Answer to Question #268959 in Optics for sai

Question

A Young’s-double-slit experiment is performed with two wavelengths - 400 and 800

nm. Assume that both sources are perfectly coherent with each other and have

equal intensity. Calculate the light-intensity pattern on a screen placed 1 m from

the slits, which are separated by a distance of 1 cm.

Answer 1

Consider two plane waves with different frequencies:

$\vec E_1=\vec E_{10}\exp i{(\omega_1t+\vec k_1\vec r_1+\phi_1)}$

$\vec E_2=\vec E_{20}\exp i{(\omega_2t+\vec k_2\vec r_2+\phi_2)}$

$\vec E=\vec E_1+\vec E_2$

$I=<\vec E>_{\tau}$

...

So, we will get

$I=I_1+I_2+2\vec E_{10}\vec E_{20}\cos[\Delta\omega(t_0+\tau/2)+\Delta\vec k \vec r+\Delta\phi]\cdot\sin c\Delta\omega\tau/2$ ,

where

$\Delta\omega=\omega_1-\omega_2$

$\Delta\vec k\vec r=\vec k_1\vec r_1-\vec k_2\vec r_2$

$\Delta\phi=\phi_1-\phi_2$