Temporal Coherence

The Michelson interferometer measures the temporal coherence of the wave. Here a single wave front $\tilde{E}(t)$ is split into two $\tilde{E}_1 (t)$ and $\tilde{E}_2(t)$ at the beam splitter. This is referred to as division of amplitude. The two waves are then superposed, one of the waves being given an extra time delay $\tau$ through the difference in the arm lengths. The intensity of the fringes is

$$I = \frac{1}{2} \langle [\tilde{E}_1(t)+\tilde{E}_2(t+\tau)] \, [\tilde{E}_1(t)+\tilde{E}_2(t+\tau)]^* \rangle$$ (11.9)

$$= I_1 + I_2 + \frac{1}{2} \langle \tilde{E}_1(t) \, \tilde{E}_2^*(t+\tau) + \tilde{E}_1^*(t) \, \tilde{E}_2(t+\tau)\rangle$$

where it is last term involving $\tilde{E}_1(t) \, \tilde{E}_2^*(t+\tau)...$ which is responsible for interference. In our analysis of the Michelson interferometer in the previous chapter we had assumed that the incident wave is purely monochromatic ie. $\tilde{E}(t)= \tilde{E}e^{i \omega t}$ whereby

$$\frac{1}{2} \langle \tilde{E}_1(t) \, \tilde{E}_2^*(t+\tau) ... ...tilde{E}_2(t+\tau)\rangle = 2 \sqrt{I_1 I_2} \cos(\omega \tau)$$ (11.10)

The above assumption is an idealization that we adopt because it simplifies the analysis. In reality we do not have waves of a single frequency, there is always a finite spread in frequencies. How does this affect eq. 11.10?

As an example let us consider two frequencies $\omega_1=\omega - \Delta \omega/2$ and $\omega_2=\omega + \Delta \omega/2$ with $\Delta \omega \ll \omega$

$$\tilde{E}(t)=\tilde{a}\left[ e^{i \omega_1 t} + e^{i \omega_2 t} \right]\,.$$ (11.11)

This can also be written as

$$\tilde{E}(t)=\tilde{A}(t) e^{i \omega t}$$ (11.12)

which is a wave of angular frequency $\omega$ whose amplitude varies slowly with time. We now consider a more realistic situation where we have many frequencies in the range $\omega - \Delta \omega/2$ to $\omega + \Delta \omega/2$. The resultant will again be of the same form as eq. (11.12) where there is a wave with angular frequency $\omega$ whose amplitude $\tilde{A}(t)$ varies slowly on the timescale

$$T\sim \frac{2 \pi}{\Delta \omega}\,.$$

Figure 11.4: Variation of E with time for monochromatic and polychromatic light

Note that the amplitude $A(t)$ and phase $\phi(t)$ of the complex amplitude $\tilde{A}(t)$ both vary slowly with timescale T. Figure 11.4 shows a situation where $\Delta \omega/\omega=0.2$, a pure sinusoidal wave of the same frequency is shown for comparison. What happens to eq. (11.10) in the presence of a finite spread in frequencies? It now gets modified to

$$\frac{1}{2} \langle \tilde{E}_1(t) \, \tilde{E}_2^*(t+\tau) ... ...\tau)\rangle = 2 \sqrt{I_1 I_2} C_{12}(\tau) \cos(\omega \tau)$$ (11.13)

where $C_{12}(\tau) \le 1$. Here $C_{12}(\tau)$ is the temporal coherence of the two waves $\tilde{E}_1 (t)$ and $\tilde{E}_2(t)$ for a time delay $\tau$. Two waves are perfectly coherent if , partially coherent if $0<C_{12}(\tau)<1$ and incoherent if $C_{12}(\tau)=0$. Typically the coherence time $\tau_c$ of a wave is decided by the spread in frequencies

(11.14)

The waves are coherent for time delays $\tau$ less than $\tau_c$ ie. $C_{12}(\tau) \sim 1$for $\tau < \tau_c$, and the waves are incoherent for larger time delays ie. $C_{12}(\tau) \sim 0$ for $\tau > \tau_c$. Interference will be observed only if $\tau < \tau_c$. The coherence time can be converted to a length-scale $l_c=c \tau_c$ called the coherence length.

