NLS-Glossary

is a thin dielectric layer coated on an optical device or component to reduce the reflection of light and increase the transmitted light intensity. Consider a thin layer of a dielectric material such as $S{i}_3{N}_4$ (silicon nitride) on the surface of a semiconductor optoelectronic device such as a solar cell. If this antireflection coating has an intermediate refractive index then the thin dielectric coating can reduces the reflected light intensity. In this case n1(air) = 1, n2(coating) ≈ 1.9 and n3(Si) = 3.5. Light is first incident on the air/coating surface and some of it becomes reflected. Suppose that this reflected wave is A. Wave A has experienced a 180° phase change on reflection as this is an external reflection. The wave that enters and travels in the coating then becomes reflected at the coating/semiconductor surface. This wave, say B, also suffers a 180° phase change since n3 > n2. When wave B reaches A, it has suffered a total delay of traversing the thickness d of the coating twice. The phase difference is equivalent to $K_{c }\left(2d\right)$ where $K_c=2\pi /{\lambda }_c$ is the wavevector in the coating and is given by $2\pi /{\lambda }_c$ where ${\lambda }_c$ c is the wavelength in the coating. Since ${\lambda }_c = \lambda /n_2$ where λ is the free-space wavelength, the phase difference Δφ between A and B is ( $2\pi n_2/\lambda$ )(2d). To reduce the reflected light, A and B must interfere destructively and this requires the phase difference to be π or oddmultiples of π, mπ where m = 1,3,5,… is an odd-integer. Thus $m\pi =2d \left(\frac{2\pi n_2}{\lambda }\right)$, $d=m \left(\frac{\lambda }{4n_2}\right)$- Thus, the thickness of the coating must be multiples of the quarter wavelength in the coating and depends on the wavelength. To obtain a good degree of destructive interference between waves A and B, the two amplitudes must be comparable. It turns out that we need $n2=\sqrt{\left(n1n3\right)}$ When $n2=\sqrt{\left(n1n3\right)}$ then the reflection coefficient between the air and coating is equal to that between the coating and the semiconductor. In this case we would need $\sqrt{(3.5)}$ or 1.87. Thus, $S{i}_3{N}_4$ is a good choice as an antireflection coating material on Si solar cells. Generally an AR coating operates at one or over a narrow range of wavelengths