Naked Science Forum
On the Lighter Side => New Theories => Topic started by: hamdani yusuf on 21/08/2017 09:16:13

How many times would a light ray be reflected inside a circular mirror?
Let's say we have a circular mirror which is reflective in inner side. A laser beam is shone through a small hole on the mirror. How many times the laser beam would be reflected, if the incoming angle is 10 degrees?

This is effectively how a fibre optic works (https://www.thenakedscientists.com/getnaked/experiments/waterfibreoptics).

This is effectively how a fibre optic works (https://www.thenakedscientists.com/getnaked/experiments/waterfibreoptics).
I think the answer for the question needs math technique/tricks which are not readily available in most school textbooks.
There are more to explore, which I'll add to the playlist later as I have the time.
Sorry for the wrong link I put in the opening post. It has been corrected now.

I think the answer will depend upon the quality of the mirror; even tiny losses will quickly turn all of the energy in the photons into heat.

How many times would a light ray be reflected inside a circular mirror?
Let's say we have a circular mirror which is reflective in inner side. A laser beam is shone through a small hole on the mirror. How many times the laser beam would be reflected, if the incoming angle is 10 degrees?
There is no angle using a circle, the light remains linear.

I think the answer will depend upon the quality of the mirror; even tiny losses will quickly turn all of the energy in the photons into heat.
Let's assume that the loss from the mirror is negligible, at least until the light exit again through incoming hole.
First step to answer the question about reflection of light in a circle. Introducing a new variable : order of reflection. Using incoming angle to calculate ratio between order of reflection and number of reflection.

Next step to answer the question. Discovering a hidden variable implicitly stated in the question, which is greatest common factor between order of reflection and number of reflection.
After the value of s is obtained, we need to subtract by 1 to get the number of reflection, because when the head of the ray comes back to incoming point, it will leave the circle.

If the beam has a finite width (and diffraction means it will have) them you need to consider the "focussing" effect of the curved mirror.
Essentially the light will be scattered in every direction.
Incidentally, unless the mirror has infinite mass, the wavelength will also be scrambled.

I see this as an idealized math problem instead of realistic physics problem.
The effects from finite width of the light beam can be addressed/minimized by making the circle large enough.

So, not actual science then...

So, not actual science then...
It's a math problem. If you don't think that math is science, then you're right.

So, not actual science then...
It's a math problem. If you don't think that math is science, then you're right.
It would be a math problem if the mirror was a polygon.

It would be a math problem if the mirror was a polygon.
Why does it stop being math problem when the number of sides of the polygon is increased toward infinity, i. e. becomes a circle?

It would be a math problem if the mirror was a polygon.
Why does it stop being math problem when the number of sides of the polygon is increased toward infinity, i. e. becomes a circle?
Because it stops working.
If a beam of light hits a curved surface, it's no longer a beam.

Because it stops working.
If a beam of light hits a curved surface, it's no longer a beam.
What prevents a mathematical ray from having infinitesimally narrow width?

Diffraction stops it being scientific.

In geometry, a ray can be defined as a part of a line that has a fixed starting point but no end point.

How many times would a
light geometric ray be reflected inside a circular mirror?

Diffraction stops it being scientific.
How does a curved surface create diffraction?

Diffraction stops it being scientific.
How does a curved surface create diffraction?
I wasn't aware that anyone had said that it did.

Diffraction stops it being scientific.
How does a curved surface create diffraction?
I wasn't aware that anyone had said that it did.
How should I interpret this?
It would be a math problem if the mirror was a polygon.
Why does it stop being math problem when the number of sides of the polygon is increased toward infinity, i. e. becomes a circle?
Because it stops working.
If a beam of light hits a curved surface, it's no longer a beam.

How should I interpret this?
Carefully.
And you should interpret this
What prevents a mathematical ray from having infinitesimally narrow width?
Diffraction stops it being scientific.
as saying that science specifically diffraction means that you can not have an infinitesimally narrow beam of light.

