You are Here: Home >< Physics

# What is light scattering? watch

1. In my quest to begin studying particle physics, I began studying about photons and, consequently, about the Rayleigh scattering. However, I don't quite seem to be able to understand the concept of light scattering and as I best understand things by asking questions, I would like to request the help of The Student Room.

Firstly, I watched this video by Michael van Biezen (great resource, by the way): https://www.youtube.com/watch?v=gYee...J#t=126.891397. I then watched this video by The Curious Engineer: https://www.youtube.com/watch?v=0b1fqodmZJ0. That is as far as my knowledge regarding Rayleigh scattering goes.

Allow me to present my confusion here, as the comment section for the second video is disabled. At 1:48 of the second video, it says that blue light interacts with atmospheric molecules more than red light does, which means that the first overpowers the latter, leading to an overall blue colour in the sky. However, if that's the case, then how come at 3:23, the light is red? Shouldn't that effect simply be scaled up, thus leading to an even greater difference between how much blue light can be seen and how much red light can be seen? Shouldn't this lead to an even more blue colour of the sky?

I also fail to understand what light scattering actually means (probably the reason for my previous bit of confusion). Why does light have to be scattered before it reaches the eye? I come to the conclusion that it does, because red light is not visible in the sky while blue light is, and the first is scattered less than the second, meaning that the more the light is scattered, the more visible it is to the eye.

This, to me, is self-contradictory. If light is scattered by the molecules, then shouldn't it be less likely to reach the eye of people on earth? Thus, if blue light is scattered more than red light, shouldn't the sky actually be red as opposed to being blue, because red light is more likely to reach the eye than blue light?

Hopefully the outrageous amount of questions I threw at everyone reading this post will make some understand the root of my confusion and, ideally, point it out to me. Thank you very much.

This is what I understand until now:
Spoiler:
Show

Blue light being scattered makes it more visible up to a certain point.

Red light being scattered less makes it less visible up to that same point.

At sunshine, the light has to go through more atmosphere.

As a result, blue light keeps being scattered to the point where it can't be seen, while red light is scattered much less, meaning it doesn't reach that point, ultimately leading to it being able to reach the eye.

For the sake of ease of expression, let's say light being seen is a "good" thing.

Why is light being scattered good for blue light and bad for red light when there's little atmosphere between the receptor and the light source, but bad for blue and good for red when there's a lot of atmosphere?
2. the blue of the sky during daytime is from scattered light going from the sky into your eye - the reddish light from a setting sun is the light that hasn't been scattered. You don't see the red light going straight through the atmosphere *because* it's going straight through and not being scattered into your eye.

if light wasn't scattered in the atmosphere the sky would be black in the daytime, like the sky appears in photos taken on the moon.
3. Blue light scattering more than red means that more indirect blue light enters your eye compared to red. So, on average, more blue light ends up at your eye (from all parts of the sky).

You're thinking of it as from your direct view that if one ray of light has its blue light scattered away from you, you will see more red, which is true. But you have to extend it to the whole region of the sky, so there is way more blue light scattering towards you, than away, simply because there is more light that isn't directly incident on your eye compared to that which is directly incident on your eye.
4. (Original post by Joinedup)
the blue of the sky during daytime is from scattered light going from the sky into your eye - the reddish light from a setting sun is the light that hasn't been scattered. You don't see the red light going straight through the atmosphere *because* it's going straight through and not being scattered into your eye.

if light wasn't scattered in the atmosphere the sky would be black in the daytime, like the sky appears in photos taken on the moon.
(Original post by Protoxylic)
Blue light scattering more than red means that more indirect blue light enters your eye compared to red. So, on average, more blue light ends up at your eye (from all parts of the sky).

