# Interference Phenomena in a new form of Refractometer

Interference Phenomena in a new form of Refractometer  (1882)
by Albert Abraham Michelson

American Journal of Science, Vol. XXIII, pp. 395-400

Art. XLVI.—Interference Phenomena in a new form of Refractometer; by Albert A. Michelson

In an experiment undertaken with a view to detecting the relative motion of the earth and the luminiferous ether (Am. Journal of Science, No. 128, vol. xxii,) it was necessary to produce interference of two pencils of light, which had traversed paths at right angles with each other. This was accomplished as follows: The light from a lamp at ${\displaystyle a}$, fig. 1, was separated into two pencils at right angles, ${\displaystyle bc}$, ${\displaystyle bd}$, by the plane-parallel glass ${\displaystyle b}$, and these two pencils were returned to ${\displaystyle b}$ by the mirrors ${\displaystyle c}$ and ${\displaystyle d}$, whence they coincided along ${\displaystyle be}$, where they were viewed by the eye, or by a small telescope at ${\displaystyle e}$.

It is evident that, so far as the interference is concerned, the apparatus may be replaced by a film of air whose thickness is ${\displaystyle bc-cd}$, and whose angle is that formed by the image of ${\displaystyle d}$ in ${\displaystyle b}$, with ${\displaystyle c}$.

The problem of interference in thin films has been studied by Feussner, but his equations do not appear to give the explanation of the phenomena observed. In particular, in the "Annalen der Physik und Chemie," No. 12, 1881, on page 558, Feussner draws the conclusion that the interference fringes are straight lines, whereas, in the above described apparatus they are in general curves: and there is but one case—that of the central fringe in white light—which is straight.

 Fig. 1 Fig. 2

I have therefore thought it worth while to attempt the solution of the problem for a film of air, for small angles of incidence and neglecting successive reflections; and though the solution is not perhaps adapted to the general problem, it accounts for all the phenomena observed in the special case.

Let ${\displaystyle Om_{1}}$, ${\displaystyle Om_{2}}$, fig. 2, be two plane mirrors whose intersection is projected at ${\displaystyle O}$, and whose mutual inclination is ${\displaystyle \varphi }$. The illumination at any point, ${\displaystyle P}$ (not necessarily in the plane of the figure), will depend on the mean difference of phase of all the pairs of rays starting from the source and reaching ${\displaystyle P}$, after reflection from the mirrors; a pair of rays signifying two rays which have originated at the same point of the source.

If the area of the luminous surface is sufficiently large the illumination at ${\displaystyle P}$ will be independent of the distance, form, or position of the surface. Suppose, therefore, that the luminous surface coincides with the surface ${\displaystyle Om_{1}}$. Its image in ${\displaystyle Om_{1}}$ will also coincide with ${\displaystyle Om_{1}}$, and its image in ${\displaystyle Om_{2}}$ will be a plane surface symmetrical with ${\displaystyle Om_{1}}$ with respect to the surface ${\displaystyle Om_{2}}$, and for every point, ${\displaystyle p}$, of the first image there is a corresponding point, ${\displaystyle p'}$, of the second, symmetrically placed and in the same phase of vibration. Suppressing, now, the source of light and the mirrors, and replacing them by the two images, the effect on any point, ${\displaystyle P}$, is unaltered.

Consider now a pair of points ${\displaystyle pp'}$. Let ${\displaystyle \vartheta }$ be the angle formed by the line joining ${\displaystyle P}$ and ${\displaystyle p}$ (or ${\displaystyle p'}$) with the normal to the surface; ${\displaystyle \vartheta }$ and ${\displaystyle \varphi }$ being both supposed small,

${\displaystyle \Delta =Pp'-Pp=pp'\cdot \cos \vartheta }$.

