Why angle of reflection and incident are equal??

[Mera Bharat Mahan]

That's my only question.

I am a newbie here, so I don't know if this is the right section to post this question.

I am expecting a non-mathematical answer with minimum science terminologies, as I won't understand it at all.

 

If an atom is hit by a photon, its electron excites and changes its shell and then again falls back to its previous shell, giving a photon.

My question is, why the given out photon goes in perfect opposite direction of the incident photon to make angle i = angle r.

 

[swansont]

That's my only question.

I am a newbie here, so I don't know if this is the right section to post this question.

I am expecting a non-mathematical answer with minimum science terminologies, as I won't understand it at all.

 

If an atom is hit by a photon, its electron excites and changes its shell and then again falls back to its previous shell, giving a photon.

My question is, why the given out photon goes in perfect opposite direction of the incident photon to make angle i = angle r.

 

 

Reflection does not involve an excitation to another shell.

 

Classically, one would apply the requirements of electric and magnetic fields at the surface (called boundary conditions). The only condition where they hold is if the angles are the same. Or, one could apply Fermat's principle, that the light takes the path of least time. http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fermat.html

 

Quantum-mechanically, the absorption/emission is by a virtual state, and the only solution that conserves energy and momentum is equal angles.

[Mera Bharat Mahan]

can you please elaborate??

[swansont]

Light is an electromagnetic wave, i.e. it has an electric and magnetic field. At an interface, the field components parallel to the surface are continuous, because for at least one of the fields, the parallel component depends on the angle. If you have 100% reflection, the only way for that to be true is if the angles are equal. Showing this requires math, which you said you didn't want.

[EdEarl]

The incident and reflection angles of a ball are equal, except for spin (English for billiards players). Though, friction makes it hard to move a ball without spin. The surface it hits must be smooth, too.

 

Photons have spin, but I think it is around the direction of the path of the photon, not like typical spin of a ball.

[Mera Bharat Mahan]

Showing this requires math, which you said you didn't want.

I didn't say I don't want math. I said it because I won't understand it.

At an interface, the field components parallel to the surface are continuous, because for at least one of the fields, the parallel component depends on the angle.

Please explain this.

[studiot]

 

I didn't say I don't want math. I said it because I won't understand it.

 

 

Hello ajinkya, you might be suprised what most people would understand.

Much maths is only a posh way of stating what people know anyway.

 

Do you understand resolution of forces, velocities, distances etc into components?

 

By this I mean, do you understand where I would end up if I walked paces west then 100 paces north, relative to my starting point?

 

You answer will determine how I phrase my explanation of your question about the mechanism of reflection.

[Mera Bharat Mahan]

yes!! that is what I am being tought in class for last 3 days.

go on...

[studiot]

I'm gald you have been studying resolution of vectors. You should be able to follow this explanation.

 

With reference to the sketches.

 

Let us say we have a single particle moving at steady velocity V towards a wall at angle alpha. It hits the wall and bounces off at angle beta as shown in fig1.

 

What you are asking is why does beta = alpha.

 

In fig2 I have resolved V into two components V_y at right angles to the wall and V_x, parallel to it.

V is obviously the vector sum of V_y and V_x

 

In fig3 I consider the instant of impact. Now the velocity parallel to the wall, V_x is unaffected by the impact. The particle continues travelling to the right at the same velocity. That is V_x after the impact is the same as V_x before the impact.

 

However in fig3 we also see that the particle is brought to a complete stop at the instant of impact in the y direction. That is

 

At impact V_y = 0 momentarily.

 

If you have learned about momentum we say that all the momentum in the y direction becomes the force generated by the particle colliding with the wall.

 

If you know Newton's laws you can write this more mathematically, if not it doesn't matter, just accept that this is what generates the force of the particle hitting the wall.

 

I have called this force F_y in fig4.

 

Now one of Newton's law says that action (the impact) by a force of one object on another is always returned by a reaction force from the suffering body (the wall) which is equal in magnitude and opposite in direction to the action.

 

So the wall exerts a force -F_y on the particle

This must accelerate the particle to -V_y since it took +F_y to stop the particle from +V_y

 

I have shown this in fig5

 

So the particle now moves off the wall (ie it reflects) with a velocity V' whose components are

 

V_x and -V_y at angle beta.

 

Can you see that this angle must be the same as angle alpha, since the components have the same magnitude, but only differ in direction?

 

Edited July 16, 2013 by studiot

[Mera Bharat Mahan]

hmm.....

thank you...

 

does a photon reacts in same manner with wall of atoms??

[studiot]

 

hmm.....

thank you...

 

does a photon reacts in same manner with wall of atoms??

 

 

Hopefully you understood that.

 

You may have noted that I did not try to calculate the force of impact. Doing that brings in some difficult mathematics.

 

Happily it doesn't matter what the value is - It is the fact the that action and react are equal and opposite that counts.