Given the light beam is small enough compared to diameter of the mirror, and the incoming angle is selected to be a rational number of degree with small enough denominator, and the hole on the mirror surface is large enough, the light can exit from the mirror before diverging too much. In scientific education, we often find simplified problems with idealized conditions, such as negligible surface friction and air resistance, zero mass string and pulley to study classical mechanics. In electronics, we often see problems involving ideal conductors, resistors, inductors and capacitors.

I first encountered this problem in 1996 when I joined a scientific research competition for teenagers. A competitor came up with the question regarding reflection of a line inside a circle. By seeing his diagrams at a glance, I knew that he hadn't got the right answer.
After I got back to my hometown, I revisited the problem and finally got the answer. I also combined it with simple geometry to get general solutions for length of the lines and areas covered by those lines (like the pentagon inside a pentagram) for any incoming angle.
I presented the problems as a prized open contest in an exhibition representing my highschool. That's when I realized that it was not a widely known math problem. At least, it wasn't a part of highschool curriculum.

Further exploration brought me to incomplete reflection problem, which I finally solved two years later. The problem goes like this.
A system of circular light reflector creates a pattern with s vertices and t order of reflection. How many times the light is reflected before it reaches nth point?
The key to solve it is an algorithm to calculate inverse of modulo.

" How many times would a light ray be reflected inside a circular mirror?"
Once, after that, it's no longer a ray.

" How many times would a light ray be reflected inside a circular mirror?"
Once, after that, it's no longer a ray.
Will it stop being reflected?

Will it stop being reflected?
In that question you use the word "it".
to what does the word refer?
You can't be referring to the light ray because that no longer exists.

Will it stop being reflected?
In that question you use the word "it".
to what does the word refer?
You can't be referring to the light ray because that no longer exists.
Something that doesn't exist can't be reflected.

Something that doesn't exist can't be reflected.
So why did you ask if it was?
Will it stop being reflected?

Something that doesn't exist can't be reflected.
So why did you ask if it was?
Will it stop being reflected?
Because I don't think it would stop existing. Otherwise we would have to reject all of optics.

Just draw what happens to a fairly narrow beam of light when it enters a hole in a hollow spherical mirror and is reflected.
How big is when it has crossed the sphere twice (i.e. through and back)?

Just draw what happens to a fairly narrow beam of light when it enters a hole in a hollow spherical mirror and is reflected.
How big is when it has crossed the sphere twice (i.e. through and back)?
It depends on the width of the light beam and the curvature of the mirror. But you can restrict the calculations for the center of the light beam.

It depends on the width of the light beam and the curvature of the mirror
OK for a beam of width W and a mirror of radius R

Caution, optics is not my area of expertise(assuming I have any!). I would imagine a narrow beam through a small hole would cause diffraction about the edge of said hole.

Caution, optics is not my area of expertise(assuming I have any!). I would imagine a narrow beam through a small hole would cause diffraction about the edge of said hole.
The hole is larger than the light beam. The edges are not shone.

It depends on the width of the light beam and the curvature of the mirror
OK for a beam of width W and a mirror of radius R
Say W= 0.1 mm, while R=10 m

It depends on the width of the light beam and the curvature of the mirror
OK for a beam of width W and a mirror of radius R
Say W= 0.1 mm, while R=10 m
OK, now you have taken a few days to decide what sizes you want, you can get back to what you were asked.
Just draw what happens to a fairly narrow beam of light when it enters a hole in a hollow spherical mirror and is reflected.
How big is when it has crossed the sphere twice (i.e. through and back)?

mirror ball.jpg (37.2 kB . 615x618  viewed 1448 times)Here's a very bad sketch, but it illustrates the point.
I drew the incoming beam. The thin lines represent the outside edges of the outgoing beam
If you shine a beam of light at a concave mirror it is brought to a focus.
The focal length is half the radius of curvature.
So the distance from the focal point (F) to the point (P) where the light strikes the mirror is half the radius of the sphere and, the distance to the other side of the sphere (G) is 3 times as big. (It's 3/4 times the diameter as opposed to 1/4 times the diameter)
That means the edges of the light form (roughly) two similar triangles , one 3 times as big as the other .
So the width of the beam when it strikes G is about 3 times W.
Now that (divergent) light beam is bounced back across the mirror.
If it was a parallel beam then the same thing would happen to it as happened to the original beam. It would be 3 times as big when it hit the mirror for a third time.
So it would be 9 times W.
But it was already diverging after the first reflection, so the width will be even bigger.
But to a rough approximation, the width of the beam, after n reflections is (at least) 3^n times bigger than the original beam.
This is essentially why integrating spheres work.