The difference between this value of ${\displaystyle \Delta }$ and the true value is ${\displaystyle 2Pp\cdot \sin ^{2}{\tfrac {\psi }{2}}}$, where ${\displaystyle \psi }$ is the angle subtended by ${\displaystyle pp'}$ at ${\displaystyle P}$. If ${\displaystyle \vartheta }$ is a small quantity, ${\displaystyle \psi }$ is a small quantity of the second order, and ${\displaystyle \sin ^{2}{\tfrac {\psi }{2}}}$ is a small quantity of the fourth order; consequently ${\displaystyle 2Pp\cdot \sin ^{2}{\tfrac {\psi }{2}}}$ may be neglected. We have therefore, to a very close approximation, ${\displaystyle \Delta =pp'\cos \vartheta }$; or, substituting ${\displaystyle 2t}$ for ${\displaystyle pp'}$, ${\displaystyle 2t}$ being the distance between the images, at the point where they are cut by the line ${\displaystyle Pp}$,

${\displaystyle \Delta =2t\cos \vartheta }$

Let ${\displaystyle cdef}$, ${\displaystyle c'd'e'f'}$, fig. 3, represent the two images, and let their intersection be parallel with ${\displaystyle cf}$, and their inclination be

Fig. 3

${\displaystyle 2\varphi }$. Let ${\displaystyle P}$ be the point considered; ${\displaystyle P'}$, the projection of ${\displaystyle P}$ on the surface ${\displaystyle cdef}$; and ${\displaystyle PB}$, the line forming with ${\displaystyle PP'}$ the angle ${\displaystyle \vartheta }$. Draw ${\displaystyle P'D}$ parallel to ${\displaystyle cf}$, and ${\displaystyle P'C}$, at right angles, and complete the rectangle ${\displaystyle BDP'C}$. Let ${\displaystyle P'PC=i}$ and ${\displaystyle DPP'=\theta }$. Let ${\displaystyle PP'=P}$, and the distance between the surfaces at ${\displaystyle P'=2t_{0}}$. We have then

 {\displaystyle {\begin{aligned}t=t_{0}+CP'\cdot \tan \varphi =&t_{0}+P\tan \varphi \cdot \tan i,\ {\text{and}}\\\Delta =&2\left(t_{0}+P\tan \varphi \tan i\right)\cos \vartheta ,\ {\text{or}}\\\Delta =&{\frac {2\left(t_{0}+P\tan \varphi \tan i\right)}{\sqrt {1+\tan ^{2}i+\tan ^{2}\theta }}}\end{aligned}}} (1)

We see that in general ${\displaystyle \Delta }$ has all possible values, and therefore all phenomena of interference would be obliterated. If, however, we observe the point ${\displaystyle P}$ through a small aperture, ${\displaystyle ab}$, the pupil of the eye, for instance, the light which enters the eye from the surfaces will be limited to the small cone whose angle is ${\displaystyle bPa}$, and if the aperture be sufficiently small the differences in ${\displaystyle \Delta }$ may be reduced to any required degree.

It is proposed to find such a distance ${\displaystyle P}$, that with a given aperture these differences shall be as small as possible, which is equivalent to finding the distance from the mirrors at which the phenomena of interference are most distinct. The change of ${\displaystyle \Delta }$ for a change in ${\displaystyle \theta }$, is

 ${\displaystyle {\frac {\delta \Delta }{\delta \theta }}=-{\frac {2\left(t_{0}+P\tan \varphi \tan i\right){\frac {\tan \theta }{\cos ^{2}\theta }}}{\left(1+\tan ^{2}i+\tan ^{2}\theta \right)^{\frac {3}{2}}}}}$ (2)

The change of ${\displaystyle \Delta }$ for a change in ${\displaystyle i}$, is

 ${\displaystyle {\frac {\delta \Delta }{\delta i}}=2{\frac {\left(1+\tan ^{2}i+\tan ^{2}\theta \right){\frac {P\tan \varphi }{\cos ^{2}i}}-\left(t_{0}+P\tan \varphi \tan i\right){\frac {\tan i}{\cos ^{2}i}}}{\left(1+\tan ^{2}i+\tan ^{2}\theta \right)^{\frac {3}{2}}}}}$ (3)

For ${\displaystyle {\tfrac {\delta \Delta }{\delta \theta }}=0}$ we have ${\displaystyle \theta =0}$ (or ${\displaystyle \Delta =0}$).