You're thinking of it as from your direct view that if one ray of light has its blue light scattered away from you, you will see more red, which is true. But you have to extend it to the whole region of the sky, so there is way more blue light scattering towards you, than away, simply because there is more light that isn't directly incident on your eye compared to that which is directly incident on your eye.
Let me see if I get this right. The blue light from the sun gets scattered most across the atmospheric molecules. This, in turn, means that said molecules vibrate at such frequency that they literally appear blue?
5. (Original post by Nichita)
Let me see if I get this right. The blue light from the sun gets scattered most across the atmospheric molecules. This, in turn, means that said molecules vibrate at such frequency that they literally appear blue?
I mean at a qualitative level yes, it's much more complicated than that in reality. Scattering here means coupling of the electric field to the electric dipole (or rather the charges on/of) the molecule. This oscillates the molecule, changing it's polarity and the frequency of EM radiation scattered predominantly varies depending on how easy/hard it is to polarise the molecule, for our composition of gases in the atmosphere, this happens to be blue light (as for the visible spectrum, blue has the lowest wavelength). So the field couples to the EM waves in the blue spectrum most and so these oscillations gives off the same radiation at different angles and whose intensity is slightly different to the incident intensity. Overall the photon energy is conserved because this is an elastic scattering process. There are different scattering processes that happen in the atmosphere like Raman scattering which is completely inelastic in that the wavelength of EM radiation out is lower than the incident radiation. Since this excites vibrational (or excited rotational) modes of the molecule which then de-excite to give lower energy EM radiation out.
6. (Original post by Protoxylic)
I mean at a qualitative level yes, it's much more complicated than that in reality. Scattering here means coupling of the electric field to the electric dipole (or rather the charges on/of) the molecule. This oscillates the molecule, changing it's polarity and the frequency of EM radiation scattered predominantly varies depending on how easy/hard it is to polarise the molecule, for our composition of gases in the atmosphere, this happens to be blue light (as for the visible spectrum, blue has the lowest wavelength). So the field couples to the EM waves in the blue spectrum most and so these oscillations gives off the same radiation at different angles and whose intensity is slightly different to the incident intensity. Overall the photon energy is conserved because this is an elastic scattering process. There are different scattering processes that happen in the atmosphere like Raman scattering which is completely inelastic in that the wavelength of EM radiation out is lower than the incident radiation. Since this excites vibrational (or excited rotational) modes of the molecule which then de-excite to give lower energy EM radiation out.
That is a really detailed explanation that I really do not have the necessary education to fully interpret, but I feel like I understand it for the most part. I will certainly carry on my studying to be able to fully understand this phenomenon.

I feel like I'm getting really close to understanding this scattering process. If red light is scattered as well but less frequently than blue light, does this mean that the sky has red "pigment" as well? Just like a deep blue marker has red to give its specific colour, does that same concept apply to this scenario?
7. (Original post by Nichita)
That is a really detailed explanation that I really do not have the necessary education to fully interpret, but I feel like I understand it for the most part. I will certainly carry on my studying to be able to fully understand this phenomenon.

I feel like I'm getting really close to understanding this scattering process. If red light is scattered as well but less frequently than blue light, does this mean that the sky has red "pigment" as well? Just like a deep blue marker has red to give its specific colour, does that same concept apply to this scenario?
Red light is scattered yes, but less strongly i.e it's intensity for any given scattering angle is reduced for red light (intensity inversely proportional to the 4th power of wavelength) I guess you could think about it that way. To be honest, I haven't studied this in detail neither so I can only comment what I already have, my knowledge on scattering is rather limited.
8. (Original post by Nichita)

I feel like I'm getting really close to understanding this scattering process. If red light is scattered as well but less frequently than blue light, does this mean that the sky has red "pigment" as well? Just like a deep blue marker has red to give its specific colour, does that same concept apply to this scenario?
Let me see if I get this right. The blue light from the sun gets scattered most across the atmospheric molecules. This, in turn, means that said molecules vibrate at such frequency that they literally appear blue?
It's not really like pigment.

e.g. if you look at sunlight reflected by a leaf, the leaf looks green but if you look at the light being transmitted through the leaf it also looks green. The explanation here is that wavelengths of light apart from green are being absorbed and giving up their energy so that only green remains. The energy is used by the plant to do useful work running it's metabolism and also some also ends up heating up it's leaf.

