Michelson Interferometer

Figure 10.7 shows a typical Michelson interferometer setup. A ground glass plate G is illuminated by a light source. The ground glass plate has the property that it scatters the incident light into all directions. Each point on the ground glass plate acts like a source that emits light in all directions.

Figure 10.7: Michelson Interferometer

The light scattered forward by G is incident on a beam splitter B which is at $45^{\circ}$. The beam splitter is essentially a glass slab with the lower surface semi-silvered to increase its reflectivity. It splits the incident wave into two parts $\tilde{E}_1$ and $\tilde{E}_2$, one which is transmitted ($\tilde{E}_1$) and another ($\tilde{E}_2$) which is reflected. The two beams have nearly the same intensity. The transmitted wave $\tilde{E}_1$ is reflected back to B by a mirror M$_1$. and a part of it is reflected into the telescope T. The reflected wave $\tilde{E}_2$ travels in a perpendicular direction. The mirror M$_2$ reflects this back to B where a part of it is transmitted into T.

Figure 10.8: Effective set-up for Michelson Interferometer

An observer at T would see two images of G, namely G$_1$ and G$_2$ (shown in Figure 10.8) produced by the two mirrors M$_1$ and M$_2$ respectively. The two images are at a separation $2 d$ where $d$ is the difference in the optical paths from B to G$_1$ and from B to G$_2$. Note that $\tilde{E}_2$ traverses the thickness of the beam splitter thrice whereas $\tilde{E}_1$ traverses the beam splitter only once. This introduces an extra optical path for $\tilde{E}_2$ even when M$_1$ and M$_2$ are at the same radiation distance from B. It is possible to compensate for this by introducing an extra displacement in M$_1$, but this would not serve to compensate for the extra path over a range of frequencies as the refractive index of the glass in B is frequency dependent. A compensator C, which is a glass block identical to B (without the silver coating) , is introduced along the path to M$_1$ to compensate for this.

S$_1$ and S$_2$ are the two images of the same point S on the ground glass plate. Each point on the ground glass plate acts as a source emitting radiation in all directions. Thus S$_1$ and S$_2$ are coherent sources which emit radiation in all direction. Consider the wave emitted at an angle $\theta$ as shown in Figure 10.8. The telescope focuses both waves to the same point. The resultant electric field is

$$\tilde{E}= \tilde{E}_1 + \tilde{E}_2$$ (10.19)

and the intensity is

(10.20)

The phase difference arises because of the path difference in the two arms of the interferometer. Further, there is an additional phase difference of $\pi$ because $\tilde{E}_2$ undergoes internal reflection at B whereas $\tilde{E}_1$ undergoes external reflection. We then have

$$\phi_2-\phi_1=\pi + 2 d \, \cos \theta \, \frac{2 \pi}{\lambda}$$ (10.21)

So we have the condition

$$2 d \, \cos \theta_m = m \lambda \hspace{1 cm} (m=0,1,2,...)$$ (10.22)

for a minima or a dark fringe. Here $m$ is the order of the fringe, and $\theta_m$ is the angle of the $m^{\rm th}$ order fringe. Similarly, we have

$$2 d \, \cos \theta_m = \left(m + \frac{1}{2}\right) \lambda \hspace{1 cm} (m=0,1,2,...)$$ (10.23)

as the condition for a bight fringe. The fringes will be circular as shown in Figure 10.9. When the central fringe is dark, the order of the fringe is

$$m=\frac{2 d}{\lambda} \,.$$ (10.24)

Figure 10.9: Michelson fringes

Le us follow a fringe of a fixed order, say $m$, as we increase $d$ the difference in the length of the two arms. The value of $\cos \theta_m$ has to decrease which implies that $\theta_m$ increases. As $d$ is increased, new fringes appear at the center, and the existing fringes move outwards and finally move out of the field of view. For any value of $d$, the central fringe has the largest value of $m$, and the value of $m$ decreases outwards from the center.

