Question #268959

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.


1
Expert's answer
2021-11-21T17:27:10-0500

Consider two plane waves with different frequencies:


E1=E10expi(ω1t+k1r1+ϕ1)\vec E_1=\vec E_{10}\exp i{(\omega_1t+\vec k_1\vec r_1+\phi_1)}

E2=E20expi(ω2t+k2r2+ϕ2)\vec E_2=\vec E_{20}\exp i{(\omega_2t+\vec k_2\vec r_2+\phi_2)}


E=E1+E2\vec E=\vec E_1+\vec E_2


I=<E>τI=<\vec E>_{\tau}

...


So, we will get


I=I1+I2+2E10E20cos[Δω(t0+τ/2)+Δkr+Δϕ]sincΔωτ/2I=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


Δω=ω1ω2\Delta\omega=\omega_1-\omega_2

Δkr=k1r1k2r2\Delta\vec k\vec r=\vec k_1\vec r_1-\vec k_2\vec r_2

Δϕ=ϕ1ϕ2\Delta\phi=\phi_1-\phi_2









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