I have a monochromatic light source (wavelength ~ 420 nm), which will be incident on the interface of two different media. Could someone please explain if the Fresnel equations applies with monochromatic light when estimating the reflectance and transmittance?

Fresnel equation:

\begin{align} R_s &= \left| \frac{n_1 \cos(\Theta_\mathrm{in}) - n_2 \cos(\Theta_\mathrm{out})}{n_1 \cos(\Theta_\mathrm{in}) + n_2 \cos(\Theta_\mathrm{out})} \right|^2 = \left| \frac{n_1 \cos(\Theta_\mathrm{in}) - n_2\sqrt{1-\left(\frac{n_1}{n_2}\sin(\Theta_\mathrm{in})\right)}}{n_1 \cos(\Theta_\mathrm{in}) + n_2 \sqrt{1-\left(\frac{n_1}{n_2}\sin(\Theta_\mathrm{in})\right)}} \right|^2 \\ R_p &= \left| \frac{n_1 \cos(\Theta_\mathrm{out}) - n_2 \cos(\Theta_\mathrm{in})}{n_1 \cos(\Theta_\mathrm{out}) + n_2 \cos(\Theta_\mathrm{in})} \right|^2 = \left| \frac{n_1 \sqrt{1-\left(\frac{n_1}{n_2}\sin(\Theta_\mathrm{in})\right)} - n_2 \cos(\Theta_\mathrm{in})}{n_1 \sqrt{1-\left(\frac{n_1}{n_2}\sin(\Theta_\mathrm{in})\right)} + n_2 \cos(\Theta_\mathrm{in})} \right|^2 \end{align}

  • $\begingroup$ Please do not post formulae as screenshots, but use MathJax instead. I have transcribed your images for now (including a mistake in a missing square), but you should use direct MathJax typesetting for future posts. $\endgroup$ Sep 3, 2019 at 14:46
  • $\begingroup$ I've also removed your final paragraph, as it asks a separate question -- threads here should contain one question per question (or, at most, a tightly-related set of queries). If you're still curious, post a separate thread. $\endgroup$ Sep 3, 2019 at 14:48

2 Answers 2


Monochromatic simply means that the light is of constant frequency. The frequency only affects the value of the refractive index for dispersive media. It has no affect on the formulae for the Fresnel coefficients. If your monochromatic light is unpolarized, then you need to average over all possible polarizations to properly estimate the reflectance.

  • $\begingroup$ It’s a long way of saying yes, but cannot fault your logic ;) $\endgroup$
    – boyfarrell
    Aug 27, 2019 at 21:48
  • $\begingroup$ Thank you for your explanations, now I got a better view of the case I am dealing with. $\endgroup$
    – Xin
    Aug 28, 2019 at 7:52
  • $\begingroup$ BTW, when a constant monochromatic light (λ≈420 nm) passes through a ø 2.5 mm and 15 mm collimator to produce parallel light beams, is this light unpolarized? $\endgroup$
    – Xin
    Aug 28, 2019 at 8:00
  • $\begingroup$ Depends what type of light source you are using. It sounds like you are using some kind of blue or green laser source, which may be emitting partially polarized light. To achieve linear polarization, you need to use a polarizer after the collimator as well. $\endgroup$ Aug 28, 2019 at 8:22

Yes. Fresnel equation as written as valid for some fixed wavelength since the index of refraction $n$ is usually wavelength-dependent. If your source is not monochromatic, you need to compute $R_\lambda$ for each wavelength (using the appropriate $n$ is each case), multiply each $R_\lambda$ by its proportion in the incident beam, and sum over all $\lambda$ (or integrate if your distribution is continuous).

Usually $n$ will vary slowly with $\lambda$ most of the time. This results in dispersion, as illustrated (presumably for glass) by this iconic album cover:

enter image description here

  • 1
    $\begingroup$ (+1) for the answer and the iconic album cover! ;-) $\endgroup$
    – Ed V
    Nov 29, 2021 at 1:55

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