Post by: hamdani yusuf on 21/08/2017 09:16:13
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?
Post by: chris on 21/08/2017 10:43:40
Post by: hamdani yusuf on 21/08/2017 13:53:26
This is effectively how a fibre optic works (https://www.thenakedscientists.com/get-naked/experiments/water-fibre-optics).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.
Post by: chris on 22/08/2017 20:01:47
Post by: guest39538 on 25/08/2017 18:30:46
How many times would a light ray be reflected inside a circular mirror?There is no angle using a circle, the light remains linear.
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?
Post by: hamdani yusuf on 18/03/2022 07:59:43
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.
Post by: hamdani yusuf on 18/03/2022 08:04:15
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.
Post by: Bored chemist on 18/03/2022 08:54:39
Essentially the light will be scattered in every direction.
Incidentally, unless the mirror has infinite mass, the wavelength will also be scrambled.
Post by: hamdani yusuf on 18/03/2022 10:17:05
The effects from finite width of the light beam can be addressed/minimized by making the circle large enough.
Post by: Bored chemist on 18/03/2022 13:45:43
Post by: hamdani yusuf on 20/03/2022 04:57:58
So, not actual science then...It's a math problem. If you don't think that math is science, then you're right.
Post by: Bored chemist on 20/03/2022 09:44:24
It would be a math problem if the mirror was a polygon.So, not actual science then...It's a math problem. If you don't think that math is science, then you're right.
Post by: hamdani yusuf on 20/03/2022 10:09:21
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?
Post by: Bored chemist on 20/03/2022 10:14:32
Because it stops working.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?
If a beam of light hits a curved surface, it's no longer a beam.
Post by: hamdani yusuf on 20/03/2022 12:16:10
Because it stops working.What prevents a mathematical ray from having infinitesimally narrow width?
If a beam of light hits a curved surface, it's no longer a beam.
Post by: Bored chemist on 20/03/2022 13:11:00
Post by: hamdani yusuf on 20/03/2022 21:45:41
Post by: Bored chemist on 20/03/2022 22:05:30
Post by: hamdani yusuf on 20/03/2022 22:24:44
Diffraction stops it being scientific.How does a curved surface create diffraction?
Post by: Bored chemist on 20/03/2022 22:34:13
I wasn't aware that anyone had said that it did.Diffraction stops it being scientific.How does a curved surface create diffraction?
Post by: hamdani yusuf on 21/03/2022 12:17:57
How should I interpret this?I wasn't aware that anyone had said that it did.Diffraction stops it being scientific.How does a curved surface create diffraction?
Because it stops working.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?
If a beam of light hits a curved surface, it's no longer a beam.
Post by: Bored chemist on 21/03/2022 12:27:52
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.
Post by: hamdani yusuf on 22/03/2022 09:20:40
Post by: hamdani yusuf on 02/05/2022 15:55:41
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.
Post by: hamdani yusuf on 03/05/2022 08:24:09
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.
Post by: Bored chemist on 03/05/2022 08:41:24
Once, after that, it's no longer a ray.
Post by: hamdani yusuf on 05/05/2022 05:16:15
" How many times would a light ray be reflected inside a circular mirror?"Will it stop being reflected?
Once, after that, it's no longer a ray.
Post by: Bored chemist on 05/05/2022 08:34:57
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.
Post by: hamdani yusuf on 06/05/2022 11:01:51
Something that doesn't exist can't be 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.
Post by: Bored chemist on 06/05/2022 13:00:45
Something that doesn't exist can't be reflected.So why did you ask if it was?
Will it stop being reflected?
Post by: hamdani yusuf on 06/05/2022 23:03:50
Because I don't think it would stop existing. Otherwise we would have to reject all of optics.Something that doesn't exist can't be reflected.So why did you ask if it was?Will it stop being reflected?
Post by: Bored chemist on 08/05/2022 10:22:01
How big is when it has crossed the sphere twice (i.e. through and back)?
Post by: hamdani yusuf on 09/05/2022 23:31:35
Just draw what happens to a fairly narrow beam of light when it enters a hole in a hollow spherical mirror and is reflected.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.
How big is when it has crossed the sphere twice (i.e. through and back)?
Post by: Bored chemist on 10/05/2022 08:38:31
It depends on the width of the light beam and the curvature of the mirrorOK for a beam of width W and a mirror of radius R
Post by: paul cotter on 10/05/2022 10:35:31
Post by: hamdani yusuf on 10/05/2022 14:20:06
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.
Post by: hamdani yusuf on 10/05/2022 14:21:57
Say W= 0.1 mm, while R=10 mIt depends on the width of the light beam and the curvature of the mirrorOK for a beam of width W and a mirror of radius R
Post by: Bored chemist on 10/05/2022 18:03:45
OK, now you have taken a few days to decide what sizes you want, you can get back to what you were asked.Say W= 0.1 mm, while R=10 mIt depends on the width of the light beam and the curvature of the mirrorOK for a beam of width W and a mirror of radius R
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)?
Post by: Bored chemist on 10/05/2022 18:20:40
mirror ball.jpg (37.2 kB . 615x618 - viewed 3323 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.
Post by: hamdani yusuf on 11/05/2022 12:19:21
If you shine a beam of light at a concave mirror it is brought to a focus.What would I get in this case?
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.
Say W= 0.1 mm, while R=10 m
Post by: Bored chemist on 11/05/2022 13:31:27
You need my help to multiply 0.1 by 3?If you shine a beam of light at a concave mirror it is brought to a focus.What would I get in this case?
