If you read the QED from Feynman he says that it is possible to get a 100% transmission of light for a specific thickness of glass. He explains the phenomenon perfectly but I don't know how to calculate the thickness of glass for a given frequency of light.

Just assume that you send photons of the same color perpendicular to a plane of glass with a thickness $d$. To get zero reflection, light being reflected from the both surfaces of glass must cancel out because of the interference.

How can I calculate the thickness of the glass for zero reflection?


1 Answer 1


As you identified, this is an elementary problem in interferometry. You may find a solution in most textbooks studying the Fabry-Perot etalon. One such book is the classic "Principles of Optics" by Born and Wolf. I'll reproduce the treatment if you do not have access to the book.

Consider an electric planewave impinging normally upon a thin slab of glass of thickness $d$. Upon propagating a round-trip on the slab, the planewave acquires a phase $\delta$ given by

$$\delta = 2nk_0d,$$

where $k_0$ is the wavenumber in vacuum and $n$ is the glass's refractive index. Let $r$ and $t$ be the reflection and transmission coefficients from the medium of incidence (vacuum or air) to the glass, and let $r'$, $t'$ be the reflection and transmission coefficients from the glass back to the medium of incidence.

If the incident electric field has amplitude $E_i$ at the air-glass boundary, it is straightforward to see that the amplitude of the wave reflected without traveling in the glass is $rE_i$. The amplitude of the wave returned by the glass ("reflected") after one round-trip is $tt'r'\exp(i\delta)E_i$. The one returned after two round-trips is $tt'(r')^3\exp(2i\delta)$. The one returned after $n$ round-trips is $tt'(r')^{2n-1}\exp(in\delta)E_i$. Thus, to find the total amplitude of the reflected wave $E_r$, we need to sum all these contributions. Hence,

$$ E_r = E_i\left(r + tt'\sum_{n = 1}^\infty(r')^{2n-1}\exp(in\delta)\right)\\ = E_i\left(r + \frac{tt'r'\exp(i\delta)}{1 - r'^2\exp(i\delta)}\right). $$

The last part used the well-known formula for an infinite geometric series.

From Frensel's formulae, we find that

$$r = -r',\\ r^2 = r'^2 = R,\\ tt' = T,\\ R + T = 1, $$ for a slab of a dielectric material such as glass. Hence, the reflected amplitude can be simplified to

$$ E_r = E_i\frac{1 - \exp(i\delta)\sqrt{R}}{1 - R\exp(i\delta)}. $$

To find an expression for the reflected intensity, we use $I \propto |E|^2$ and find

$$ I_r = \frac{2R\sin^2(\delta/2)}{(1 - R)^2 + 4R\sin^2(\delta/2)}I_i. $$ Hence, the reflected intensity is zero when $\delta/2$ is one of the zeros of the $\sin$-function.

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    $\begingroup$ Note this only describes losses due to surface reflections there will still be finite amount of absorption from the actual bulk glass and scattering from any imperfections $\endgroup$ Oct 28, 2017 at 2:48

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