one of the classic benchtop demos of rayleigh scattering uses a mixture of milk and water - most people would say milk is a white liquid but when you mix it with water you get scattering of blue(ish) light off at an angle to the incident beam and transmission of orange and red light. Energy is not being removed in any wavelength.

but note I think the effect looks less pronounced in a lot of videos because video cameras sense when they're getting a coloured light source and try correcting it back to white (this is so consumers can use them indoors and outdoors without it looking weird because light from filament bulbs is redder than sunlight. With old style film cameras you had to pick indoor or outdoor film or use a coloured filter).

you could try reproducing a similar milk water experiment yourself.
9. This is what I understand until now:

Blue light being scattered makes it more visible up to a certain point.

Red light being scattered less makes it less visible up to that same point.

At sunshine, the light has to go through more atmosphere. As a result, blue light keeps being scattered to the point where it can't be seen, while red light is scattered much less, meaning it doesn't reach that point, ultimately leading to it being able to reach the eye.

For the sake of ease of expression, let's say light being seen is a "good" thing.

Why is light being scattered good for blue light and bad for red light when there's little atmosphere between the receptor and the light source, but bad for blue and good for red when there's a lot of atmosphere?
10. (Original post by Nichita)
This is what I understand until now:

Blue light being scattered makes it more visible up to a certain point.

Red light being scattered less makes it less visible up to that same point.

At sunshine, the light has to go through more atmosphere. As a result, blue light keeps being scattered to the point where it can't be seen, while red light is scattered much less, meaning it doesn't reach that point, ultimately leading to it being able to reach the eye.

For the sake of ease of expression, let's say light being seen is a "good" thing.

Why is light being scattered good for blue light and bad for red light when there's little atmosphere between the receptor and the light source, but bad for blue and good for red when there's a lot of atmosphere?
This doesn't seem to be a correct understanding and I don't think it'd get many points if it came up as an exam question.

Your eye is a light detector and you only see light that ends up going into your eye - if the light doesn't go into your eye you don't see it.

you can see light that is going directly from a light source into your eye or you can see indirect light that has come from a light source then been scattered into your eye.

if you look directly at the setting sun (taking precautions not to damage your eyes or cause pain) you're looking at the direct light from the sun. if you're looking at the sky you're looking at light that has taken an indirect route from the sun into your eye.

now it turns out that the atmosphere doesn't scatter all frequencies of sunlight equally - blue frequencies are scattered more.

what this means for viewing the direct light from the sun is that it appears reddened because the blue light has been scattered off onto paths that don't end up in your eyeball.

otoh what this means for viewing the sky is that it appears blue because you're seeing the blue light that IS being scattered - there's still plenty of red light going through the atmosphere but you don't see it because it doesn't get scattered and doesn't end up going into your eyeball.
11. (Original post by Joinedup)
This doesn't seem to be a correct understanding and I don't think it'd get many points if it came up as an exam question.

Your eye is a light detector and you only see light that ends up going into your eye - if the light doesn't go into your eye you don't see it.

you can see light that is going directly from a light source into your eye or you can see indirect light that has come from a light source then been scattered into your eye.

if you look directly at the setting sun (taking precautions not to damage your eyes or cause pain) you're looking at the direct light from the sun. if you're looking at the sky you're looking at light that has taken an indirect route from the sun into your eye.

now it turns out that the atmosphere doesn't scatter all frequencies of sunlight equally - blue frequencies are scattered more.

what this means for viewing the direct light from the sun is that it appears reddened because the blue light has been scattered off onto paths that don't end up in your eyeball.

otoh what this means for viewing the sky is that it appears blue because you're seeing the blue light that IS being scattered - there's still plenty of red light going through the atmosphere but you don't see it because it doesn't get scattered and doesn't end up going into your eyeball.
This is making no sense to me. Blue light being scattered at midday makes it more observable than red, but blue light being scattered at sunset makes it less observable than red. Why?
12. Light scattering is a process in which photons of particular energies are absorbed by an atom/molecule, promoting one of their electrons to higher energy states and are then emitted in random directions when the atom/molecule returns to it's ground state. The light that shines on the atom/molecule is white light, but since only one or a few specific wavelengths are removed, the remaining light still appears white. The emitted light is of a particular frequency and it's emitted in a random direction, so it's no longer incorporated into the white light and you see it as colour. In Rayleigh, Raman and Mie scattering this same process occurs. The differences lie in the number/range of frequencies absorbed and in whether multiple photons are emitted for each photon absorbed or not.