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.Say W= 0.1 mm, while R=10 m
The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)
Post by: hamdani yusuf on 12/05/2022 05:06:56
You need my help to multiply 0.1 by 3?I want to make sure that your approximation still makes sense under your own assumptions, and it's not contaminated with my assumptions.
The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)
Post by: Bored chemist on 12/05/2022 08:44:30
Post by: hamdani yusuf on 13/05/2022 07:05:05
You need my help to multiply 0.1 by 3?You can calculate a complex equation and come out with correct value. But it doesn't necessarily represent the system in question.
The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)
Post by: Bored chemist on 13/05/2022 11:08:57
Did you notice my use of the word "about"?You need my help to multiply 0.1 by 3?You can calculate a complex equation and come out with correct value. But it doesn't necessarily represent the system in question.
The "beam" will be about the same size as the sphere by the 10th reflection
(Because 3^10 is 59049)
Also, do you really think that 3*3*3*3*3*3*3*3*3*3 is "a complex equation"?
Post by: hamdani yusuf on 13/05/2022 12:22:43
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.
Post by: hamdani yusuf on 13/05/2022 12:28:41
The "beam" will be about the same size as the sphere by the 10th reflectionIn my calculation, the width of the beam increases linearly to the number of reflection, instead of exponentially like in your approximation.
(Because 3^10 is 59049)
Post by: Bored chemist on 13/05/2022 13:03:13
In my calculationI 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?
Post by: hamdani yusuf on 13/05/2022 14:39:35
What is the width of the beam after it's reflected at G?
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.
Post by: Bored chemist on 13/05/2022 16:22:45
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.
Post by: hamdani yusuf on 13/05/2022 16:48:52
Does it increase exponentially by the number of reflection?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.
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.
Quote
"When an honest man discovers he is mistaken, he will either cease being mistaken, or cease being honest." - Anonymous.
Post by: Bored chemist on 13/05/2022 17:59:59
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.
Post by: Bored chemist on 13/05/2022 18:01:51
So, did you change the thread title to " How many times would THE
Post by: hamdani yusuf on 14/05/2022 03:20:35
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.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.
Post by: hamdani yusuf on 14/05/2022 03:21:16
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?
Post by: Bored chemist on 14/05/2022 10:46:02
All light beams are (eventually) divergent- because of diffraction.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?
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.
Post by: hamdani yusuf on 19/05/2022 12:44:21
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?
Post by: Bored chemist on 19/05/2022 13:01:48
Why do you call it diffraction?It was called diffraction before I was born.
How much is the divergence caused by diffraction,It depends.
Post by: hamdani yusuf on 20/05/2022 15:14:27
This article shows the difference between Reflection,-Refraction,-and-Diffraction.Why do you call it diffraction?It was called diffraction before I was born.How much is the divergence caused by diffraction,It depends.
https://www.physicsclassroom.com/class/waves/Lesson-3/Reflection,-Refraction,-and-Diffraction
(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)
Post by: Bored chemist on 20/05/2022 16:49:03
This article shows the difference between Reflection,-Refraction,-and-Diffraction.Why did you post it?
Post by: Bored chemist on 20/05/2022 16:50:03
And in this article, no mention of diffraction is found.Does it mention beer?
Are you surprised that not all articles mention all words?
Post by: hamdani yusuf on 21/05/2022 11:35:37
To remind you that not every kind of wave spreading is called diffraction.This article shows the difference between Reflection,-Refraction,-and-Diffraction.Why did you post it?
Post by: Bored chemist on 21/05/2022 12:06:21
How did you come to the mistaken conclusion that I had said, or even thought, that it was?To remind you that not every kind of wave spreading is called diffraction.This article shows the difference between Reflection,-Refraction,-and-Diffraction.Why did you post it?
Post by: hamdani yusuf on 23/05/2022 03:23:06
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.
Post by: Bored chemist on 23/05/2022 12:10:59
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.
Post by: hamdani yusuf on 23/05/2022 14:27:39
The spread is determined both by the mirror and by how narrow the beam is.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.
If you make the beam narrow enough it will result in more spreading than the mirror.
Quote
https://en.wikipedia.org/wiki/DiffractionYou seem to refer to this formula when saying that narrower light beam spread more.
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.
(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.
Post by: Bored chemist on 23/05/2022 17:55:29
But you are the first one I know to confuse between diffraction and curved mirror reflection.THE ONLY ONE WHO IS CONFUSED IS YOU.
Post by: Bored chemist on 23/05/2022 18:26:47
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.
Post by: hamdani yusuf on 25/05/2022 05:20:07
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.
Because it stops working.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?
If a beam of light hits a curved surface, it's no longer a beam.
Post by: Bored chemist on 25/05/2022 09:08:59
I started off by pointing out the most obvious reason that you are wrong.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.Because it stops working.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?
If a beam of light hits a curved surface, it's no longer a beam.
Have you accepted that you are wrong yet?
Post by: hamdani yusuf on 25/05/2022 11:13:28
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.
Post by: hamdani yusuf on 25/05/2022 11:14:57
perfectly flat mirror exactly 1 cm in diameterI'm not sure if you refer to mathematical abstractions or real physical model.
Post by: hamdani yusuf on 25/05/2022 11:20:39
Post by: Bored chemist on 25/05/2022 13:25:01
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.was to get you to realise that diffraction also applies to flat mirrors.
