1
$\begingroup$

I believe the topic of the present question fully explains the issue I would you to talk with me.

It is well known that the higher the electromagnetic frequency, the more difficult is its shielding.

I have seen many x-rays devices at university covered just with silver paper in order to prevent that any radiation outlay could affect the experimentalists. So I tried to isolate my phone with a silver paper, but still I could call it with another phone.

Why do radio waves overtake that shielding while x-rays do not?

$\endgroup$
6
  • 1
    $\begingroup$ Can you say what you mean by silver paper. A thin sheet of silver (say 10-20 microns) should certainly shield a phone call, but may not block a powerful digital signal (wifi). $\endgroup$ – ProfRob Apr 19 at 16:57
  • $\begingroup$ If you mean aluminium foil then I demonstrate every year to students how wrapping a phone in a single layer prevents it receiving phone calls (but not WiFi). $\endgroup$ – ProfRob Apr 19 at 19:25
  • $\begingroup$ @ProfRob: Are you saying that WiFi isn't blocked by ordinary aluminum foil? (Not arguing, I just would have thought it would be blocked.) $\endgroup$ – tom10 Apr 19 at 20:59
  • 1
    $\begingroup$ physics.stackexchange.com/questions/518906/… @tom10 $\endgroup$ – ProfRob Apr 19 at 21:32
  • $\begingroup$ @ProfRob: OK, this makes sense. Thanks. In short, it's not reduced attenuation for WiFi, but that the receiver is much closer. I can believe that. $\endgroup$ – tom10 Apr 19 at 22:17
1
$\begingroup$

The X-ray part of the electromagnetic spectrum covers quite a large spectrum. I assume medical imaging uses x-rays at the low-frequency end of the x-ray spectrum, and low intensity, to minimize tissue damage. (To make up for that the sensor must be very sensitive.) I expect medical imaging x-ray machines will be thwarted by aluminum foil.

In industrial setting x-ray technology is used for inspection, and then higher frequencies and higher intensities are used. If you make the beam intensity high enough then a sufficient amount will make it through. Aluminum is comparitively low density, and hence comparitively easy to obtain imaging of something inside.


In a laboratorium setting, for a device to leaks x-rays is unacceptable, that is a health hazard. Surely a device that leaks x-rays is not allowed.

My guess is that any actual leaking of electromagnetic energy is in the range of radio communication frequencies. If a device is inadvertently emitting waves in the radio communication spectrum it may inadvertendly be disrupting radio communication. As a precaution against that it makes sense to wrap it in aluminum foil, to contain such emissions. Also, the wrapping of the device may be to shield the device. This would then be shielding against stray electromagnetic energy in the lab that may affect some internal electronics.


The radio communication between a cell phone and a cell tower is designed to be quite robust. There is redundancy, there is technology in place to deal with the signal having been reflected from a rough reflector, which introduces distortion.

I have on occasion noticed that while in an elevator cabin, which is presumably all metal, my cell phone did not completely lose connection. I infer: as long as some reflection can still leak through some slit the cell phone is able to reconstruct the content of the original signal.

$\endgroup$
2
$\begingroup$

I personnally wrapped my phone with silver paper (aluminium foil to be precise) and the phone was shielded. However, the phone has to be completely covered in foil for it to work, a few mm hole will leak enough for the experiement to fail.

$\endgroup$
-1
$\begingroup$

I think it is because shorter wavelengths are higher energy and more easily interact with matter. Similar to how UV is more easily absorbed than visible and visible is more easily absorbed than near IR.

The same thing happens with 5.8GHz vs 2.4GHz WiFi.

Longer wavelengths also diffract around objects more easily. A longer wavelength won't cast a very clean shadow.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.