The permittivity of an ionic crystal is \\(\\epsilon(\\omega)=1+\\omega_{p}^{2}/

Question

The permittivity of an ionic crystal is      (epsilon(omega)=1+omega_{p}^{2}/(omega_{o}^{2}-omega^{2}))    where      (omega_{o}=0.05eV)    is the optical phonon frequency and      (omega_{p}=0.1eV)    is the plasma frequency. a) Plot the reflection coefficient      (R(omega))    for the electromagnetic radiation incident on the crystal. b) Identify the frequency band where the crystal is opaque (R=100%). Give the values of the bound frequencies      (omega_{1})    and      (omega_{2})    c) Show that the reflection coefficient for static electric fields      ((omega=0))    can be written as      (R(0)=[(omega_{2}-omega_{1})/(omega_{2}+omega_{1})]^{2})     

Explanation / Answer

?

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.