How big is the spot of light?
Is it
(1) smaller than 1cm
(2) exactly 1 cm
(3) bigger than 1cm?
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?
Post by: Bored chemist on 25/05/2022 13:25:56
I'm not sure if you refer to mathematical abstractions or real physical model.
Imagine...
Post by: hamdani yusuf on 27/05/2022 14:52:23
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.Your last sentence above is especially suspicious.
Essentially the light will be scattered in every direction.
Incidentally, unless the mirror has infinite mass, the wavelength will also be scrambled.
Post by: hamdani yusuf on 27/05/2022 14:55:18
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"?
Post by: Bored chemist on 27/05/2022 15:51:15
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.
Post by: Bored chemist on 27/05/2022 15:52:18
Then why did you write this?I was making the point that maths problems don't have to obey the laws of physics.
Post by: hamdani yusuf on 28/05/2022 07:39:47
That's a logical fallacy called appeal to authority.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.
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.
Post by: hamdani yusuf on 28/05/2022 07:41:42
I was making the point that maths problems don't have to obey the laws of physics.That's the point of this thread.
Post by: Bored chemist on 28/05/2022 11:59:04
It implies that the mirror has infinite massThat's exactly wrong.
Post by: Bored chemist on 28/05/2022 12:06:16
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 black-body 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.)
Post by: hamdani yusuf on 28/05/2022 17:05:53
It's not a fallacy if they are an authority.It seems like you haven't learned about Galileo.
Post by: Bored chemist on 28/05/2022 17:09:54
Galileo did a thought experiment and overturned the views of Aristotle.It's not a fallacy if they are an authority.It seems like you haven't learned about Galileo.
And then, because he was trying to explain it to people who were unaccustomed to actually thinking, he did the practical experiment.
So what?
Post by: hamdani yusuf on 30/05/2022 14:38:51
Authorities can still make mistakes. Blindly following them is a logical fallacy.Galileo did a thought experiment and overturned the views of Aristotle.It's not a fallacy if they are an authority.It seems like you haven't learned about Galileo.
And then, because he was trying to explain it to people who were unaccustomed to actually thinking, he did the practical experiment.
So what?
Post by: Bored chemist on 30/05/2022 17:35:09
Ignoring them for no reason is also foolhardy.Authorities can still make mistakes. Blindly following them is a logical fallacy.Galileo did a thought experiment and overturned the views of Aristotle.It's not a fallacy if they are an authority.It seems like you haven't learned about Galileo.
And then, because he was trying to explain it to people who were unaccustomed to actually thinking, he did the practical experiment.
So what?
However, if you bounce photons round inside a circular mirror, you end up scrambling their energies.
Post by: Origin on 30/05/2022 19:01:49
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.
Post by: hamdani yusuf on 30/05/2022 21:50:51
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.
Post by: Origin on 30/05/2022 22:07:53
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.
Post by: hamdani yusuf on 30/05/2022 22:13:41
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.
Post by: Bored chemist on 30/05/2022 22:57:14
, 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".
Post by: hamdani yusuf on 01/06/2022 11:47:53
Read history of scientific progress., 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".
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.
Post by: Bored chemist on 01/06/2022 12:52:12
What I was dismissing was thisHowever, 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.
Your last sentence above is especially suspicious.It isn't "suspicious", it's true.
Post by: hamdani yusuf on 25/07/2022 23:49:05
Post by: hamdani yusuf on 01/08/2022 11:51:05
Post by: hamdani yusuf on 01/08/2022 11:53:04
Post by: hamdani yusuf on 02/08/2022 03:17:52
Post by: hamdani yusuf on 05/08/2022 08:06:26
Post by: hamdani yusuf on 16/08/2022 02:40:33
Post by: hamdani yusuf on 16/08/2022 07:31:02
Pattern simplification process to calculate number of reflection needed to get to first position.
Post by: hamdani yusuf on 19/08/2022 03:11:20
Describing positive simplification process to calculate number of reflection to reach first position.
Post by: Sheilataylor on 23/08/2022 12:27:24
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.
Besides, equal mirrors cannot be created.
Or have we already reached such technologies and I am not right?
Post by: hamdani yusuf on 06/09/2022 07:47:11
This video explains an algorithm to calculate required reflection times to reach first position using negative simplification method.
Post by: hamdani yusuf on 19/10/2022 17:00:33
This video explains an algorithm to calculate required reflection times to reach first position using a combination of positive and negative simplification method.
Post by: hamdani yusuf on 19/10/2022 17:03:03
Post by: hamdani yusuf on 31/08/2025 12:49:45
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?
The hole is large enough for the light ray to enter the circular mirror at various angle.
The last sentence is a clarification because a chatbot claimed that if the hole is too small, the light can only enter at 0 degree, and refuse to calculate.
ChatGPT gives a correct answer
Quote
Short answer: it depends what your ?10?? is measured against.
For a circular mirror, the ray?s angle to the tangent stays constant after every bounce. Each bounce advances the hit-point around the rim by a central angle , where is the entry angle measured to the tangent. The ray returns to the hole after the smallest with
m\,\alpha \equiv 0 \pmod{180^\circ}.
If your ?incoming angle = 10?? is to the tangent (i.e., grazing entry), then
.
.
Reflections = .
If your ?incoming angle = 10?? is the usual optics angle of incidence (to the normal), then the angle to the tangent is .
.