In Rayleigh scattering the same frequency absorbed is emitted. In Raman scattering you get photons being emitted at lower energies than the photon which was initially absorbed. Finally, in Mie scattering you get a whole load of frequencies being scattered and they tend to be scattered in one particular direction, as opposed to the random scattering you get from the other two.

In particular, oxygen allotropes tend to absorb frequencies in the blue end of the spectrum, with the notable exception of O8 rings which absorb yellow/orange/red frequencies. The air is about 21% O2 and in the stratosphere you've got the ozone (O3) layer, both of which are real suckers for blue frequencies. Nitrogen, which makes up about 78% of the air is also able to absorb some blue frequencies. Overall, you get a lot of blue frequencies being scattered, much more so than lower frequency light.

(Original post by Nichita)
This is making no sense to me. Blue light being scattered at midday makes it more observable than red, but blue light being scattered at sunset makes it less observable than red. Why?
http://hyperphysics.phy-astr.gsu.edu...atm/raymie.gif
13. (Original post by Peroxidation)
Spoiler:
Show
Light scattering is a process in which photons of particular energies are absorbed by an atom/molecule, promoting one of their electrons to higher energy states and are then emitted in random directions when the atom/molecule returns to it's ground state. The light that shines on the atom/molecule is white light, but since only one or a few specific wavelengths are removed, the remaining light still appears white. The emitted light is of a particular frequency and it's emitted in a random direction, so it's no longer incorporated into the white light and you see it as colour. In Rayleigh, Raman and Mie scattering this same process occurs. The differences lie in the number/range of frequencies absorbed and in whether multiple photons are emitted for each photon absorbed or not.

In Rayleigh scattering the same frequency absorbed is emitted. In Raman scattering you get photons being emitted at lower energies than the photon which was initially absorbed. Finally, in Mie scattering you get a whole load of frequencies being scattered and they tend to be scattered in one particular direction, as opposed to the random scattering you get from the other two.

In particular, oxygen allotropes tend to absorb frequencies in the blue end of the spectrum, with the notable exception of O8 rings which absorb yellow/orange/red frequencies. The air is about 21% O2 and in the stratosphere you've got the ozone (O3) layer, both of which are real suckers for blue frequencies. Nitrogen, which makes up about 78% of the air is also able to absorb some blue frequencies. Overall, you get a lot of blue frequencies being scattered, much more so than lower frequency light.

http://hyperphysics.phy-astr.gsu.edu...atm/raymie.gif
I THINK I GOT IT!

Right, so when there are many molecules in the medium between the source and the receptor (i.e. when the amount of atmosphere is greater because of the angle of emission), Mie scattering takes the lead, which is not wavelength dependent. This, in turn, means that the light observed is composed of the wavelengths that are most abundant in sunlight - oranges, yellows, reds etc. - while the rest are less observable because they are not as abundant in sunlight - blues, purples etc.

When there are fewer molecules in the medium, Rayleigh scattering takes the lead. This, however, is wavelength dependent, and it scatters the shorter wavelengths (i.e. blues) more than it scatters longer wavelengths (i.e. reds).
14. (Original post by Nichita)
This is making no sense to me. Blue light being scattered at midday makes it more observable than red, but blue light being scattered at sunset makes it less observable than red. Why?
Think about the angle made between the path of the incoming light and the path of the scattered light. Blue light is scattered at larger angles, and red light is not scattered at larger angles as much. That's the fundamental reason that you see different colours.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: July 24, 2016
Today on TSR

### Edexcel GCSE Maths Unofficial Markscheme

Find out how you've done here

### AQA Maths Paper 1 exam discussion

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE