# Laser vs Transmiting antenna

Ok so this will probably will sound stupid to you but please put an effort into elaborating with my train of thought thank you :)

For no particular reason this question popped in my head:

People can burn small wooden plates with a 1Watt laser.

There are places in a city were you can find cellular antennas that are like 3 Watt

Why dont those antennas burn my skin for example since they have thrice the power of the laser?

Well "thats easy" I said to my self...

The laser concetrates the light into one point (idially) and the antenna spreads out the light to all directions in cover an entire area

and now the problems begin...

So my initial thought because of this was that a laser could burn me because I would receive 1watt/1point=1 watt on a point on my skin

And I dont get burned by a cellular antenna because I will receive only a part of its spreaded power so I calculated 3watt/4πr^2~=0.2389watts (asuming that I am 1 meter away from the transmitter, I also assumed that that I should divite the power using the surface area of a sphere because thats how impagine that the signal spreads out like a sphere that gets bigger and bigger :P )

but now things started to get blur for me... shouldnt this be the amount of power given in a specific point of surface the so even if only a part of that surface area falls on me wouldnt that be a great amount of energy?

how many things am I missing? :P

• You are missing a few correct units. :-) The 1W laser will burn things when focused on less than $1mm^2$, or so, i.e. the power density is $10^6W/m^2$. You (almost) correctly calculate the power density of the 3W transmitter, except for the units, which also have to be [$W/m^2$]. When you compare the two values with each other, then you can see that the laser has a four million times higher specific power in the focal spot than the transmitter and you can probably ratchet that up by another three orders of magnitude (getting close to a factor of a billion) with a really well focused laser. Commented May 6, 2016 at 23:14

$$I = \frac{1W}{\pi (1\text{mm})^2}$$ $$I = 3.2\times 10^5\text{Wm}^{-2}$$
$$I' = \frac{3W}{4 \pi (1\text{m})^2}$$ $$I' = 2.4\times 10^{-1} \text{Wm}^{-2}$$