Reflections = 9-1 = 8.
In general: with entry angle (to the tangent), the path is a regular polygon with N = 180 - gcd(α, 180) vertices, and the ray reflects N-1 times before exiting through the hole.
Post by: hamdani yusuf on 31/08/2025 12:55:09
Quote
The problem involves a laser beam entering a circular mirror through a hole, with an incoming angle of 10 degrees.
It's like a concave mirror, fully circular, with the hole on the circumference.
The beam reflects inside until it possibly exits through the hole again.
The "incoming angle" might be the angle between the ray and the radius at entry.
For the beam to exit, it must hit the hole again from inside, assuming the hole is a small point.
At each reflection, the ray's direction changes, affecting its path inside the mirror.
The incoming angle of 10 degrees is likely the angle with the normal, meaning it's almost along the radius.
In a circular billiard, the trajectory is periodic, depending on the initial angle.
The angle between the ray and the tangent at entry might be key, as a 0-degree angle would mean it's tangent and not entering.
Let's set coordinates with the center at (0,0) and the hole at (R,0) for clarity.
Hit points form a sequence: 160?, 320?, 120?, 280?, 80?, 240?, 40?, 200?.
After 8 reflections, the beam returns to the starting point and exits through the hole.
The number of reflections is 8, as it exits after reaching the hole again.
For α=10?, the central angle φ=160?, and 160/360=4/9, so the period n=9, confirming reflections=8.
It seems to use brute force to get the answer. So I asked again with incoming angle of 11 degrees.
Quote
To determine the number of times a laser beam is reflected inside a circular mirror when the incoming angle is 11 degrees measured from the normal line, we follow a similar geometric analysis as in the previous case, adjusting for the new angle.
Problem Setup
Mirror Geometry: The mirror is a perfect circle, reflective on its inner side, forming a concave cavity. We treat this as a 2D billiard problem in the plane of the circle.
Incoming Beam: The laser beam enters through a small hole on the mirror?s circumference at an angle of 11 degrees relative to the normal (the radius at the point of entry).
Reflection: The beam reflects off the inner surface according to the law of reflection (angle of incidence equals angle of reflection relative to the normal at each point).
Goal: Calculate the number of reflections before the beam exits through the same hole, assuming the hole is small but allows exit when the beam returns to the hole?s position with an appropriate angle.
Geometric Analysis
Coordinate Setup:
Place the center of the circular mirror at \( (0, 0) \).
Set the radius of the circle to \( R = 1 \) for simplicity (the number of reflections is scale-invariant).
Position the hole at \( (1, 0) \) (polar angle \( \theta = 0^\circ \)).
The normal at the hole is the radial line pointing toward the center, along \( (-1, 0) \).
Incoming Angle:
The angle of incidence is \( \theta = 11^\circ \), measured from the normal.
The incoming ray?s direction is determined relative to the normal. For an angle of 11? from the inward normal \( (-1, 0) \), the ray?s direction vector needs to be computed, but we focus on the reflection geometry.
Reflection in a Circular Billiard:
In a circular billiard, the ray reflects off the inner surface, and the central angle between consecutive reflection points depends on the angle of incidence.
The central angle \( \phi \) between two consecutive reflection points is given by:
\[
\phi = 180^\circ - 2\theta
\]
where \( \theta \) is the angle of incidence relative to the normal.
For \( \theta = 11^\circ \):
\[
\phi = 180^\circ - 2 \times 11^\circ = 180^\circ - 22^\circ = 158^\circ
\]
Periodicity of the Path:
After each reflection, the ray hits the circle at a point rotated by \( \phi = 158^\circ \) relative to the previous point.
Starting at the hole (\( \theta = 0^\circ \)), the angular positions of the reflection points are:
\[
0^\circ, 158^\circ, 316^\circ, 474^\circ, \ldots
\]
Reduce angles modulo \( 360^\circ \):
\( 0^\circ \)
\( 158^\circ \)
\( 316^\circ \)
\( 474^\circ = 474^\circ - 360^\circ = 114^\circ \)
\( 632^\circ = 632^\circ - 360^\circ = 272^\circ \)
\( 790^\circ = 790^\circ - 360^\circ = 70^\circ \)
\( 948^\circ = 948^\circ - 360^\circ = 228^\circ \)
\( 1106^\circ = 1106^\circ - 360^\circ = 86^\circ \)
\( 1264^\circ = 1264^\circ - 360^\circ = 244^\circ \)
\( 1422^\circ = 1422^\circ - 360^\circ = 102^\circ \)
\( 1580^\circ = 1580^\circ - 360^\circ = 260^\circ \)
\( 1738^\circ = 1738^\circ - 360^\circ = 118^\circ \)
\( 1896^\circ = 1896^\circ - 360^\circ = 276^\circ \)
\( 2054^\circ = 2054^\circ - 360^\circ = 134^\circ \)
\( 2212^\circ = 2212^\circ - 360^\circ = 292^\circ \)
\( 2370^\circ = 2370^\circ - 360^\circ = 150^\circ \)
\( 2528^\circ = 2528^\circ - 360^\circ = 308^\circ \)
\( 2686^\circ = 2686^\circ - 360^\circ = 166^\circ \)
\( 2844^\circ = 2844^\circ - 360^\circ = 24^\circ \)
\( 3002^\circ = 3002^\circ - 360^\circ = 182^\circ \)
\( 3160^\circ = 3160^\circ - 360^\circ = 340^\circ \)
\( 3318^\circ = 3318^\circ - 360^\circ = 198^\circ \)
\( 3476^\circ = 3476^\circ - 360^\circ = 56^\circ \)
\( 3634^\circ = 3634^\circ - 360^\circ = 274^\circ \)
Continue until the angle returns to \( 0^\circ \mod 360^\circ \).
