# How come a mobile phone signal is blocked by aluminium foil, but Wi-Fi gets through?

Major puzzle for me. I have done a demonstration in lectures for years now, where I show that a mobile phone call can be blocked by wrapping the phone in a single sheet of standard Aluminium kitchen foil.

Today, I showed that whilst the above is still true, that if the phone is enabled to use Wi-Fi to receive calls, then the call gets through!

I believe that the phone signals are operating at about 800 Mhz, whilst the Wi-Fi might be operating at frequencies as high as 5 GHz (this in the UK).

I would have thought that the higher frequency of the Wi-Fi would lead to less reflection of the signal (since impedance $$\propto \sqrt{f}$$), but the skin depth would be decreased by the same factor, leading to much greater attenuation of any signal that makes it into the foil. In both cases, I think the total power attenuation factors should be at least $$10^{13}$$.

Does anybody have any ideas (or experiments I could perform) to figure out why Wi-Fi gets through but mobile signals don't?

• How do you know that this is a difference in attenuation? Wifi is a digital protocol that has error correction. The noise background could be different in these two frequency ranges. The antennas might be different types.
– user4552
Dec 11 '19 at 0:31
• It is all about the receiver sensitivity in the phone itself. You cannot shield against wifi because the receiver is so sensitive in the cellphone. The receiver for the regular radiosignal 4G is much less sensitive. Dec 11 '19 at 0:38
• Related to but not a duplicate of Why does my cell phone still work inside a metal box? because while the cellular communication stops the WiFi continues to work.
– uhoh
Dec 11 '19 at 3:10
• @BenCrowell I don't. These all sound interesting. I also don't know how much signal needs to get through to make it ring, or if you could answer it, whether it would work. Dec 11 '19 at 7:15

## A matter of thresholds

The reality of spread-spectrum is complicated but let's imagine that the WiFi router and cell phone tower have both allocated 1 Watt to transmit to your phone, and it in turn can transmit 1 Watt back in both cases.

If the WiFi router is 10 meters away and the cell phone tower is 1 km away, then it's possible to imagine a four order of magnitude difference right from the start.

If your makeshift implementation of a Faraday cage is imperfect then the WiFi may work simply because there's so much more signal strength received at both ends.

Why would your implementation of a Faraday cage be imperfect, and why do I keep saying Faraday cage?

Unlike picometer-scale gamma ray photons which macroscopically speaking tend to travel in straight lines until scattering events, the wavelengths here are a few to ten centimeters, so this is a complex RF problem, not a shielding problem. Until you build a continuous and closed conductive surface that completely surrounds the phone, skin depth thinking is incomplete.

Aluminum has a native oxide (1, 2), so even though it may be folded over itself, it may not be forming a proper Faraday cage. There may be some points of contact with fairly low resistance, but they would have to be closely spaced, below a small fraction of a wavelength, before they could be considered continuous. They may have an electrical resistance of a fraction of an Ohm, but that's still enough to allow a bit of RF to squeeze through.

Consider trying something that doesn't have such an insulating native oxide. Copper foil or copper flashing, if cleaned of its oxide might work better. The oxide grows more slowly and is not as insulating as alumina.

And if your phone is old, you could consider soldering the foil pouch shut for good measure and dramatic effect, or make a pouch pre-soldered on three sides and alligator-clip it closed.

Ever so slightly related: Wouldn't putting an electronic key inside a small Faraday cage render it completely useless?

The penetrating capacity of radio waves is higher when their frequency is smaller. This is a fundamental principle, and this principle answers the question, because the question contains details about frequency.

Particularly, 5 GHz is higher than 800 MHz, and hence 800 MHz wave penetrates the Aluminum kitchen foil sheet more easily than 5GHz wave. However, we should also consider practical variations in frequencies: as it is given in this link.

• Please read our site's policy on self-promotion. It is alright to link to your own papers occasionally, but only if they are germane to the post and you explicitly state that it is your own paper.
– Chris
Mar 23 '21 at 8:10
• As to the content of your paper, I'm concerned, for a start, that you have somehow concluded that wifi signals are ultraviolet rays when in reality they are microwaves. It doesn't seem to help answer the question, either, so I'm removing the reference.
– Chris
Mar 23 '21 at 8:16
• As you can see from my question - the exact opposite is the case experimentally. Apr 5 '21 at 11:33