If you shine a beam of light at a concave mirror it is brought to a focus.
The focal length is half the radius of curvature.
So the distance from the focal point (F) to the point (P) where the light strikes the mirror is half the radius of the sphere and, the distance to the other side of the sphere (G) is 3 times as big. (It's 3/4 times the diameter as opposed to 1/4 times the diameter)
That means the edges of the light form (roughly) two similar triangles , one 3 times as big as the other .
So the width of the beam when it strikes G is about 3 times W.
Now that (divergent) light beam is bounced back across the mirror.
If it was a parallel beam then the same thing would happen to it as happened to the original beam. It would be 3 times as big when it hit the mirror for a third time.
So it would be 9 times W.
But it was already diverging after the first reflection, so the width will be even bigger.
But to a rough approximation, the width of the beam, after n reflections is (at least) 3^n times bigger than the original beam.
What would I get in this case?
Say W= 0.1 mm, while R=10 m

If you shine a beam of light at a concave mirror it is brought to a focus.
The focal length is half the radius of curvature.
So the distance from the focal point (F) to the point (P) where the light strikes the mirror is half the radius of the sphere and, the distance to the other side of the sphere (G) is 3 times as big. (It's 3/4 times the diameter as opposed to 1/4 times the diameter)
That means the edges of the light form (roughly) two similar triangles , one 3 times as big as the other .
So the width of the beam when it strikes G is about 3 times W.
Now that (divergent) light beam is bounced back across the mirror.
If it was a parallel beam then the same thing would happen to it as happened to the original beam. It would be 3 times as big when it hit the mirror for a third time.
So it would be 9 times W.
But it was already diverging after the first reflection, so the width will be even bigger.
But to a rough approximation, the width of the beam, after n reflections is (at least) 3^n times bigger than the original beam.
What would I get in this case?
Say W= 0.1 mm, while R=10 m
You need my help to multiply 0.1 by 3?
The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)

You need my help to multiply 0.1 by 3?
The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)
I want to make sure that your approximation still makes sense under your own assumptions, and it's not contaminated with my assumptions.

My assumptions are consistent with my assumptions and unaffected by your errors.

You need my help to multiply 0.1 by 3?
The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)
You can calculate a complex equation and come out with correct value. But it doesn't necessarily represent the system in question.

You need my help to multiply 0.1 by 3?
The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)
You can calculate a complex equation and come out with correct value. But it doesn't necessarily represent the system in question.
Did you notice my use of the word "about"?
Also, do you really think that 3*3*3*3*3*3*3*3*3*3 is "a complex equation"?

My assumptions are consistent with my assumptions and unaffected by your errors.
If you assume that your assumptions are already correct, you become blind to see your own errors. That's what happened to most religions and pseudoscience cults.

The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)
In my calculation, the width of the beam increases linearly to the number of reflection, instead of exponentially like in your approximation.

