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For example a tiny antenna in the phone, why em wave do not pass through it, but pass through concrete walls?

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closed as unclear what you're asking by Alfred Centauri, Jon Custer, user191954, Cosmas Zachos, ZeroTheHero Nov 10 '18 at 1:08

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For an antenna to pick up a sufficient signal, it does not need to stop the wave - it just needs to slightly attenuate it, which it does.

In general, when radio waves pass through obstacles, they get attenuated and, given the same level of conductivity, larger obstacles will cause greater attenuation, but even a small fraction of the absorbed energy could be sufficient for an adequate reception.

I would add that an antenna of a phone is tuned to a particular frequency range and, because of that, everything else the same, it will absorb more energy in that frequency range than a randomly shaped conductor of a similar size.

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  • $\begingroup$ "...fraction of the absorbed energy..." Weeeeell.. But energy of wave is it's frequency. If obstacle absorbes em wave energy, than the em wave that got out from obstacle should have smaller frequency? Maybe You are misconcepting, because another guy said, that em wave loses it's intensity instead of energy $\endgroup$ – user211997 Nov 6 '18 at 7:39
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    $\begingroup$ @Artur Your comment does not make sense. The energy of a wave is $n\hbar\omega $. Absorbing energy means in general that the number of quanta goes down, not the energy per quantum. $\endgroup$ – my2cts Nov 6 '18 at 13:02
  • $\begingroup$ @my2cts, well, agree $\endgroup$ – user211997 Nov 6 '18 at 15:59
  • $\begingroup$ Is there equation, that describes how does the penetration of em wave depends on it's frequency and propereties of state\obstacle? Better if You put it to Your answer $\endgroup$ – user211997 Nov 6 '18 at 16:01
  • $\begingroup$ @Artur The penetration of em waves through materials depends on many different factors and different mechanisms are involved, so there is no single equation describing how it depends on frequency. This is a separate question, which has been answered before. For instance, you can check this or other similar links. $\endgroup$ – V.F. Nov 6 '18 at 16:53
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Simply put, cell phone antennas are made of metal, and are designed to absorb radio waves at cell phone frequencies. Concrete does not interact with radio waves as strongly as metal does and although concrete buildings may have some big pieces of metal in them, at cell phone wavelengths that metal tends to reflect and sometimes bend the radio waves instead of absorbing them.

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  • $\begingroup$ a dielectric rod/resonator can also act as an antenna and is widely used in end-fire mode. $\endgroup$ – hyportnex Nov 6 '18 at 14:31
  • $\begingroup$ Whatever radiates can also absorb, and one does not need tuned metal pieces to resonate and hence radiate. To some extent everything can and does radiate but to make it a better radiator one must balance the dissipative loss against the radiative loss. This balancing act can be had without conductive surfaces, and a dielectric lens is an obvious example: its focal point is where its driving point impedance is defined and any source at its focus will be projected, ie., radiated to infinity with a beam spread caused by edge diffraction is commensurate with the lens diameter. $\endgroup$ – hyportnex Nov 6 '18 at 14:39
  • $\begingroup$ my reply was simplified in response to my assessment of the OP's knowledge level. This would have been a good question to post on the amateur radio SE because of the number of antenna physics experts over there. $\endgroup$ – niels nielsen Nov 6 '18 at 20:48