Finding the Period:
The ray returns to the starting point when the total angular displacement is a multiple of \( 360^\circ \):
\[
N \cdot 158^\circ = k \cdot 360^\circ
\]
\[
N = k \cdot \frac{360}{158} = k \cdot \frac{180}{79}
\]
Since \( \gcd(180, 79) = 1 \) (79 is prime, and 180 = \( 2^2 \cdot 3^2 \cdot 5 \)), the smallest integer \( N \) occurs when \( k = 79 \):
\[
N = 79 \cdot \frac{180}{79} = 180
\]
After 180 reflections, the total angle is:
\[
180 \cdot 158^\circ = 28440^\circ = 79 \cdot 360^\circ
\]
This confirms the ray returns to \( \theta = 0^\circ \) after 180 reflections.
Number of Reflections:
The ray hits the mirror at each reflection point. After 180 reflections, it returns to the hole?s position (\( \theta = 0^\circ \)).
In billiard problems, the number of reflections is typically the number of hits before the ray exits. Assuming the ray exits on the 180th hit (when it returns to the hole with an angle allowing exit), the number of reflections is:
\( 180 - 1 = 179 \), since the final hit is the exit.
Exit Condition:
When the ray returns to \( (1, 0) \), its direction must allow it to pass through the hole. In a circular billiard, the path is periodic, and the angle \( 2\theta = 2 \cdot 11^\circ = 22^\circ \) determines the chord angle.
Check if the path is closed:
\[
\frac{2\theta}{\pi} = \frac{22^\circ}{180^\circ} = \frac{11}{90}
\]
Since \( \frac{11}{90} \) is rational, the path is periodic. The denominator (90) and the number of reflections (180) suggest a star polygon with 90 points, but since \( \gcd(11, 90) = 1 \), the path visits 90 distinct points over two full cycles (180 reflections).
The ray exits after 179 reflections, as the 180th hit allows it to leave the hole.
Comparison with Previous Cases:
For \( \theta = 10^\circ \), the answer was 17 reflections, where:
\[
\phi = 180^\circ - 2 \cdot 10^\circ = 160^\circ
\]
\[
N = k \cdot \frac{360}{160} = k \cdot \frac{9}{4}, \quad N = 4 \cdot \frac{9}{4} = 9
\]
This gave 9 hits, with 8 reflections before exiting, but the provided answer (17) suggests a different interpretation or possible error.
For \( \theta = 22.5^\circ \), we calculated 7 reflections:
\[
\phi = 180^\circ - 2 \cdot 22.5^\circ = 135^\circ
\]
\[
N = k \cdot \frac{360}{135} = k \cdot \frac{8}{3}, \quad N = 3 \cdot \frac{8}{3} = 8
\]
This gave 7 reflections before exiting.
For \( \theta = 11^\circ \), the larger number (179) is consistent with a less commensurate angle, leading to a longer periodic path.
Final Answer
The laser beam reflects 179 times before exiting the circular mirror when the incoming angle is 11 degrees measured from the normal line.
\[
\boxed{179}
\]
Post by: hamdani yusuf on 31/08/2025 13:05:42
Quote
This is an interesting optics problem! Let me work through it step by step.For a circular mirror with the reflective surface on the inside (like a circular cavity), we need to consider the geometry of reflections.Key considerations:The mirror is circular, so the normal at any point is radial (pointing toward/away from the center)Each reflection follows the law of reflection: angle of incidence equals angle of reflectionThe incoming angle is 10? (I'll assume this means 10? from the normal at the entry point)For a circular mirror, there's an important relationship: if a ray enters at angle θ from the normal, it will continue to reflect at the same angle θ from the normal at each reflection point, but the chord it traces will gradually rotate around the circle.The number of reflections depends on the rational relationship between the angle and the circle's geometry. Specifically, if we consider the angular displacement between successive reflection points, the ray will form a regular polygon inside the circle.For an incoming angle of 10? from the normal:The ray will trace a path that creates equal angular segments around the circleThe angular separation between reflection points is related to 180? - 2?10? = 160?The ray returns to its starting point when it has completed an integer number of these 160? rotations around the 360? circleSince 360?/160? = 2.25, the ray doesn't close after a small number of reflections. Instead, it will create a complex pattern.However, for practical purposes, the beam will eventually exit through the hole after a finite number of reflections. The exact number depends on:The precise geometry of the entry holeThe exact entry angle and positionImperfections in the real systemFor a theoretical perfect system with a 10? entry angle, the ray would reflect many times (potentially dozens) before eventually aligning with the exit hole, but the exact number requires more specific geometric parameters about the hole size and position.In real systems, factors like absorption, scattering, and hole geometry typically limit this to a much smaller number of observable reflections.
Please assume the shape is perfectly circular and perfectly reflective.