In my calculation
I showed why it is exponential (actually, why it's faster than that).
You just made a claim with no supporting working.
You just assumed your assumptions were correct.
If you assume that your assumptions are already correct, you become blind to see your own errors. That's what happened to most religions and pseudoscience cults.
It's also often what I find has happened to people who are arguing with me on the internet.
Did you not realise you were doing exactly the same thing you accused me of?

mirror ball.jpg (37.2 kB . 615x618  viewed 1448 times)Here's a very bad sketch, but it illustrates the point.
I drew the incoming beam. The thin lines represent the outside edges of the outgoing beam
If you shine a beam of light at a concave mirror it is brought to a focus.
The focal length is half the radius of curvature.
So the distance from the focal point (F) to the point (P) where the light strikes the mirror is half the radius of the sphere and, the distance to the other side of the sphere (G) is 3 times as big. (It's 3/4 times the diameter as opposed to 1/4 times the diameter)
That means the edges of the light form (roughly) two similar triangles , one 3 times as big as the other .
So the width of the beam when it strikes G is about 3 times W.
Now that (divergent) light beam is bounced back across the mirror.
If it was a parallel beam then the same thing would happen to it as happened to the original beam. It would be 3 times as big when it hit the mirror for a third time.
So it would be 9 times W.
But it was already diverging after the first reflection, so the width will be even bigger.
But to a rough approximation, the width of the beam, after n reflections is (at least) 3^n times bigger than the original beam.
This is essentially why integrating spheres work.
What is the width of the beam after it's reflected at G?

What is the width of the beam after it's reflected at G?
Immediately after it is reflected (before it has gone back across the sphere) its width is still about 3W.
But it's strongly divergent. Even if the mirror at G was flat, it would reach about 3 or 4 W by the time it reached the opposite side of the sphere.
But the real question here is why do you have to ask me?
If you can't work it out for yourself, go and learn science.

What is the width of the beam after it's reflected at G?
Immediately after it is reflected (before it has gone back across the sphere) its width is still about 3W.
But it's strongly divergent. Even if the mirror at G was flat, it would reach about 3 or 4 W by the time it reached the opposite side of the sphere.
But the real question here is why do you have to ask me?
If you can't work it out for yourself, go and learn science.
Does it increase exponentially by the number of reflection?
The mirror at G is concave, which reduces the divergence.
Because you're the one who said
Diffraction stops it being scientific.
But to a rough approximation, the width of the beam, after n reflections is (at least) 3^n times bigger than the original beam.
While I suggested to consider only the center of the beam.
It depends on the width of the light beam and the curvature of the mirror. But you can restrict the calculations for the center of the light beam.
"When an honest man discovers he is mistaken, he will either cease being mistaken, or cease being honest."  Anonymous.

The mirror at G is concave, which reduces the divergence.
No.
A parallel beam striking a concave mirror is brought to a focus, but after that the beam diverges strongly.
I see you still haven't actually drawn what happens.

So, did you change the thread title to " How many times would THE CENTRE EDGE OF a light ray be reflected inside a circular mirror?" or did you cease being honest about it?

The mirror at G is concave, which reduces the divergence.
No.
A parallel beam striking a concave mirror is brought to a focus, but after that the beam diverges strongly.
I see you still haven't actually drawn what happens.
It's reduced compared to flat mirror. That's why I said that the width increases linearly to the number of reflection, instead of exponentially.

"When an honest man discovers he is mistaken, he will either cease being mistaken, or cease being honest."  Anonymous.
So, did you change the thread title to " How many times would THE CENTRE OF a light ray be reflected inside a circular mirror?" or did you cease being honest about it?
It also applies to the edge of a light ray.

"When an honest man discovers he is mistaken, he will either cease being mistaken, or cease being honest."  Anonymous.
So, did you change the thread title to " How many times would THE CENTRE OF a light ray be reflected inside a circular mirror?" or did you cease being honest about it?
It also applies to the edge of a light ray.
All light beams are (eventually) divergent because of diffraction.
So, if the centre of a beam enters the circular mirror at 10 degrees, the edge of it doesn't.
And that means your puzzle is poorly defined.

All light beams are (eventually) divergent because of diffraction.
Why do you call it diffraction?
How much is the divergence caused by diffraction, compared to the curvature of the mirror?

Why do you call it diffraction?
It was called diffraction before I was born.
How much is the divergence caused by diffraction,
It depends.