Considering the situation where there is a central dark fringe as shown in the left of Figure 10.9, let us estimate $\theta$ the radius of the first dark fringe. The central dark fringe satisfies the condition

$$2 d \, = m \lambda$$ (10.25)

and the first dark fringe satisfies

$$2 d \, \cos \theta = (m-1) \lambda$$ (10.26)

For small $\theta$ ie. $\theta \ll 1$ we can write eq. (10.26) as

$$2 d \,(1-{\theta^2\over 2}) = (m-1) \lambda$$ (10.27)

which with eq. (10.25) gives

$$\theta =\sqrt{\frac{\lambda}{d}}$$ (10.28)

Compare this with the Young's double slit where the fringe separation is $\lambda/d$.

The Michelson interferometer can be used to determine the wavelength of light. Consider a situation where we initially have a dark fringe at the center. This satisfies the condition given by eq. 10.25 where $\lambda$, $d$ and $m$ are all unknown. One of the mirrors is next moved so as to increase $d$ the difference in the lengths of the two arms of the interferometer. As the mirror is moved, the central dark fringe expands and moves out while a bright fringe appears at the center. A dark fringe reappears at the center if the mirror is moved further. The mirror is moved a distance $\Delta d$ so that $N$ new dark fringes appear at the center. Although initially $d$ and $m$ were unknown for the central dark fringe, it is known that finally the difference in lengths is $d + \Delta d$ and the central dark fringe is of order $N+m$ and hence it satisfies

$$2 (d + \Delta d)= (m+N) \lambda$$ (10.29)

Subtracting eq. 10.25 from this gives the wavelength of light to be

$$\lambda= \frac{2 \, \Delta d}{N}$$ (10.30)

We next consider a situation where there are two very close spectral lines $\lambda_1$ and $\lambda_1+ \Delta \lambda$. Each wavelength will produce its own fringe pattern. Concordance refers to the situation where the two fringe patterns coincide at the center

$$2 d = m_1 \lambda_1 = m_2 (\lambda_1 + \Delta \lambda)$$ (10.31)

and the fringe pattern is very bright. As $d$ is increased, $m_1$ and $m_2$ increase by different amounts with $\Delta m_2 < \Delta m_1$. When $m_2=m_1-1/2$, the bright fringes of $\lambda_1$ coincide with the dark fringes of $\lambda_1+ \Delta \lambda$ and vice-versa, and consequently the fringe pattern is washed away. The two set of fringes are now said to be discordant.

It is possible to measure $\Delta \lambda$ by increasing $d$ to $d + \Delta d$ so that the two sets of fringes that are initially concordant become discordant and are finally concordant again. It is clear that if $m_1$ changes to $m_1 + \Delta m$, $m_2$ changes to $m_2 + \Delta m -1$ when the fringes are concordant again. We then have

$$2 (d + \Delta d)=(m_1 + \Delta m) \lambda_1 = (m_2 + \Delta m-1 ) (\lambda_1 + \Delta \lambda)$$ (10.32)

which gives

$$\lambda_1=\left( \frac{2 \, \Delta d}{\lambda_1}-1\right) \Delta \lambda$$ (10.33)

where on assuming that $2 \Delta d /\lambda_1=m_1 \gg 1$ we have

$$\delta \lambda =\frac{\lambda_1^2}{\Delta d} \,.$$ (10.34)

The Michelson interferometer finds a variety of other application. It was used by Michelson and Morley in 1887 to show that the speed of light is the same in all directions. The armlength of their interferometer was $11 \, {\rm m}$. Since the Earth is moving, we would expect the speed of light to be different along the direction of the Earth's motion. Michelson and Morley established that the speed of light does not depend on the motion of the observer, providing a direct experimental basis for Einstein's Special Theory of Relativity.

Figure 10.10: Laser Interferometer Gravitational-Wave Observatory

The fringe patter in the Michelson interferometer is very sensitive to changes in the mirror positions, and it can be used to measure very small displacements of the mirrors. A Michelson interferometer whose arms are $4 \, {\rm km}$ long (Figure 10.10) is being used in an experiment called Laser Interferometer Gravitational-Wave Observatory (LIGO^10.1) which is an ongoing effort to detect Gravitational Waves, one of the predictions of Einstein's General Theory of Relativity. Gravitational waves are disturbances in space-time that propagate at the speed of light. A gravitational wave that passes through the Michelson interferometer will produce displacements in the mirrors and these will cause changes in the fringe pattern. These displacements are predicted to be extremely small. LIGO is sensitive enough to detect displacements of the order of $10^{-16} \, {\rm cm}$ in the mirror positions.

