Mumbai university > FE > SEM 2 > Applied Physics II

Marks: 8M

Year: Dec 2013

An anti-reflection coating is applied in order to reduce the intensity of reflected light and increase the transmission.

It is done by using the phenomenon of destructive interference. In order to achieve this, a thin film of refractive index less than that of glass on which it is applied, is deposited.

Let $\bar{λ_m}$ be the mean wavelength of incident light rays, in the medium.

As both the rays BC and DE, are reflected from a denser medium, they both have a phase change of π, thus not contributing to path difference.

$$Δ = (2n -1) \frac{λ}{2}$$

In order to get a destructive interference, path difference,

For Normal incidence,

$$Δ = 2 × t = \frac{\bar{λ_m}}{2}$$ (For destructive interference)

$$t = \frac{\bar{λ_m}}{4}$$

If $\bar{λ_a}$ is the mean wavelength in air or vacuum,

$$μ = \frac{\bar{λ_a}}{λ_m}$$

$$t = \frac{\bar{λ_a}}{4μ}$$

Hence, in order to get a non-reflective coating, the thickness of coating should be,

$$t = \frac{\bar{λ_a}}{4μ}$$ OR $$t = \frac{\bar{λ_m}}{4}$$