Why do you call it diffraction?
It was called diffraction before I was born.
How much is the divergence caused by diffraction,
It depends.
This article shows the difference between Reflection,Refraction,andDiffraction.
https://www.physicsclassroom.com/class/waves/Lesson3/Reflection,Refraction,andDiffraction
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.physicsclassroom.com%2FClass%2Fwaves%2Fu10l3b6.gif&hash=36016a28e23bdaff08fecd6165c08e9a)
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.physicsclassroom.com%2FClass%2Fwaves%2Fu10l3b8.gif&hash=67a9a39cd49a225d006b3b5c9803e5b7)
And in this article, no mention of diffraction is found.
Reflection and the Ray Model of Light  Lesson 3  Concave Mirrors
The Anatomy of a Curved Mirror
https://www.physicsclassroom.com/Class/refln/u13l3a.cfm
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.physicsclassroom.com%2FClass%2Frefln%2Fu13l3a3.gif&hash=dbc01069282b7fd7008e3cca7d5ca763)

This article shows the difference between Reflection,Refraction,andDiffraction.
Why did you post it?

And in this article, no mention of diffraction is found.
Does it mention beer?
Are you surprised that not all articles mention all words?

This article shows the difference between Reflection,Refraction,andDiffraction.
Why did you post it?
To remind you that not every kind of wave spreading is called diffraction.

This article shows the difference between Reflection,Refraction,andDiffraction.
Why did you post it?
To remind you that not every kind of wave spreading is called diffraction.
How did you come to the mistaken conclusion that I had said, or even thought, that it was?

How did you come to the mistaken conclusion that I had said, or even thought, that it was?
From your previous posts.
Why do you call it diffraction?
It was called diffraction before I was born.
How much is the divergence caused by diffraction,
It depends.
All light beams are (eventually) divergent because of diffraction.

There is no logical path from what I said to what you seem to think it means.
Diffraction was called diffraction long before I was born.
So it isn't me who called it diffraction.
Whether a beam of light is spread more by diffraction than by the curvature of the mirror will depend on circumstances.
The spread is determined both by the mirror and by how narrow the beam is.
If you make the beam narrow enough it will result in more spreading than the mirror.
Any beam of light starts off converging, diverging or parallel.
But a converging beam comes to a focus (like I showed in the picture) and then diverges.
A diverging beam is already diverging.
So the only possibility that is left is a parallel beam.
but that diverges too because of diffraction.
So, diffraction means that all beams diverge in the end.
You need to get to grips with the fact that it is you who doesn't understand science because you refuse to study it.

The spread is determined both by the mirror and by how narrow the beam is.
If you make the beam narrow enough it will result in more spreading than the mirror.
I've seen many people confused between diffraction and interference. But you are the first one I know to confuse between diffraction and curved mirror reflection.
https://en.wikipedia.org/wiki/Diffraction
Diffraction refers to various phenomena that occur when a wave encounters an obstacle or opening. It is defined as the interference or bending of waves around the corners of an obstacle or through an aperture into the region of geometrical shadow of the obstacle/aperture. The diffracting object or aperture effectively becomes a secondary source of the propagating wave. Italian scientist Francesco Maria Grimaldi coined the word diffraction and was the first to record accurate observations of the phenomenon in 1660.
You seem to refer to this formula when saying that narrower light beam spread more.
(https://wikimedia.org/api/rest_v1/media/math/render/svg/1fe6158f4b3a6fe55c7ad8ecf30fa29f959cba86)
But you need to remember that d here is the width of the slit, not the width of the light beam itself. In my example above, the width of the incoming slit is much wider than the width of the light beam. The light beam doesn't interact with the edges of the slit.

But you are the first one I know to confuse between diffraction and curved mirror reflection.
THE ONLY ONE WHO IS CONFUSED IS YOU.