Problems

1. An electromagnetic plane wave with $\lambda=1 \, {\rm mm}$ is normally incident on a screen with two slits with spacing $d= 3\, {\rm mm}$.

   a.

   How many maxima will be seen, at what angles to the normal?

   b.

   Consider the situation where the wave is incident at $30^{circ}$ to the normal.

2. Two radio antennas separated by a distance $d=10 \, {\rm m}$ emit the same signal at frequency $\nu$ with phase difference $\phi$. Determine the values of $\nu$ and $\phi$ so that the radiation intensity is maximum in one direction along the line joining the two antennas while it is minimum along exactly the opposite direction. How do the maxima and minima shift of $\phi$ is reduced to half the earlier value?
3. A lens of diameter $5.0 \, {\rm cm}$ and focal length $20 \, {\rm cm}$ is cut into two identical halves. A layer $1 \, {\rm mm}$ in thickness is cut from each half and the two lenses joined again. The lens is illuminated by a point source located at the focus and a fringe pattern is observed on a screen $50 \, {\rm cm}$ away. What is the fringe spacing and the maximum number of fringes that will be observed?

4. Two coherent monochromatic point sources are separated by a small distance, find the shape of the fringes observed on the screen when, a) the screen is at one side of the sources and normal to the screen is along the line joining the two sources and b) when the normal to the screen is perpendicular to the line joining the sources.

5. The radiation from two very distant sources A and B shown in the Figure 10.11 is measured by the two antennas 1 and 2 also shown in the figure. The antennas operate at a wavelength $\lambda$. The antennas produce voltage outputs $\tilde{V}_1$ and $\tilde{V}_2$ which have the same phase and amplitude as the electric field $\tilde{E}_1$ and $\tilde{E}_2$ incident on the respective antennas. The voltages from the two antennas are combined
   
   $$\tilde{V}=\tilde{V}_1+\tilde{V}_2$$
   
   
   and applied to a resistance. The average power $P$ dissipated across the resistance is measured. In this problem you can assume that $\theta \ll 1$ (in radians).

   Figure 10.11:

   

   a.

   What is the minimum value of $d$ (separation between the two antennas) at which $P=0$?

   b.

   Consider a situation when an extra phase $\phi$ is introduced in $\tilde{V}_1$ before the signals are combined. For what value of $\phi$ is $P$ independent of $d$?

6. Lloyd's mirror: This is one of the realisations of Young's double slit in the laboratory. Find the condition for a dark fringe at P on the screen from the Figure 10.12. Also find the number of fringes observed on the screen. Assume source wavelength to be $\lambda$.

   Figure 10.12: Lloyd's mirror

7. Calculate the separation between the secondary sources if the primary source is placed at a distance $r$ from the mirror-joint and the tilt angle is $\theta$.

8. Two coherent plane waves with wave vectors and $\vec{k}_1=k [\sin 30^{\circ} \hat{i} + \cos 30^{\circ} \hat{j}]$ with $k=1.2 \times 10^{-6} {\rm m}^{-1}$ are incident on a screen which is perpendicular to the $x$ axis to produce straight line fringes. Determine the spacing between two successive dark lines in the fringe pattern.
9. Starting from a central dark fringe, eigth successive bright and dark fringes are are observed at the center when one of the mirrors of a Michelson interferometer is moved $2.2 \, \mu \, {\rm m}$. Determine the wavelength of the light which is being used. (5.5 A)
10. A Sodium lamp emits light at two neighbouring wavelengths $5890 {\rm A}$ and . A Michelson interferometer is adjusted so that the fringes are in concordance. One of the mirrors is moved a distance $\Delta d$ so that the fringes become discordant and concordant again. For what displacement $\Delta d$ are the fringes most discordant ie. the fringe pattern becomes the faintest, and for what $\Delta d$ does it become concordant again?
11. A Michelson interferometer illuminated by sodium light is adjusted so that the fringes are concordant with a central dark fringe. What is the angular radius of the first dark fringe if the order of the central fringe is $m=100$ and $m=1000$?
12. What happens if a Michelson interferometer is illuminated by white light? Also consider the situation where $d=0$ ie. the two arms have the same length.