# When are the reflectance R and the transmittance T equal to each other at normal incidence?

Really struggling with this one.

Under what conditions would the reflectance R and the transmittance T be equal to each other at normal incidence at an interface? What would be the values of R and T under these conditions?

So

$R= (n1-n2)^2/(n1+n2)^2$

$T=4n1n2/(n1+n2)^2$

Where $n1$ and $n2$ are the refractive indices either side of the interface.

No attenuation, satisfying the conservation of energy across a boundary of infinitesimal distance, $R+T=1$

$R=T=1/2$

Setting $R=T$ gives the relation $n1=$$3 (n2) + 2 sqrt(2)(n2$)

This is where I'm stuck. I don't find this answer satisfying, as it is only valid for refractive indices well beyond the usual range.

What am I missing?

Your math seems correct; for other solutions including complex values for $n_1$ and $n_2$, see e.g. Wolfram Alpha.
The discrepancy relative to values of n for everyday materials is because R is normally much smaller. For example, most glass-air surfaces have $R\approx 0.04$. For $R=0.5$ you would need non-everyday materials.
• you have $n_1 = 3 n_2 + 2 \sqrt{2} n_2$ which is one equation in two unknowns, $n_1$ and $n_2$. If you take $n_2 = 1$ (so that the medium 2 is a vacuum) then this gives $n_1 \approx. 5.828$. Looking at en.wikipedia.org/wiki/List_of_refractive_indices shows a list of materials with refractive indices. The largest value has $n \approx 4$ for germanium for $\lambda \approx 3000 - 16000$ nm (refractive indices depend on frequency). – jim Jun 18 '17 at 21:58