Quote
Thank you for the clarification! Let me work through this more precisely for a perfect circular mirror.For a perfectly circular mirror with perfectly reflective inner surface, I need to analyze the geometric pattern of reflections.When a ray enters at 10? from the normal, the key insight is that after each reflection, the ray maintains the same angle with respect to the local normal (due to the law of reflection), but the reflection points trace out a regular pattern around the circle.The critical parameter is the chord angle - the angle subtended at the center between successive reflection points. For a ray entering at angle θ from the normal, this chord angle is 2θ.For θ = 10?:Chord angle = 2 ? 10? = 20?Number of reflections before returning to start = 360?/20? = 18This means the laser beam will undergo exactly 18 reflections before returning to its starting point and exiting through the same hole.Here's why: The ray traces out a regular 18-sided polygon (18-gon) inscribed in the circle. After 18 reflections, it returns to the original entry point at the same angle, allowing it to exit through the hole.This is a beautiful geometric property of circular mirrors: the number of reflections is always 360?/(2θ) when this results in an integer, where θ is the angle from the normal at entry.So for a 10? entry angle in a perfect circular mirror: 18 reflections before the beam exits.
Post by: hamdani yusuf on 31/08/2025 13:07:59
Quote
From the search results, here are the relevant points about laser beam reflections inside a circular or spherical mirror:
A laser beam inside a perfect reflective spherical or circular mirror theoretically can reflect infinitely many times because the light can keep bouncing inside without loss if the mirror is ideal (100% reflective). In practice, mirrors are not perfect, and the beam eventually dissipates or escapes through openings like the hole where the laser entered .
The number of reflections depends on the angle of incidence. If the angle of incidence is very low, like 10 degrees, the beam will reflect multiple times while traveling around inside the circular boundary.
The geometry of the reflections inside a circle means the beam will follow a chord path between reflection points on the mirror's inner edge. It will reflect repeatedly along this path until it either escapes through the hole or the mirror imperfections absorb the light.
A related physics problem mentions that at certain specific incidence angles (e.g. 45?), the beam satisfies conditions leading to many reflections. At lower or different angles, the path and number of reflections vary. The beam will continue reflecting as long as it does not hit the hole or lose intensity .
Since the hole size allows the laser beam to enter at various angles, the exact number of reflections before the beam either escapes or is absorbed would depend on the exact entry angle and hole position on the circular mirror.
The exact number of reflections for a 10-degree incoming angle cannot be given as a fixed number without knowing the mirror radius, exact hole position, and size, as well as considering the beam divergence and mirror imperfections. But in theory, the beam will reflect multiple times around the circle following the law of reflection with angle of incidence equal to angle of reflection .
If the hole were very small and the mirror nearly perfect, the beam could reflect many times (potentially dozens or hundreds) before escaping or absorption. If the hole is larger or the mirror less reflective, the number of reflections would be fewer.
Would a calculated estimate or further exploration into the geometric optics path calculation be helpful here?
Please assume the shape is perfectly circular and perfectly reflective.
Quote
Assuming the mirror is perfectly circular and perfectly reflective, and considering the laser enters through a small hole at an incoming angle of 10 degrees inside this circular boundary:
The laser beam will reflect infinitely many times inside the mirror without losing intensity because:
A perfectly reflective circular boundary causes the beam to follow a predictable path, reflecting at equal angles of incidence and reflection.
The beam traces chords inside the circle. With no energy loss, it never dissipates.
The hole allows the beam in only once, but after entering, since the mirror is perfect and circular, the beam cannot escape through the hole again unless it hits the hole directly on a subsequent reflection path.
Given an angle of 10 degrees, the beam path will likely form a closed or quasi-periodic polygonal pattern inside the circle, reflecting indefinitely.
This ideal scenario results in an infinite number of reflections unless the beam's internal path aligns exactly to exit the hole, which is an unlikely geometric coincidence.
In summary, the laser beam will reflect infinitely many times inside the perfect circular mirror with 10-degree incidence, as energy losses and escape through the hole are neglected under ideal assumptions .
Post by: hamdani yusuf on 31/08/2025 13:09:10
Quote
Let's analyze the problem step by step:
Given:
? A circular mirror (reflective on the inner side).
? A laser beam enters through a small hole at an angle of 10 degrees (with respect to the tangent at the point of entry).
? The hole is large enough for the light ray to enter at various angles (but we are given a specific angle of 10 degrees).
We need to find how many times the laser beam is reflected before it exits (possibly through the same hole or another, but typically we assume it might exit through the same hole if it returns).
Key Observations:
1. The path of the light inside the circular mirror will be a polygon inscribed in the circle (since each reflection follows the law of reflection: angle of incidence equals angle of reflection).
2. The angle between the incoming ray and the tangent at the point of entry is 10 degrees. This is the same as the angle between the ray and the radius (since the radius is perpendicular to the tangent). So, the angle with the radius is 90 - 10 = 80 degrees? Actually, careful:
? At the point of entry, the tangent is perpendicular to the radius.
? The incoming ray makes an angle of 10 degrees with the tangent. Therefore, the angle with the normal (which is the radius) is 90 - 10 = 80 degrees. So the angle of incidence is 80 degrees? Wait, that seems large. Actually, let's clarify:
Typically, the angle of incidence is measured from the normal. So if the laser enters at 10 degrees to the tangent, then the angle to the normal is 90 - 10 = 80 degrees. So the angle of incidence is 80 degrees.
But actually, for the reflection inside the circle, the geometry is such that the angle between the chord (the path between two reflections) and the tangent is constant. In fact, the path is symmetric and the angle between the chord and the tangent is the same at every reflection.