But you need to remember that d here is the width of the slit, not the width of the light beam itself. In my example above, the width of the incoming slit is much wider than the width of the light beam. The light beam doesn't interact with the edges of the slit.
The problem isn't what I need to remember. The thing here isn't that I have forgotten that you said that the beam is smaller than the hole in the mirror.
The problem is that you have forgotten what that means. How did the light get made into a beam?
Imagine a beam of visible light from a distant star comes in through my large open window and hits a perfectly flat mirror exactly 1 cm in diameter and is reflected to form a circular spot of light on a screen at a distance of 1 metre.
How big is the spot of light?
Is it
(1) smaller than 1cm
(2) exactly 1 cm
(3) bigger than 1cm?
The thing here isn't that I have forgotten that you said that the beam is smaller than the hole in the mirror.
The problem is that you have forgotten what that means.

But you are the first one I know to confuse between diffraction and curved mirror reflection.
THE ONLY ONE WHO IS CONFUSED IS YOU.
Somehow you think that flat mirrors don't have the problem.
It would be a math problem if the mirror was a polygon.
Why does it stop being math problem when the number of sides of the polygon is increased toward infinity, i. e. becomes a circle?
Because it stops working.
If a beam of light hits a curved surface, it's no longer a beam.

But you are the first one I know to confuse between diffraction and curved mirror reflection.
THE ONLY ONE WHO IS CONFUSED IS YOU.
Somehow you think that flat mirrors don't have the problem.
It would be a math problem if the mirror was a polygon.
Why does it stop being math problem when the number of sides of the polygon is increased toward infinity, i. e. becomes a circle?
Because it stops working.
If a beam of light hits a curved surface, it's no longer a beam.
I started off by pointing out the most obvious reason that you are wrong.
Have you accepted that you are wrong yet?

I started off by pointing out the most obvious reason that you are wrong.
What's obvious is that you are being inconsistent. You bring up diffraction problem to curved mirror to dismiss the solution, while at the same time ignoring diffraction for flat mirror.

perfectly flat mirror exactly 1 cm in diameter
I'm not sure if you refer to mathematical abstractions or real physical model.

My research 25 years ago was presented as mathematical problem instead of physical one. At the end of the paper I mentioned the connection with complex numbers. The video above reminds me of that.

while at the same time ignoring diffraction for flat mirror.
The whole point of this
Imagine a beam of visible light from a distant star comes in through my large open window and hits a perfectly flat mirror exactly 1 cm in diameter and is reflected to form a circular spot of light on a screen at a distance of 1 metre.
How big is the spot of light?
Is it
(1) smaller than 1cm
(2) exactly 1 cm
(3) bigger than 1cm?
was to get you to realise that diffraction also applies to flat mirrors.
Unfortunately, you failed to answer it.
Because you didn't do the work, you now still don't understand.
If you don't want to understand, why do you keep asking?
Are you a troll?

I'm not sure if you refer to mathematical abstractions or real physical model.
Imagine...

was to get you to realise that diffraction also applies to flat mirrors.
Then why did you write this?
It would be a math problem if the mirror was a polygon.
And this?
If the beam has a finite width (and diffraction means it will have) them you need to consider the "focussing" effect of the curved mirror.
Essentially the light will be scattered in every direction.
Incidentally, unless the mirror has infinite mass, the wavelength will also be scrambled.
Your last sentence above is especially suspicious.

There is no angle using a circle, the light remains linear.
It seems I missed reading this comment. What do you mean by "the light remains linear"?

Your last sentence above is especially suspicious.
Did it lead you to suspect that I know more than you?
If someone who makes a living as a scientist says something about science, it is worth considering that it is correct, rather than assuming that it isn't.

Then why did you write this?
I was making the point that maths problems don't have to obey the laws of physics.

Your last sentence above is especially suspicious.
Did it lead you to suspect that I know more than you?
If someone who makes a living as a scientist says something about science, it is worth considering that it is correct, rather than assuming that it isn't.
That's a logical fallacy called appeal to authority.
It is possible that you're overrated.
Incidentally, unless the mirror has infinite mass, the wavelength will also be scrambled.
It implies that the mirror has infinite mass, the wavelength will not be scrambled.

I was making the point that maths problems don't have to obey the laws of physics.
That's the point of this thread.

It implies that the mirror has infinite mass
That's exactly wrong.

