Interferential devices

Glass Blade

The simplest wavefront-splitting interferometric system is given by a glass lamp or a glass corner observed in reflection. This paragraph is strongly based on Chapter 6 of reference [1].

During the refraction on an air-glass type diopter, about 4% of the light energy is reflected. So the light reflected or transmitted can be the cause of an interferences phenomenon. In this paragraph we will only deal with the interferences by reflection, as the case transmission is similar. An extended and monochromatic source located in the air lights a blade with parallel sides of index , of thickness  (figure 5) put down on a third medium of index . As the source is extended, we determine the localization area of the interferences fringes.

    Figure 5 : Glass Blade
Figure 5 : Glass Blade [zoom...]Info

Incident beam stemming from primary source partially reflects in following the direction while a part of refracted beam is reflected following then refracted in the direction . Contributions of beam and following are negligeable because these beams energy light decreases fast. Indeed, if there is 4% of energy light for the first reflected beam , there is only 0.0059% for the third beam . Both beams et stemming from the same incident beam , emerges parallely between them, they “interfere at infinity” . If a screen is located in the image focus plane of a convergent lens, the lens emergent beams cross in M, thus the interference figure is projected on the screen.

As in the case of Young slits, we can express the course difference  in function of the characteristics of the interferential device, which is to say of the blade, as well as the geometrical shape of interferences fringes.

Incident beam gives two reflected beams and . Beyond points and both reflected beams cover the same optical path. On the other hand, between and beam covers the distance in the air and beam covers path in the index medium . The difference of optical paths between both beams and equals:

Let us considerer triangle  :


Let Descartes Law be applied for the refraction  :

For triangle we have the following two trigonometric relationships:




Replacing , and by their expressions according to , and in the first equation:

Two different cases have to be considered:

  • If the indexes are such as:

Both reflections in and are of the same kind, which is to say that each time reflection occurs from a less refringent medium on a more refringent medium. So the course difference  is equal to the difference of optical path:

  • If the indexes are such as:

Reflections are not the same, so we will recognize that, in this case, to add to the course difference [1 ] we have to add to the difference of optical path:

The points set for which the course difference is the same are in the same state of interference. The geometrical aspect of the interferences fringes is given searching the conditions for which .

In the case of light fringes, interferences are constructive and the course difference   equals a whole number times the wavelength (see lesson “Interferences Principle”)

For a given device, the wavelength, the blade index and thickness are constant and the points in the same state of interference verify:

As the refraction and incidence angles are linked by Descartes Law, that leads to . The observation of interferences figure on a screen located in the focus image plane of the lens alternately shows bright and dark concentric rings (Figure 6).

All emergent beams that interfere at the level of a same ring correspond to incident beams that have the same angle of incidence. These interferences fringes are called “same inclination rings”.

    Figure 6 : Same inclination rings
Figure 6 : Same inclination rings [zoom...]Info

Now we will study angular rays of the same inclination rings for a blade thickness . We position ourselves in the case of a bright rings center. In the center, the course difference written corresponds to a zero angle of reflection , it is equal to a odd number times the half-length wave:

in the case where:

For the peripheral black rings , raises, the course difference is smaller than and:

The interval of course difference between a peripheral black ring of order and the central ring equals:

And with a limited development of cosine for same angles  :


Descartes Law applied to small angles makes the deduction of angle of incidence  possible:

The angular rays of the rings corresponding to the same state of interferences as the center vary as the square root of the successive whole numbers. If the observation is made in the lens focus plane of the image focal distance , these rings linear beams are (figure 7).

    Figure 7 : Observation of the same inclination rings at a lens focus
Figure 7 : Observation of the same inclination rings at a lens focus [zoom...]Info

In the air and for small angles, the interfringe is expressed as:

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