For ${\displaystyle {\frac {\delta \Delta }{\delta i}}=0}$ we have ${\displaystyle \left(1+\tan ^{2}i+\tan ^{2}\theta \right)P\tan \varphi =\left(t_{0}+P\tan \varphi \tan i\right)\tan i}$, or

${\displaystyle \left(1+\tan ^{2}\theta \right)P\tan \varphi =t_{0}\tan i}$, whence ${\displaystyle P={\frac {t_{0}}{\tan \varphi }}\tan i\cos ^{2}\theta }$

Hence the fringes will be most distinct when ${\displaystyle \theta =0}$ and when

 ${\displaystyle P={\frac {t_{0}}{\tan \varphi }}\tan i}$ (4)

This condition coincides nearly with that found by Feussner.

If the thickness of the film is zero, or if the angle of incidence is zero, the fringes are formed at the surface of the mirrors. If the film is of uniform thickness, the fringes appear at infinity. If at the same time ${\displaystyle \varphi =0}$, and ${\displaystyle t_{0}=0}$, or ${\displaystyle i=0}$ and ${\displaystyle \varphi =0}$, the position of the fringes is indeterminate. If ${\displaystyle i}$ has the same sign as ${\displaystyle \varphi }$, the fringes appear in front of the mirrors; if ${\displaystyle i}$ has the opposite sign, the fringes appear behind the mirrors.

To find the form of the curves as viewed by the eye at ${\displaystyle E}$, let ${\displaystyle SE=D}$; call ${\displaystyle T}$ the distances between the surfaces at ${\displaystyle E'}$, the projection of ${\displaystyle E}$. From ${\displaystyle P}$ draw ${\displaystyle PR}$ parallel to ${\displaystyle DP'}$, and ${\displaystyle RS}$ at right angles, and let ${\displaystyle RS=c}$. We have then ${\displaystyle t_{0}=T+c\tan \varphi }$, whence, substituting for ${\displaystyle c}$ its value ${\displaystyle D\tan i}$,

 ${\displaystyle \Delta =2{\frac {T+(D+P)\tan \varphi \tan i}{\sqrt {1+\tan ^{2}i+\tan ^{2}\theta }}}}$ (5)

If on a plane perpendicular to ${\displaystyle EE'}$ at distance ${\displaystyle D}$ from ${\displaystyle E}$, we call ${\displaystyle x}$ distances parallel to ${\displaystyle P'C}$, and ${\displaystyle y}$ distances parallel to ${\displaystyle P'D}$, reckoned from the projection of ${\displaystyle E}$ on this plane, then, putting ${\displaystyle \varphi =K}$ and ${\displaystyle D+P=S}$, we have for the equation to the curves, as they would appear on this surface to an eye at ${\displaystyle E}$,

${\displaystyle \Delta =2{\frac {DT+SKx}{\sqrt {D^{2}+x^{2}+y^{2}}}}}$

or
 ${\displaystyle \Delta ^{2}y^{2}=\left(4S^{2}K^{2}-\Delta ^{2}\right)x^{2}+8TSKDx+\left(4T^{2}-\Delta ^{2}\right)D^{2}}$ (6)

If, numerically,

 ${\displaystyle \Delta <2SK}$ the curve is a hyperbola, ${\displaystyle \Delta =2SK}$ the curve is a parabola, ${\displaystyle \Delta >2SK}$ the curve is an ellipse, ${\displaystyle K=0}$ the curve is a circle, ${\displaystyle \Delta =0}$ the curve is a straight line,

All the deductions from equations (4) and (6) have been approximately verified by experiment.

It is to be observed that in the most important case, and that most likely to occur in practice, namely, in the case of the central fringe in white light, we have ${\displaystyle \Delta =0}$, and therefore also ${\displaystyle t_{0}=0}$; and in this case the central fringe is a straight line formed on the surface of the mirrors. Practically, however, it is impossible to obtain a perfectly straight line, for the surface of the mirrors is never perfect.

It is to be noticed that the central fringe is black, for one of the pencils has experienced an external, the other an internal reflection from the surface ${\displaystyle b}$, fig. 1. This will not however be true unless the plate ${\displaystyle g}$ (which is employed to compensate the effect of the plate ${\displaystyle b}$) is of exactly the same thickness as ${\displaystyle b}$, and placed parallel with ${\displaystyle b}$. When these conditions are not fulfilled, the true result is masked by the effect of "achromatism" investigated by Cornu (Comptes Rendus, vol. xciii, Nov. 21st, 1881). This remark leads naturally to the investigation of the effect of a plate of glass with plane parallel surfaces, interposed in the path of one of the pencils.