An estimate of the frequency spread $\Delta \nu=\Delta \omega/2\pi$ can be made by studying the intensity distribution of a source with respect to frequency. Full width at half maximum (FWHM) of the intensity profile gives a good estimate of the frequency spread.

The Michelson interferometer can be used to measure the temporal coherence $C_{12}(\tau)$. Assuming that $I_1=I_2$, we have $V=C_{12}(\tau)$. Measuring the visibility of the fringes varying $d$ the difference in the arm lengths of a Michelson interferometer gives an estimate of the temporal coherence for $\tau=d/c$. The fringes will have a good contrast $V \sim 1$ only for $d < l_c$. The fringes will be washed away for $d$ values larger than $l_c$.

Problems

1. Consider a situation where Young's double slit experiment is performed using light of wavelength $550 \, {\rm nm}$ and $d=1 \, {\rm m}$. Calculate the visibility assuming a source of angular width $1^{'}$ and . Plot $I(\theta)$ for both these cases.
2. A small aperture of diameter $0.1 \, {\rm mm}$ at a distance of $1 {\rm m}$ is used to illuminate two slits with light of wavelength $\lambda=550 \, {\rm nm}$. The slit separation is $d=1 \, {\rm mm}$. What is the fringe spacing and the expected visibility of the fringe pattern? ( $5.5 \times 10^{-4} \, {\rm rad}$, $V=0.95$)
3. A source of unknown angular extent $\alpha$ emitting light at $\lambda=550 \, {\rm nm}$ is used in a Young's double slit experiment where the slit spacing $d$ can be varied. The visibility is measured for different values of $d$. It is found that the fringes vanish $(V=0)$ for $d=10\, {\rm cm}$. [a.] What is the angular extent of the source? ( $5.5 \times 10^{-6}$)
4. Estimate the coherence time $\tau_c$ and coherence length $l_c$ for the following sources

   Source	$\lambda$ nm	$\Delta \lambda$ nm
   White light	550	300
   Mercury arc	546.1	1.0
   Argon ion gas laser	488	0.06
   Red Cadmium	643.847	0.0007
   Solid state laser	785	$10^{-5}$
   He-Ne laser	632.8	$10^{-6}$

5. Assume that Kr$^{86}$ discharge lamp has roughly the following intensity distribution at various wavelengths, $\lambda$ (in $nm$),
   
   $$I(\lambda)={{36I_0}\over{36+({\lambda-605.616})^2\times 10^8}}~~.$$
   
   
   Estimate the coherence length of Kr$^{86}$ source.(Ans. 0.3m)

6. An ideal Young's double slit (i.e. identical slits with negligible slit width) is illuminated with a source having two wavelengths, $\lambda_1=418.6~ nm$ and $\lambda_2=421.4~ nm$. The intensity at $\lambda_1$ is double of that at $\lambda_2$.

   a) Compare the visibility of fringes near order $m=0$ and near order $m=50$ on the screen [visibility = ${({I_{max} - I_{min}})/({I_{max} + I_{min}})}$].(Ans. 1:0.5)

   b) At what order(s) on the screen visibility of the fringes is poorest and what is this minimum value of the visibility. (Ans. 75, 225 etc. and 1/3)

7. An ideal Young's double slit (separation $d$ between the slits) is illuminated with two identical strong monochromatic point sources of wavelength $\lambda$. The sources are placed symmetrically and far away from the double slit. The angular separation of the sources from the mid point of the double slit is $\theta_s$. Estimate $\theta_s$ so that the visibility of the fringes on the screen is zero. Can one have visibility almost 1 for a non zero $\theta_s$.

   Hint: See the following figure 11.5,

   Figure 11.5: Two source vanishing visibility condition

   (Further reading: Michelson's stellar interferometer for estimating angular separation of double stars and diameters of distant stars)