A monochromatic plane wave is incident normally from a dielectric medium $A$ onto another dielectric medium $B$. The indices of refraction satisfy $n_A < n_B$. One-fourth of the incident energy is reflected back into medium $A$. Let $E$ be the resultant electric field due to the superposition of the incident wave and the reflected wave. Then, the ratio of the two time-averages $\langle \vec{E}^2 \rangle_{\text{min}}/\langle \vec{E}^2 \rangle_{\text{max}}$ is
Dielectric Interface Wave Reflection Analysis
This problem involves a monochromatic plane wave normally incident on a dielectric interface. We are given the energy reflection coefficient and asked to find the ratio of minimum to maximum time-averaged squared electric field in the first medium.
Energy Reflection Coefficient
The problem states that one-fourth of the incident energy is reflected back into medium $A$. This gives the energy reflection coefficient, $R_E$:
$ R_E = \frac{1}{4} $
Amplitude Reflection Coefficient
The amplitude reflection coefficient, $r$, is related to the energy reflection coefficient by $R_E = |r|^2$. Therefore:
$ |r| = \sqrt{R_E} = \sqrt{\frac{1}{4}} = \frac{1}{2} $
Resultant Electric Field Amplitude
In medium $A$, the resultant electric field $E$ is the superposition of the incident electric field ($E_i$) and the reflected electric field ($E_r$). Let the amplitude of the incident wave be $E_0$. The amplitude of the reflected wave is $|r|E_0$. The resultant amplitude varies spatially due to interference:
- Maximum Amplitude ($E_{\text{max\_amp}}$): Occurs during constructive interference.
$ E_{\text{max\_amp}} = E_0 + |r|E_0 = E_0 + \frac{1}{2}E_0 = \frac{3}{2}E_0 $
- Minimum Amplitude ($E_{\text{min\_amp}}$): Occurs during destructive interference.
$ E_{\text{min\_amp}} = |E_0 - |r|E_0| = |E_0 - \frac{1}{2}E_0| = \frac{1}{2}E_0 $
Time-Averaged Squared Electric Field Ratio
The time-averaged squared electric field, $\langle \vec{E}^2 \rangle$, is proportional to the square of the electric field amplitude ($E_{\text{amp}}^2$).
- $ \langle \vec{E}^2 \rangle_{\text{max}} \propto (E_{\text{max\_amp}})^2 = \left(\frac{3}{2}E_0\right)^2 = \frac{9}{4}E_0^2 $
- $ \langle \vec{E}^2 \rangle_{\text{min}} \propto (E_{\text{min\_amp}})^2 = \left(\frac{1}{2}E_0\right)^2 = \frac{1}{4}E_0^2 $
The ratio is:
$ \frac{\langle \vec{E}^2 \rangle_{\text{min}}}{\langle \vec{E}^2 \rangle_{\text{max}}} = \frac{\frac{1}{4}E_0^2}{\frac{9}{4}E_0^2} = \frac{1}{9} $
Important Questions from Reflection and Refraction
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