Question

1. Suppose two plane waves of light with electric field amplitudes E1 and E2 arrive at a detector at time t having travelled different distances z1 and z2. Assume the field vectors have the same polarization direction p. The fields of the two plane waves at the detector individually are Eipcos(ot - kzi) and E.pcos(at-kz2 ) . Here the wavenumber is k-2πίλ and the angular frequency is ω ck-2πο:/v , where λ is the wavelength, с is the speed, and v is the frequency of the light. (a) What is are the time-averaged intensities /1 and /2 of the two plane waves? (b) If the two waves are coherent, we find the total field by summing the two fields. What is an expression for the time average of the intensity Lotal of the resulting wave field? (c) Sketch a plot of the total intensity as a function of the difference in distances z2 z (d) In the plot, what are expressions for the maximum ll and minimum Iotavalues and what is the visibility Vis Ξ (1 total,nax-I totalmun )/(1 totalmıax + 1 totalmi") ?

Answer 1

Given informations:

$\vec{E}_{2} = E_{2} \ cos(\omega t-kz_{2}) \ \hat{p}$

(a) What is the time-averaged intensites of I₁ and I₂ of two plane waves?

Time average of any function f(t) an be given by

$<f(t)>_{T} = \frac{1}{T}\int_{0}^{T} f({t}') \ d{t}'$

Now, let us first see the expression for the intensity I in general from the electric field E which is given by

   $I = \vec{E}.\vec{E} = E^{2}$

Then the time averaged intensity is given by

$<I>_{T} \ = \frac{1}{T}\int_{0}^{T} E^{2}({t}') \ d{t}'$

Now, the time-averaged intensity I₁ is given by

$<I_{1}>_{T} \ =\frac{1}{T} \int_{0}^{T} E_{1}^{2} \ cos^{2}(\omega {t}' - kz_{1}) \ d{t}'$

$=\frac{1}{T} \ E_{1}^{2} \int_{0}^{T} cos^{2}(\omega {t}' - kz_{1}) \ d{t}'$

$=\frac{1}{T} \ E_{1}^{2} \int_{0}^{T}\frac{ [1 +cos \ 2(\omega {t}' - kz_{1})]}{2}\ d{t}'$

$=\frac{E_{1}^{2}}{2T} \ \left \{ \right.\int_{0}^{T} d{t}' + \ \int_{0}^{T} cos \ 2(\omega {t}' )- kz_{1})]\ d{t}'\left. \right \}$

$=\frac{E_{1}^{2}}{2T} \left \{ T \ + \ 0 \right \}$

${\color{Red} <I_{1}>_{T} \ =\ \frac{E_{1}^{2}}{2}} \ \ \ \ - - - - \ ANSWER$

Similarlly, the time-averaged intensity I₂ is given by

      ${\color{Red} <I_{2}>_{T} \ =\ \frac{E_{2}^{2}}{2}} \ \ \ \ - - - - \ ANSWER$

(b) Expression for the resultant field E_tot is given by

$\vec{E}_{total} \ = \ \vec{E}_{1} \ + \ \vec{E}_{2}$

Expression for the resultant intensity I_total is given by

   $I_{total} \ = \ (\vec{E}_{1} \ + \ \vec{E}_{2}).(\vec{E}_{1} \ + \ \vec{E}_{2})$

$= \ \vec{E}_{1}.\vec{E}_{1} \ + \ \vec{E}_{2}.\vec{E}_{2} \ + \ 2\vec{E}_{1}.\vec{E}_{2}$

Now, the expression for the time-averaged resultant intensity is given by

  

  

  

Now we only need to find 2E₁E₂ as shown below

   $2E_{1}E_{2} = 2 \ E_{1} \ cos(\omega t-kz_{1}) \ E_{2} \ cos(\omega t-kz_{2})$

$= 2 \ E_{1} \ E_{2}\ cos(\omega t-kz_{1}) \ \ cos(\omega t-kz_{2})$

$=\ 2\ E_{1} \ E_{2} \ \left [ cos(2\omega t - k(z_{1}+z_{2})) \ + \ cos(k(z_{2}-z_{1}))\right ]$

Now, its time average can be given as

$<2 E_{1} E_{2}>_{T}\ =\ \frac{1}{T} \ \int_{0}^{T}2\ E_{1} \ E_{2} \ \left [ cos(2\omega {t}' - k(z_{1}+z_{2})) \ + \ cos(k(z_{2}-z_{1}))\right ] \ d{t}'$

  

Therefore,

   ${\color{Red} <I_{total}>_{T}\ =} \ {\color{Red} \frac{E_{1}^{2}}{2} \ + \ \frac{E_{1}^{2}}{2} \ + \ E_{1}E_{2} \ cos(k(z_{2}-z_{1}))} \ \ \ \ \ ----- ANSWER$

d)

$<I_{total}>^{max} \ = \ \frac{E_{1}^{2}}{2} \ + \ \frac{E_{2}^{2}}{2} \ + \ E_{1}\ E_{2}$

$<I_{total}>^{min} \ = \ \frac{E_{1}^{2}}{2} \ + \ \frac{E_{2}^{2}}{2} \ - \ E_{1}\ E_{2}$

Now,

$Vis \ = \ \frac{<I_{total}>^{max} - <I_{total}>^{min}}{<I_{total}>^{max} + <I_{total}>^{min}}$

          ${\color{Red} Vis \ = \ \frac{2\ E_{1}\ E_{2}}{E_{1}^{2} \ + \ E_{2}^{2}}} \ \ \ \ \ - - - - - ANSWER$

c)