 

 

Is light made of a stream of particles (we now call them photons) or is it some form wave motion?

This question has been puzzling scientists for nearly 500 years and is still not fully answered since the answer appears to be much more complicated and involve some very deep physics and mathematics.

 

The short answer is yes photons behave exactly as I've descibed above.

This was Newton's original explanation for light (although he called them corpuscles).

He also showed how a stream of corpuscles could account for the other activity of light - That of refraction.

 

However someone called Young proposed a rival theory that also accounts for both these. It was the wave theory.

Both theories correctly predict the size of angles in reflection and refraction.

However the the corpusculer theory requires that light speeds up when it enters a denser medium, such as water or glass whilst the wave theory required that light slowed down. No one at the time was able to measure the speed of light so the argument raged.

In the early eighteen hundreds Foucalt did measure the speed and showed that light slows down so everybody thought Ahh - light is a wave.

But at the end of the nineteen hundreds the photelectric effect was discovered. This cannot be a wave phenomenon. Combined with the explanation for spectroscopic lines the quantum theory was born whereby light has characteristics of both particles and waves.

In the middle of the twentieth century further, more bizare effects were discovered that we call quantum entanglement, so the story is ongoing.

[Mera Bharat Mahan]

I like to read about light. But when mathematical part comes I simply skip it. I don't know how people understand it.

I just like thinking(without calculation) and that is when this question come to my mind. I was actually looking for an answer which contained step by step process like photon comes then hits the atom then atom absorbs it, like that.

Still your answer is fully appreciated. No doubt about that.

 

I am not good at english so i can't fully tell you what i think. I am 100% greatfull to each answerer. Especially of studiot. I am telling this because my language might represent i am rude or anything. i don't know.

 

so,

can you please try to tell the step by step procedure??

Your previous answer is not waste at all, like i have mentioned above.

I am asking you answer in this way because i am a guy who prefers to imagine than doing maths etc.

So please...

[studiot]

What more did you want to know? I made it about as simple as possible.

[Mera Bharat Mahan]

I am taught that when atom receives energy i.e it is incident with EM wave it absorbs it and then releases it. You can not tell in what direction the EM wave will go.

Mirror is too made up of some kind of atoms. So what makes them to make angle i = angle r ?? And releasing all EM waves same as it has received??

[studiot]

That is a perfectly reasonable question to wonder about.

 

You need to distinguish between macroscopic and sub atomic effects.

 

Macroscopic effect are due to the combined efforts of a very large number of particles and any network of forces or bonds joining them.

 

Sub atomic effects are due (usually) to the interaction between a few (sub atomic) particles or even the interaction between one particle and its surroundings.

 

The absorbtion and subsequent emission of photons is a sub atomic effect, called the photo-electric effect.

Note that this occurs for a specific wavelength or a number of wavelengths, for example the sodium yellow spectrum lines. It does not occur for a broad range of frequencies as happens in reflection of light.

 

Reflection is a macroscopic effect that is best not thought of in terms of quantum mechanics. Note I did not offer any structure to my reflective wall. This is usual for the physics of macroscopic quantites - we do not enquire into the fine structure.

 

Note, however if you made a mirror of polished sodium, it would remove the yellow light and perhaps re emit it randomly, as you describe. Meanwhile it would reflect the rest of the wavelengths you shone onto it.

[Mera Bharat Mahan]

What is contribution of free electrons in reflection??

[studiot]

Consider a situation where there are plenty of free electrons, say in a cathode ray tube.

 

Are they reflective?

Edited July 23, 2013 by studiot

[Mera Bharat Mahan]

Then why metals are used for making mirrors??

[studiot]

What is the difference between a mirror and a sheet of paper in terms of what happens to incident light?

 

Look up the term 'specular'.

Edited July 23, 2013 by studiot

[Nabla]

Sorry, I'm still stuck on the first part of the conversation.

 

studiot, you said that At impact V_y = 0 momentarily. How can that be the case if photons always move at the speed of light?

 

surely it does not spontaneously accelerate in the x direction to compensate

[Mera Bharat Mahan]

I read Wikipedia, and got

 

reflection of mirror is specular and of white paper it is diffuse reflection.

 

Should I dig a little deeper??

[studiot]

What is the difference between the surface of a (polished) metal and a piece of paper at a microscopic (not molecular) level?

 

What is the difference between their material structure?

[Strange]

I was actually looking for an answer which contained step by step process like photon comes then hits the atom then atom absorbs it, like that.

 

You might want to look at these lectures by Richard Feynman. He explains the reflection and refraction of light in terms of photon/electron interaction. The full theory is called Quantum Electrodynamics and this is a non-technical description of it.

 

http://vega.org.uk/video/subseries/8

[Mera Bharat Mahan]

What is the difference between the surface of a (polished) metal and a piece of paper at a microscopic (not molecular) level?

 

What is the difference between their material structure?

I think, metal surface is plane and paper surface is rough.