Worked example 6.3.1: Antireflection coating of a silicon solar cell

$R=(\frac{3.9-1}{3.9+1}{)}^2=0.350$: $35\%$ of the light is lost.

Zero reflection needs $n_c=\sqrt{1\times 3.9}=1.975$, and nitride ($2.0$) is close. The thickness is a quarter of the wavelength in the layer:

$n_{\mathrm{in}}=n_c^2/n_s=4/3.9=1.026$, so

At $400$ nm ${\beta }_{c}d=3\pi /4$ ($\tan {\beta }_{c}d=-1$), at $1000$ nm ${\beta }_{c}d=0.3\pi$ ($\tan {\beta }_{c}d=1.376$). In units of ${\eta }_0$, ${\eta }_c=0.5$, ${\eta }_s=0.2564$, and

The transmitted fraction $1-R$ rises from $0.650$ to $0.9998$, a factor $1.54$.

As ${\beta }_{c}d$ moves from $\pi /2$ towards $0$ (long wavelengths) or towards $\pi$ (a half wave, $300$ nm), $R$ grows steadily from $0.016\%$ to the bare $35\%$, where the layer disappears. Both values in (d) lie in between $✓$ (Figure 6.3.21).