This is related to the question how can electromagnetic waves reach a cell phone in faraday cage?, where in the answer it was stated that the holes (=size of the mesh) would need to be smaller than that of the wavelengths of the EM radiation used for cell phone signals for attenuation to 'start'.

Also, in the question Should a Faraday cage block a radio's signal? where the focus of the question is on AM radio frequencies (which I know are not the same used by cell phones), but the idea was using baking trays, effectively making a Faraday 'box'. However, it lead to the question Building a Faraday cage for mobile networks that is transparent for optical wavelengths, which looked at blocking cell phone signals using a cage, but still allowing optical wavelengths in.

However, I am seeking to know the answer to is there is a relationship between a grounded Faraday cage mesh hole size (relative to the cellphone signal used) and cell phone reception attenuation?

  • $\begingroup$ I'm pretty sure this is a reasonably straight forward scattering problem. Perhaps try solving the problem for a single hole first and then apply the linearity of E&M to understand a grid. $\endgroup$
    – DanielSank
    Commented Dec 22, 2014 at 22:57
  • 2
    $\begingroup$ Its got nothing to do with whether the cage is grounded. The fact that there is a 20 metre cable going to an earthing plug somewhere cannot possibly affect the interactions between an EM wave a couple of centimetres across and a hole in the mesh of comparable size. Mirrors don't have to be grounded to work, and if the mesh size is much smaller than the wavelength the Faraday cage is in fact a mirror - it reflects the EM radiation back. $\endgroup$
    – Peter Webb
    Commented Mar 2, 2015 at 13:08

1 Answer 1


The most important concept relating Faraday cage hole size to cell phone signal attenuation is the idea of a cutoff frequency. For round holes, you would model them as cylindrical waveguides. For simplicity, we'll consider rectangular waveguides instead.

Matching the boundary conditions at the metal wall, you get so called transverse electric (TE) and transverse magnetic (TM) modes. These look like partially standing waves, with a traveling wave component for the third. For TE modes, they're of the form (for polarization in the y direction):

$$E = E_0 \sin(k_x x) \cos(k_y y) e^{i(k_z z-\omega t)}$$

There's a multitude of standing wave modes. These are described by different values of $k_x$ and $k_y$, which are solved by setting the above expression to zero at the walls of the waveguide (for the sine portion), or derivative zero (for the cosine portion). The solutions:

$$k_x = \frac{m\pi}{w}$$ $$k_y = \frac{n\pi}{h}$$

Where $w$ and $h$ are the width and height of the waveguide, and $m$ and $n$ are integers. Substituting the above expression into the wave equation, we get the relation between the different $k$ components and frequency.

$$\left( \frac{\omega}{c} \right)^2 = {k_x}^2+{k_y}^2+{k_z}^2 $$

The lowest possible such frequency is when $k_z = 0$ and $m=1,n=0$

$$f_c= \frac{c}{2w}$$

This is the cutoff frequency. Below this frequency, the signal exponentially decays as it propagates through the structure. To show this, solve for $k_z$, and write it in terms of cutoff frequency.

$$k_z = \frac{2 \pi}{c} \sqrt{f^2-{f_c}^2}$$

Evidently, at frequencies below cutoff, $k_z$ becomes imaginary. Substituting this into our travelling wave expression, it becomes exponential decay.

$$\alpha := \frac{2 \pi}{c} \sqrt{{f_c}^2-f^2}, f < f_c$$

$$E = E_0 \sin(k_x x) \cos(k_y y) e^{-\alpha z-i\omega t}$$

Notice that for our rectangular waveguide, the cutoff frequency depended only on the width. In general,

$$\lambda_c \approx \textrm{largest feature size}*2 $$

(This is exactly true for a rectangular waveguide, and should hold approximately for other shapes.)

For Faraday cages with openings in the size of centimeters, the cutoff frequency is around 20GHz, which is quite large compared cell phone signals in the range of 2GHz. We can approximate the decay constant $\alpha$

$$\alpha \simeq \frac{2 \pi f_c}{c} = \frac{2\pi}{\lambda _c} $$

To figure out the amount of decay, we need to assume some length $l$ to the opening (equivalently, thickness of the cage material), and then substitute $l$ for $z$ in the wave expression. Converting this to a decibel scale, we get the following power loss:

$$\frac{l}{\textrm{largest feature size}} (48.6 \textrm{dB})$$

Another important point is that the signal will mostly negatively interfere with itself in the cage, except at a few spots within the cage where it is effectively amplified (probably the center). If the cage features are fairly large, you might be able to notice this signal hotspot.

Edit: There are also complex effects where the fields in one hole can induce fields in another hole. The above analysis is a simplified description of a complex field problem, but I expect the general principles to hold.

  • $\begingroup$ Dave, what you pointing out in your answer is the interaction between the edges respectively the tubes of the holes with a radio wave "There are also complex effects where the fields in one hole can induce fields in another hole." Why not to take this observation as an explanation for the deflection of photons from slits and edges? As you pointed out the modes are discrete. For EM radiation this led to an intensity distribution behind any edge. Instead of EM radiation this has to be true also for a beam of electrons. $\endgroup$ Commented Nov 18, 2015 at 6:20