1. Actually, a standard result for a circular mirror: if a light ray enters at an angle θ (with respect to the tangent), then the ray will hit the wall after rotating by an angle of 2θ along the circle (because the geometry is symmetric). So the successive reflections occur at points separated by an arc of 2θ.
Why?
? At the point of entry, the angle with the tangent is θ.
? By the law of reflection, the angle with the tangent is preserved.
? The chord subtends an angle of 2θ at the center? Let's derive:
Consider a circle of radius R. Let the light enter at point A at an angle θ with the tangent. The tangent at A is perpendicular to the radius OA. So the angle between the incoming ray and OA is(90 - θ). The light travels in a straight line to hit the circle at point B. At B,the tangent is perpendicular to OB. We want to find the arc AB.
In triangle OAB (isosceles with OA=OB=R), the angle at O is the central angle between A and B. The angle at A in the triangle is the angle between OA and the chord AB.This is the same as the angle between the radius and the chord. But the chord makes an angle θ with the tangent, and the tangent is perpendicular to the radius, so the angle between the chord and the radius is 90 - θ. So in triangle OAB:
? Angle at A = 90 - θ
? Similarly, angle at B = 90 - θ (by symmetry? Actually, at B the incident angle is also θ with the tangent, so similarly the angle between the chord and OB is 90 - θ).
? Therefore, the angle at O = 180 - 2*(90 - θ) = 2θ.
So the central angle between A and B is 2θ.
Thus, each chord (between reflections) subtends an angle of 2θ at the center.
Therefore, after n reflections, the total angle traversed along the circle is n * (2θ).
The light will exit through the hole when the total angle is an integer multiple of 360 degrees, because then it returns to the starting point.
So we need to find the smallest n such that n * (2θ) is a multiple of 360.
That is: n * (2θ) = 360 * k, for some integer k. Then n= (360 * k) / (2θ) = (180 * k) / θ.
We want the number of reflections until it exits. Actually, it exits after n reflections? Wait: after n reflections, the light has traveled n chords and is at the starting point (the hole). So it exits after n reflections? But careful: when it returns to the entry point, it exits without reflecting again? So the number of reflections is n, and then it exits.
But if it returns exactly to the hole after n reflections, it exits without that nth reflection? Actually, the reflection occurs at the boundary. So if it hits the hole exactly, it might exit without reflecting. So we need to see when the point is exactly the hole.
In this case, the hole is at the entry point. So we want the total arc length to be a multiple of 360.
So: n * (2θ) = 360 * k => n = (180 * k) / θ.
We need n to be an integer. So we need (180 * k) / θ to be integer.
Given θ = 10 degrees.
So n = (180 * k) / 10 = 18 * k.
So for k=1, n=18. That means after 18 reflections, the light returns to the hole and exits.
But wait: does it actually reflect 18 times? Or does it exit after 18 reflections? Actually, when it returns to the hole, it is at the entry point. So it does not reflect there; it exits. Therefore, the number of reflections is 18.
But is this the first time? For k=1, n=18. So it reflects 18 times and then exits.
We can check: each chord subtends 210=20 degrees. After 18 chords, total angle = 1820=360 degrees. So it returns to the start.
Therefore, the laser beam is reflected 18 times before it exits.
Answer:
\boxed{18}
The laser beam will be reflected 18 times before exiting through the same hole.
Post by: hamdani yusuf on 31/08/2025 13:19:51
Quote
This is a classic problem involving **geometric optics and rotational symmetry** in a **circular mirror**.It still give the same answer even when it's clarified that the incoming angle is measured from normal line.
We are given:
- A **circular mirror**, reflective on the **inside** (like a circular ring with mirrored inner surface).
- A **laser beam** enters through a **small hole**.
- The **angle of incidence** with respect to the tangent (or radius) is **10 degrees**.
- We are to determine **how many times the laser reflects** before it exits through the same hole (or possibly repeats its path).
But here's the key: since the hole is small and the laser enters through it, the beam will **reflect multiple times** inside the circle. The question is: **after how many reflections does it return to the hole and exit?**Spoiler: show
In our case, because of the hole, the beam exits after completing the path, so the number of reflections is **18**, and then it would reflect, but instead it exits.
No ? it can't reflect if there's no mirror.
So the only consistent answer is that the beam reflects **17 times**, then returns to the hole and exits.
But let's calculate the number of times it hits the mirror:
- It hits at: 20?, 40?, 60?, ..., 340? ? that's 17 points.
Then it hits at 0? ? but it's the hole, so no reflection.
So 17 reflections.
The points are at k?20? for k=1 to 17.
Then k=18 is 360?=0? ? hole.
So yes.
### ✅ Final Answer:
\[
\boxed{17}
\]
The laser beam will be reflected **17 times** before it exits through the hole.
Post by: hamdani yusuf on 31/08/2025 13:22:59
Quote
Let?s break down the problem step by step to understand how many times a laser beam would be reflected inside a circular mirror when shone through a small hole at an incoming angle of 10 degrees.
....
This means the laser beam will reflect 36 times before returning to the entry point, assuming no energy loss or absorption
Post by: hamdani yusuf on 31/08/2025 13:29:25
Quote
Understanding the ProblemIt gives the correct answer after I reminded that the entry point can't have reflection.
We have a circular mirror that's reflective on its inner surface. A laser beam enters through a small hole in this mirror at an incoming angle of 10 degrees. We need to determine how many times the laser beam will be reflected inside the mirror before it exits through the same hole.
