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Spatial Coherence

The Young's double slit experiment (Figure 11.1) essentially measures the spatial coherence. The wave $\tilde{E}(t)$ at the point P on the screen is the superposition of $ \tilde{E}_1 (t) $ and $\tilde{E}_2(t) $ the contributions from slits $1$ and $2$ respectively. Let us now shift our attention to the values of the electric field $ \tilde{E}_1 (t) $ and $\tilde{E}_2(t) $ at the positions of the two slits. We define the spatial coherence of the electric field at the two slit positions as
\begin{displaymath}
C_{12}(d)=\frac{\frac{1}{2}\, \langle \tilde{E}_1(t) \tilde{...
...)+ \tilde{E}_1^*(t)
\tilde{E}_2(t) \rangle }{ 2\sqrt{I_1 I_2}}
\end{displaymath} (11.1)

Figure 11.1: Young's double slit with a point source
\begin{figure}\begin{center}
\epsfig{file=chapt11//yy.eps,height=1.2in} \end{center}\end{figure}

The waves from the two slits pick up different phases along the path from the slits to the screen. The resulting intensity pattern on the screen can be written as

\begin{displaymath}
I=I_1 + I_2 + 2 \sqrt{I_1 I_2} C_{12}(d) \cos(\phi_2-\phi_1)
\end{displaymath} (11.2)

where $\phi_2-\phi_1$ is the phase difference in the path from the two slits to the screen. The term $\cos(\phi_2 -\phi_1)$ gives rise to a fringe pattern.

The fringe visibility defined as

\begin{displaymath}
V=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}
\end{displaymath} (11.3)

quantifies the contrast of the fringes produced on the screen. It has values in the range $1 \le V \le 0$. A value $V=1$ implies very high contrast fringes, the fringes are washed away when $V=0$.
Figure 11.2: Fringe intensity for different visibilities
\begin{figure}\begin{center}
\epsfig{file=chapt11//fringe.eps,height=1.8in} \end{center}\end{figure}
Figure 11.2 shows the fringe pattern for different values of $V$. It can be easily checked that the visibility is related to the spatial coherence as
\begin{displaymath}
V=\frac{2 \sqrt{I_1 I_2}\mid C_{12}(d)\mid}{I_1+I_2}
\end{displaymath} (11.4)

and the visibility directly gives the spatial coherence $V=\mid
C_{12}(d)\mid$ when $I_1=I_2$.

Let us first consider the situation when the two slits are illuminated by a distant point source as shown in Figure 11.1. Here the two slits lie on the same wavefront, and $\tilde{E}_1(t)=\tilde{E}_2(t)$. We then have

\begin{displaymath}
\frac{1}{2} \, \langle \tilde{E}_1(t) \tilde{E}_2^*(t) \rang...
...langle \tilde{E}_1^*(t) \tilde{E}_2(t) \rangle
= I_1 = I_2 \,.
\end{displaymath} (11.5)

whereby $C_{12}(d)=1$ and the fringes have a visibility $V=1$.

Figure 11.3: Double slit with a wide source
\begin{figure}\epsfig{file=chapt11//youngs_3.eps,height=1.15in} \end{figure}
We next consider the effect of a finite source size. It is assumed that the source subtends an angle $\alpha$ as shown in Figure 11.3. This situation can be analyzed by first considering a source at an angle $\beta $ as shown in the figure. This produces an intensity
\begin{displaymath}
I ( \theta, \beta)= 2 I_1 \left[ 1+ \cos \left( \frac{2 \pi d}{\lambda} (
\theta + \beta ) \right) \right]
\end{displaymath} (11.6)

at a point at an angle $\theta$ on the screen where it is assumed that $\theta,\beta \ll 1$. Integrating $\beta $ over the angular extent of the source
$\displaystyle I (\theta)$ $\textstyle =$ $\displaystyle \frac{1}{ \alpha}\int \limits^{ \alpha/2}_{- \
\alpha/2} I(\theta, \beta) \, d \beta$  
  $\textstyle =$ $\displaystyle 2 I_1 \left[ 1 + \frac{\lambda }{ \alpha 2 \pi d } \left\{
\sin \...
...i d}{\lambda} \left(
\theta - \frac{ \alpha}{2} \right)\right] \right\} \right]$  
  $\textstyle =$ $\displaystyle 2 I_1 \left[ 1 + \frac{\lambda}{\pi d \alpha}
\cos \left( \frac{2...
...ta }{\lambda} \right) \sin \left( \frac{\pi
d \alpha}{\lambda } \right) \right]$ (11.7)

It is straightforward to calculate the spatial coherence by comparing eq. (11.7) with eq. (11.2). This has a value

\begin{displaymath}
C_{12}(d)=\sin\left(\frac{\pi d \alpha}{\lambda} \right) \big{ /}
\left(\frac{\pi d \alpha}{\lambda} \right)
\end{displaymath} (11.8)

and the visibility is $V=\mid
C_{12}(d)\mid$. Thus we see that the visibility which quantifies the fringe contrast in the Young's double slit experiment gives a direct estimate of the spatial coherence. The visibility, or equivalently the spatial coherence goes down if the angular extent of the source is increased. It is interesting to note that the visibility becomes exactly zero when the argument of the Sine term the expression (11.8) becomes integral multiple of $\pi$. So when the width of the source is equal to the visibility is zero.

Why does the fringe contrast go down if the angular extent of the source is increased? This occurs because the two slits are no longer illuminated by a single wavefront, There now are many different wavefronts incident on the slits, one from each point on the source. As a consequence the electric fields at the two slits are no longer perfectly coherent $\mid C_{12}(d) \mid <1$ and the fringe contrast is reduced.

Expression (11.8) shows how the Young's double slit experiment can be used to determine the angular extent of sources. For example consider a situation where the experiment is done with starlight. The variation of the visibility $V$ or equivalently the spatial coherence $C_{12}(d)$ with varying slit separation $d$ is governed by eq. (11.8). Measurements of the visibility as a function of $d$ can be used to determine $\alpha$ the angular extent of the star.


next up previous contents
Next: Temporal Coherence Up: Coherence Previous: Coherence   Contents
Physics 1st Year 2009-01-06