That's a logical fallacy called appeal to authority.
It's not a fallacy if they are an authority.
Anyway, here's what I said about it the last time the subject came up.
The force which the photons exert does not necessarily mean there is transfer of energy. Energy is transferred when the force moves through a distance
In order to be perfectly reflective the walls have to be infinitely massive (this causes other problems).
However, there's another way to do it.
You can imagine a nearly massless mirror.
When a photon hits it, it will move and take some energy from the photon. But that means that, when another photon hits it on the other side, it will add energy to that photon.
Overall, the sum of the energies will be conserved The wavelengths of the photons will be "scrambled" and will settle down to a blackbody distribution.
The energy (on average) imparted to the light, rigid mirror will be Boltzmann's constant times 3 times the temperature. (That's the same energy as would be carried by an electron or proton at that temperature.)

It's not a fallacy if they are an authority.
It seems like you haven't learned about Galileo.

It's not a fallacy if they are an authority.
It seems like you haven't learned about Galileo.
Galileo did a thought experiment and overturned the views of Aristotle.
And then, because he was trying to explain it to people who were unaccustomed to actually thinking, he did the practical experiment.
So what?

It's not a fallacy if they are an authority.
It seems like you haven't learned about Galileo.
Galileo did a thought experiment and overturned the views of Aristotle.
And then, because he was trying to explain it to people who were unaccustomed to actually thinking, he did the practical experiment.
So what?
Authorities can still make mistakes. Blindly following them is a logical fallacy.

It's not a fallacy if they are an authority.
It seems like you haven't learned about Galileo.
Galileo did a thought experiment and overturned the views of Aristotle.
And then, because he was trying to explain it to people who were unaccustomed to actually thinking, he did the practical experiment.
So what?
Authorities can still make mistakes. Blindly following them is a logical fallacy.
Ignoring them for no reason is also foolhardy.
However, if you bounce photons round inside a circular mirror, you end up scrambling their energies.

Authorities can still make mistakes. Blindly following them is a logical fallacy.
It is not a logical fallacy to cite an expert. If I have a spot on my skin and I tell someone it's cancer and they disagree, it is not an appeal to authority if I say my doctor told me it is cancer. Citing a politician's opinion on global warming is an appeal to authority because they aren't experts in climate.

It is not a logical fallacy to cite an expert.
I didn't say it was. But here's what is.
It's not a fallacy if they are an authority.
More over, they often disagree and contradict one another. In this case, at least one of them must be false. Selecting which one to accept should be based on their supporting evidence.

More over, they often disagree and contradict one another. At least one of them must be false. Selecting which one to accept should be based on their supporting evidence.
This is just another example of you enjoying the feeling of confusion.

However, if you bounce photons round inside a circular mirror, you end up scrambling their energies.
In my example with 10° incoming angle, the light will only be reflected 8 times before coming out of the circle through the opening. The energy loss as heat can be minimized. It's not a good reason to dismiss the problem altogether.

, they often disagree and contradict one another.
Where did you see that ?
In order for your point to be valid, you have to show that it happened "often".

, they often disagree and contradict one another.
Where did you see that ?
In order for your point to be valid, you have to show that it happened "often".
Read history of scientific progress.
If you are in the business of scientific breakthroughs, you'll have to argue a lot with your research colleagues on details of the experiments.

However, if you bounce photons round inside a circular mirror, you end up scrambling their energies.
In my example with 10° incoming angle, the light will only be reflected 8 times before coming out of the circle through the opening. The energy loss as heat can be minimized. It's not a good reason to dismiss the problem altogether.
What I was dismissing was this
Your last sentence above is especially suspicious.
It isn't "suspicious", it's true.

This video shows how to calculate the length of the light ray inside a circle.

This video shows how to calculate area of the shape formed by the light ray inside a circle.

This video shows how to deal with negative incoming angle.

This video explains about the formation of virtual reflection pattern.

Now here comes the harder part : incomplete reflection.

First position method to solve incomplete reflection problem.

Pattern simplification process to calculate number of reflection needed to get to first position.