Problem

The set-up of the model and the notations used are shown in figure 25.

Let us take the case of a two wave interferometer with perfectly coherent light, since the source is punctual and monochromatic, the distribution of illumination is:

In the case of Young's slits, the interfringe distance is expressed as:

Since the source is perfectly coherent, the maximum contrast is equal to 1.

If the source S is replaced by an infinitely narrow vertical slit, the different source points are incoherent amongst themselves and the light is partially spatially coherent. The optical path difference for a point M on the screen is given by the relation in paragraph D.3 of this course (Evaluation of the Optical Path Difference):

Here the vectors and are orthogonal (figure 26).

Consequently and the degree of coherence is equal to 1, the contrast does not diminish.

A given fringe corresponds to a given optical path difference  If we interpose a glass plate in the trajectory of F1 (1) the optical path increases because the index of the plate is superior to the index of the air. In order to compensate for this difference, the path length of the trajectory (2) must increase so that the value of the optical path difference does not change.

1. Therefore, the fringes pass through the side where the plate has been inserted.

2. The 2.9 mm shift is caused by the passing of the fringes:

   because a spatial period corresponds to an interfringe distance.

3. When the glass plate is interposed, the optical path difference varies because on a thickness of e the air index 1 is replaced by the index n of the glass. The variation is thus:

   The passage from one fringe to the next of the same nature (dark or bright) corresponds to a variation in the optical path difference of , from which:

Both slits do not have the same intensity, so we must split the current relation:

The relation between both intensities verifies:

from which:

i.e.:

which produces a contrast of:

The case of a white light source is the same as the one described in the case study in paragraph II.B "Case of white light"

The sources still have the same intensity, but the source is no longer spatially coherent because it has a width of e. The set-up and notations used are represented in 27.

The interference fringes have a weaker contrast. The intensity distribution on the screen is given by:

represents the Fourier transform of the source's brightness function.

The source's spectral distribution of luminance can be described by a rectangle function with a width of e :

for which the Fourier transform is:

where is the wave number of the source and is the angle from which the slits are seen from the source center.

Consequently, the degree of spatial coherence is equal to:

First, let's consider a single source slit that has been shifted on the plane (X , Y) by a quantity p following the X axis. For a point M on the screen the optical path difference varies by . The calculation is analogous to the one in the Case Study. If this optical path difference is equal to   then the figure of interference is identical to the preceding one. The figure is translated downwards by the equivalent of one interfringe distance. A bright fringe (respectively dark) has replaced the following bright fringe (respectively dark).

Consequently, the single source slit can be replaced by a lattice:

The source is now spatially coherent, but partially temporally incoherent.

As seen in paragraph C.3.b ("Examples" in Temporal Coherence), the energy spectral density is expressed as:

and we have (see paragraph C.3.b "Examples"):

Graph of the intensity distribution on the screen: the interferogram has beats and a similar structure to the one shown on figure 28.

The interferogram of the doublet is also written as:

The inferfringe distance is thus given by:

with .

Therefore, we have:

The contrast function cancels out when the cosine function is equal to , i.e.:

Same procedure as the previous excercise.

The value of is given by:

The energy spectral density is:

The spectral shape is thus:

The Fourier transform of is:

The degree of coherence is:

The interference signal is expressed as:

The coherence length of the source can be calculated by using the expression:

i.e.:

Taking into account that:

the result is:

When this value is applied it produces:

The standard illumination has an average value of 1/2. It is shown on figure 29.

The cosine function always has the same period , the geometry does not change so the period on the screen and the interfringe remain the same.

The value of is given by:

and the value of is given by:

When each one of the sodium doublets can be described by a Gaussian function, the caseis similar to the real situation. The energy spectral density function is written as:

The procedure is the same as in the previous exercise. We calculate the Fourier transform of the spectral energy density, which is equal to the product of the Fourier transforms of the two convoluted functions.

These Fourier transforms have been previously calculated.

The two Dirac distributions convoluted to the gaussian function are written as:

and their Fourier transform is:

From this, we can deduce the expression of the normalized interferogram:

The intensity as a function of the optical path difference is represented in figure 30.