1. Construction
Michelson's interferometer is an optical instrument used to produce interference fringes by dividing a single beam of light into two coherent beams and then recombining them.
Its main components are:
- Monochromatic Light Source (S): Provides light of a single wavelength.
- Beam Splitter (G₁): A semi-silvered glass plate placed at 45° to the incident beam. It divides the incident light into two parts.
- Compensating Plate (G₂): A plane-parallel glass plate of the same material and thickness as the beam splitter. It ensures that both beams pass through equal thicknesses of glass.
- Mirrors M₁ and M₂: Plane mirrors placed perpendicular to each other. One mirror is generally movable, allowing accurate measurement of displacement.
- Telescope or Screen: Used to observe the interference fringes.
The beam splitter divides the incident light into two coherent beams. These beams travel along different paths, are reflected by the mirrors, and finally recombine to produce interference.
2. Working
Suppose a beam of monochromatic light from source S falls on the semi-silvered plate G₁.
- A part of the light is reflected towards mirror M₁.
- The remaining part is transmitted towards mirror M₂.
After reflection:
- The beam reflected from M₁ returns towards G₁.
- The beam reflected from M₂ also returns towards G₁.
These two beams are then recombined and travel in the same direction towards the observer. Since they originate from the same source, they are coherent and produce interference.
The optical arrangement can be considered equivalent to interference between the direct image M₁ and the virtual image M₂' of M₂, formed by reflection in the beam splitter.
Thus, the interference fringes are produced by the reflected waves from two virtually parallel surfaces M₁ and M₂'.
3. Derivation of the Shape of Fringes
Case I: Fringes of Equal Inclination
When M₁ and M₂' are parallel, the interference fringes are called fringes of equal inclination.
Let the effective separation between M₁ and M₂' be d. Consider a ray making an angle θ with the normal.
The optical path difference between the two interfering rays is:
Due to the phase change of π on reflection at one of the surfaces, an additional path difference of λ/2 is introduced.
Therefore, the effective path difference is:
Condition for Dark Fringes
For dark fringes:
Therefore:
This is the condition for dark fringes.
Condition for Bright Fringes
For bright fringes:
or
Shape of Fringes
For a particular fringe order n:
Since d, n, and λ are constants:
Thus, θ is constant.
All rays making the same angle θ with the normal form a cone. When this cone is observed by a telescope or projected onto a screen, its cross-section is a circle.
Fringes of equal inclination are circular fringes.
These circular fringes are also called Haidinger's fringes.
Radius of Fringes
Let the radius of the fringe be \(r_n\) and \(OT = D\), as shown in Fig.
\[ \tan\theta_n = \frac{r_n}{D} \]
Therefore, \[ r_n = D\tan\theta_n \]
For a dark fringe,
\[ 2d\cos\theta_n = n\lambda \]
or
\[ \cos\theta_n = \frac{n\lambda}{2d} \]
Therefore, the radius of the dark fringe is
\[ \boxed{r_n = D\sqrt{\frac{4d^2}{n^2\lambda^2} - 1} } \]
Similarly the radius of the bright fringe is
\[ \boxed{r_n = D\sqrt{\frac{4d^2}{(n+1/2)^2\lambda^2} - 1} } \]
Important Characteristics
- They are concentric circular fringes.
- Their centre is usually dark or bright depending on the optical path difference.
- They are localized at infinity.
- They are observed using a telescope focused at infinity.
- The fringes are called fringes of equal inclination because every fringe corresponds to a constant angle of inclination.
Case II: Fringes of Equal Thickness
If M₁ and M₂' are slightly inclined to each other, they form a small wedge-shaped film.
Let the angle between the two surfaces be α.
At a distance x from the line of intersection, the thickness of the air film is:
For a small angle:
Hence:
For nearly normal incidence, the optical path difference is:
Considering the phase change on reflection, the condition for dark fringes is:
Substituting t = xα:
Therefore:
This shows that the position of a fringe depends linearly on the fringe order n.
Fringe Width
The distance between two successive dark fringes is:
Therefore:
where β is the fringe width.
Since the fringe width is constant, the fringes are equally spaced straight lines.
Fringes of equal thickness are straight, parallel and equally spaced.
Summary of Fringe Formation
| Arrangement of Mirrors | Type of Fringes | Shape |
|---|---|---|
| M₁ parallel to M₂' | Fringes of Equal Inclination | Circular |
| M₁ slightly inclined to M₂' | Fringes of Equal Thickness | Straight and Parallel |
Final Result
In Michelson's interferometer:
because:
and each fringe corresponds to a constant angle of inclination.
When the mirrors are slightly inclined:
which gives:
Hence, the fringes are straight, parallel and equally spaced.