Electromagnetic radiation prevalence

[Waldheri]

Hello,

Perhaps my wording is a bit off so excuse me for that. I think it best to explain by way of a thought experiment. Say there is a device that counts the amount of photons with the same wavelength for all possible wavelengths. What is then the intensity vs. wavelength curve that I would see if I put this device in space, and will it be significantly different from what I would see if I put it on earth?

I've tried searching for this but I think I use the wrong terms in Google to get what I want. I will be very thankful for anyone that can provide some good sources on this.

Cheers,
Waldheri

 

[johng23]

If you're pointing the detector at some random point in space, then the spectrum you will see just depends on what celestial bodies are in the region of space that the detector can see, and what type of radiation these bodies emit. If you're pointing it at the sun then you will see the spectrum of the sun. http://en.wikipedia.org/wiki/Sunlight

On earth, you need to consider the Earth's atmosphere both absorbing and scattering the light. That page and also the page on Rayleigh scattering are relevant to the processes for sunlight. In terms of other types of radiation from space, I'm not sure which are scattered or absorbed, but if you search for the answer to that question, it should be easier to find.

 

[Waldheri]

Thank you for your reply.

I asked this questions because I was thinking of the human eye, and how it can only detect a small slice in the electromagnetic spectrum. First I thought this had to mean we are blind to most em-radiation, but upon second thought I realized I also need to know how much em-radiation with the visible wavelengths there actually isin comparison to other em-radiation.

So it boils down to: we see only a tiny slice of the spectrum, but most radiation is in that spectrum or do we actually miss the majority of what's to be seen?

 

[johng23]

It's definitely the case that our eyes simply can't detect other wavelengths, regardless of how much radiation is present in those parts of the spectrum. But in terms of the amounts, the page for infrared says "Sunlight at zenith provides an irradiance of just over 1 kilowatt per square meter at sea level. Of this energy, 527 watts is infrared radiation, 445 watts is visible light, and 32 watts is ultraviolet radiation." So even neglecting all the rest of the spectrum, we are missing over half of what reaches us from the sun.

I don't really know the answer to your general question though, obviously.

 

[sardar]

Hello Waldheri,

Your thought experiment raises a very interesting question about the prevalence of electromagnetic radiation in different environments. The intensity vs. wavelength curve you would see from your device would indeed be different in space compared to on Earth. This is due to the varying levels of electromagnetic radiation present in these two locations.

In space, there is a higher prevalence of high-energy electromagnetic radiation, such as X-rays and gamma rays, as these are emitted by stars and other celestial bodies. On Earth, the atmosphere acts as a shield, filtering out a significant amount of this high-energy radiation and allowing mostly visible light and some infrared and radio waves to reach the surface.

Additionally, the intensity of electromagnetic radiation on Earth is also affected by factors such as the time of day, weather conditions, and location on the planet. For example, the intensity of sunlight will be different at different latitudes and altitudes.

If you are looking for sources on this topic, I suggest searching for "electromagnetic radiation prevalence on Earth vs. space" or "electromagnetic spectrum in different environments". You may also find it helpful to look into the concept of the electromagnetic spectrum, which illustrates the different types of electromagnetic radiation and their wavelengths.

I hope this helps answer your question and provides some direction for further research. Best of luck with your studies!

Sincerely,

Scientist