Fig. 4

The effect is independent of the position of the glass plate, provided its surface is kept parallel with the corresponding mirror. Suppose, therefore, that it is in contact with the latter and let ${\displaystyle cd}$, fig. 4, represent the common surface. Let ${\displaystyle t=hi=}$ thickness of the glass, ${\displaystyle i=}$ angle of incidence, ${\displaystyle r=}$ angle of refraction, ${\displaystyle n=}$ index of refraction, ${\displaystyle \lambda =}$ wave-length of light. Let ${\displaystyle ef}$ represent the image of the other mirror, and put ${\displaystyle n_{0}={\tfrac {hk}{t}}}$.

It can be readily demonstrated that the path of the rays in the instrument is equivalent to that given in the figure, where one of the rays follows the path ${\displaystyle qnmh}$, and the other the path ${\displaystyle rolh}$. Suppose the mirrors ${\displaystyle cd}$ and ${\displaystyle ef}$ parallel. Then as has been previously shown, the curves of interference are concentric circles, formed at an infinite distance. Therefore the rays ${\displaystyle qn}$, ${\displaystyle ro}$, whose path is to be traced, are parallel, and from the point ${\displaystyle h}$ they coincide. Their difference of path is ${\displaystyle 2nm-2hl-op}$, and their difference of phase is

 {\displaystyle {\begin{aligned}\varphi =&{\frac {2nm}{\lambda }}-{\frac {2hl}{\frac {\lambda }{n}}}-{\frac {op}{\lambda }}={\frac {2n_{0}\cdot t}{\lambda \cos i}}-{\frac {2t\cdot n}{\lambda \cos r}}\\&-{\frac {2t}{\lambda }}\left(n_{0}\tan i-\tan r\right)\sin i,\ {\text{whence}}\\\varphi =&{\frac {2t}{\lambda }}\left[n_{0}\cos i-n\cos r\right]\end{aligned}}} (7)

Let it be proposed to find the value of ${\displaystyle n_{0}}$ which renders any particular ring achromatic. The condition of achromatism, as given by Cornu, is ${\displaystyle {\tfrac {d\varphi }{d\lambda }}=0}$, which gives

${\displaystyle \varphi +2t\left(\cos r{\frac {dn}{d\lambda }}-n\sin r{\frac {dr}{dn}}\cdot {\frac {dn}{d\lambda }}\right)=0}$

We have

${\displaystyle n={\frac {\sin i}{\sin r}}}$ whence ${\displaystyle {\frac {dr}{dn}}=-{\frac {\sin ^{2}r}{\sin i\cos r}}}$, whence

${\displaystyle \varphi +{\frac {2t}{\cos r}}\cdot {\frac {dn}{d\lambda }}=0}$

By Cauchy's formula we have ${\displaystyle n=a_{1}+{\tfrac {\alpha _{2}}{\lambda ^{2}}}}$ whence ${\displaystyle {\tfrac {dn}{d\lambda }}=-{\tfrac {2\alpha _{2}}{\lambda ^{3}}}}$.

Substituting, we have ${\displaystyle \varphi -{\tfrac {4\alpha _{2}t}{\lambda ^{3}\cos r}}=0}$ or ${\displaystyle n_{0}\cos i=n\cos r={\tfrac {2\alpha _{2}}{\lambda ^{2}\cos r}}={\tfrac {{\tfrac {2\alpha _{2}}{\lambda ^{2}}}+n\cos ^{2}r}{\cos i\cos r}}}$, or finally,

 ${\displaystyle n_{0}={\frac {2\left(n-\alpha _{1}\right)+n\cos ^{2}r}{\cos i\cos r}}}$ (8)

If the angle ${\displaystyle i}$ is small, the value of ${\displaystyle n_{0}}$ will vary very little with ${\displaystyle i}$, consequently there will be a large number of circles all nearly achromatised. Under favorable circumstances as many as one hundred rings have been counted, using an ordinary lamp, as source of light. The difference of path of the two pencils which produce these rings in white light may exceed a thousand wave lengths.

This work is in the public domain in the United States because it was published before January 1, 1926.

The author died in 1931, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 80 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.