The Michelson interferometer is undoubtedly one of the most studied interferometers worldwide.
Its configuration is shown in Figure 1. The beam light from an extended source travels towards
the beam splitter, which divides it into two beams. Both light rays travel through arms 1 and 2
of the interferometer towards mirrors M1 and M2, respectively. Subsequently, the rays are reflected
again towards the beam splitter, where they are recombined to continue their trajectory towards the
screen S on which is observed the interference pattern formed by fringes of equal inclination.
Optical interference can be explained and demonstrated by the next treatment mathematical [1-5].
Let us consider each beam of light as plane electromagnetic waves propagating in a medium. For
this study only the electric field is considered. These waves can be written as:
E₁ = E₀₁ sen[ωt+α₁] , (1)
E₂ = E₀₂ sen[ωt+α₂] , (2)
Where E₀₁, E₀₂, α₁ and α₂ represent the amplitudes and phases of waves E₁ and E₂, respectively.
Let us consider that both waves have the same frequency, coexisting at the same point in space.
The disturbance results in the linear superposition of these waves:
E = E₁ + E₂ (3)
Assuming the superposition of the waves E₁ and E₂, the resulting irradiance I is defined as:
I = I₁ + I₂ + 2 [I₁I₂]1/2cosβ, (4)
Where I₁ and I₂ represent the irradiances related to waves E₁ and E₂, respectively. While the phase
difference between the two waves is represented by β= α2-α1. The third term on the right-hand side of
equation 4 is known as the interference term. The phase difference β is the decisive factor in achieving
constructive or destructive interference between the two waves. When β = 0,±2π,±4π,…. (or β = 2mπ,
where m is an integer), the resulting amplitude is a maximum, and total constructive interference occurs.
In another case when β = ±π,±3π,…. (or β = (2m+1)π), the result is a minimum value, and total destructive
interference occurs [1-4]. The above means that, for total constructive interference the valleys and peaks
of each wave coincide. Whereas, for total destructive interference the valleys overlap the peaks.
Considering that the light waves have the same amplitude, that is, I1=I2=I0 and using the half angle formula
for the cosine in equation 4, we can obtain the following expression:
The intensity distribution in the fringes of equal inclination observed in the interferogram is obtained from
equation 5, where the phase difference is given by the following expression [4]:
I = 2I₀(1+cosβ) = 4I₀cos²(β/2). (5)
When the optical path difference between the two waves is zero, and the mirrors are strictly perpendicular,
solely a bright or dark fringe will be visualized on the scree, due to constructive or destructive interference
[4]. When the optical path length difference between the two arms is different from zero interference fringes
with circular geometry begin to be visualized [4]. These fringes types are often called fringes of equal inclination
(circular fringes). The fringe density is higher when the optical path difference is increasing. On the other hand,
when the mirrors are not entirely perpendicular to each other, fringes made up of straight lines or curved fringes
will be observed. This fringes type is called fringes of equal thickness [3-5].
β = 4π/λdcosθ (6)
Where λ is the wavelength of the light, θ is the inclination of the beams reaching the screen, and d is the difference
in length between the arms of the interferometer.
REFERENCIAS
NUMERICALLY SIMULATED INTERFEROGRAM
The interferogram formed by fringes of equal inclination (concentric rings) or of equal thickness, as shown in the figures below, can be obtained numerically using equation 5. The number of rings can be changed by varying the phase difference d.
a)
b)
a)
b)
Matlab code to obtain fringes of equal inclination