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.
Consider two plane waves with different frequencies:
E⃗1=E⃗10expi(ω1t+k⃗1r⃗1+ϕ1)\vec E_1=\vec E_{10}\exp i{(\omega_1t+\vec k_1\vec r_1+\phi_1)}E1=E10expi(ω1t+k1r1+ϕ1)
E⃗2=E⃗20expi(ω2t+k⃗2r⃗2+ϕ2)\vec E_2=\vec E_{20}\exp i{(\omega_2t+\vec k_2\vec r_2+\phi_2)}E2=E20expi(ω2t+k2r2+ϕ2)
E⃗=E⃗1+E⃗2\vec E=\vec E_1+\vec E_2E=E1+E2
I=<E⃗>τI=<\vec E>_{\tau}I=<E>τ
...
So, we will get
I=I1+I2+2E⃗10E⃗20cos[Δω(t0+τ/2)+Δk⃗r⃗+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/2I=I1+I2+2E10E20cos[Δω(t0+τ/2)+Δkr+Δϕ]⋅sincΔωτ/2 ,
where
Δω=ω1−ω2\Delta\omega=\omega_1-\omega_2Δω=ω1−ω2
Δk⃗r⃗=k⃗1r⃗1−k⃗2r⃗2\Delta\vec k\vec r=\vec k_1\vec r_1-\vec k_2\vec r_2Δkr=k1r1−k2r2
Δϕ=ϕ1−ϕ2\Delta\phi=\phi_1-\phi_2Δϕ=ϕ1−ϕ2
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