Key Points:
1. Circular Mirror Reflective Inside: This implies the inner surface acts like the inside of a cylinder or sphere where light reflects.
2. Small Hole: Large enough for the light to enter at various angles, but small enough that we can consider it a point for exit purposes.
3. Incoming Angle of 10 Degrees: This is the angle between the incoming light ray and the tangent to the circle at the point of entry.
Visualizing the Scenario
Imagine a circle representing the mirror's cross-section. The light enters at a point on the circumference (let's call it point A) at a 10-degree angle to the tangent at A. After entering, the light travels in a straight line inside the circle, hits the inner surface at another point (B), reflects, and continues this process until it returns to A.
Determining the Reflection Pattern
The path of the light inside the circle can be understood by considering the angles:
1. Angle of Incidence: The angle between the incoming light ray and the normal (radius) at the point of entry. Since the angle with the tangent is 10 degrees, the angle with the radius (normal) is 90 - 10 = 80 degrees.
2. Reflection Law: The angle of reflection equals the angle of incidence. So, at each reflection, the light ray reflects at the same angle relative to the normal.
Calculating the Number of Reflections
The light will travel in a series of chords inside the circle. The angle between successive chords can be calculated based on the geometry.
Each time the light reflects, the angle it makes with the radius is the same (80 degrees), but the direction changes. The light will continue to reflect until it returns to the starting point (hole).
The number of reflections can be determined by how many times the light needs to reflect to complete a full 360-degree rotation around the circle.
The angle between two successive reflection points (as seen from the center) is twice the angle between the chord and the radius. Here's why:
- At the first reflection point (B), the angle between the incoming ray and the radius is 80 degrees. The reflected ray will make the same angle with the radius, so the total angle between the two chords (from A to B and from B to C) as seen from the center is 2 * (90 - angle with tangent) = 2 * (90 - 10) = 160 degrees.
However, this might not be the most straightforward approach. An alternative is to consider the central angle subtended by each chord.
Each chord corresponds to a central angle θ. The relationship between the angle of incidence (with the tangent) and the central angle is:
θ = 180? - 2 * (angle with tangent) = 180? - 2 * 10? = 160?.
This means that each reflection point is 160? around the circle from the previous one.
Starting at 0?, the sequence of reflection points will be at 160?, 320?, 480? (which is 480 - 360 = 120?), 280?, 440? (440 - 360 = 80?), 240?, 400? (400 - 360 = 40?), 200?, 360? (which is 0? - back to the start).
Counting these steps:
1. 0? to 160? - first reflection
2. 160? to 320? - second reflection
3. 320? to 120? - third reflection
4. 120? to 280? - fourth reflection
5. 280? to 80? - fifth reflection
6. 80? to 240? - sixth reflection
7. 240? to 40? - seventh reflection
8. 40? to 200? - eighth reflection
9. 200? to 0? - ninth reflection (returns to start)
Thus, the light completes its path and exits the hole after 9 reflections.
Verifying with a Smaller Angle
To ensure this method is correct, let's test with a smaller angle where the answer is known. Suppose the incoming angle is 0 degrees (light enters along the tangent).
- θ = 180? - 20? = 180?.
- This means the light goes straight across the diameter, reflects once at the opposite side, and returns to the start: 1 reflection. This matches our expectation.
Another test: incoming angle of 30 degrees.
- θ = 180? - 230? = 120?.
- Sequence: 0?, 120?, 240?, 360? (back to 0?).
- Reflections at 120? and 240?, then exits: 2 reflections. This seems correct because the light completes a triangle inside the circle.
Similarly, for 10 degrees, the sequence covers all points until it returns, resulting in 9 reflections.
Mathematical Generalization
The number of reflections (N) before the light returns to the starting point can be calculated by:
N = 360? / θ - 1, where θ = 180? - 2 * (angle with tangent).
But since θ = 180? - 2 * 10? = 160?,
360? / 160? = 2.25, which is not an integer, indicating that the light will complete an integer number of steps after which the total angle is a multiple of 360?.
We need to find the smallest integer N such that N * 160? is a multiple of 360?.
Find the least common multiple (LCM) of 160 and 360.
160 = 2^5 * 5
360 = 2^3 * 3^2 * 5
LCM is 2^5 * 3^2 * 5 = 32 * 9 * 5 = 1440.
Then N = 1440 / 160 = 9.
Thus, after 9 steps (reflections), the total angle is 9 * 160? = 1440?, which is 4 full rotations (1440 / 360 = 4), bringing the light back to the starting point.
Final Answer
After carefully analyzing the geometry and verifying with smaller angles, the laser beam will be reflected 9 times before exiting through the same hole when the incoming angle is 10 degrees.
Post by: hamdani yusuf on 31/08/2025 13:31:31
Quote
We'll assume the laser beam is a single light ray. The number of reflections of a light ray inside a circular mirror is dependent on the angle of incidence and the diameter of the circle. Since the angle is 10 degrees, the number of reflections is determined by the number of times 10 degrees can be added before reaching 180 degrees. Calculation The ray of light will follow a path that forms a star inside the circle. The ray will be reflected each time it hits the mirror. The angle of incidence is given as 10^\circ. According to the law of reflection, the angle of reflection is equal to the angle of incidence. This means that each time the laser beam hits the mirror, it will be reflected at an angle of 10^\circ relative to the normal at that point. The normal to the circle at any point is a line that passes through the center of the circle. Therefore, the path of the light ray will form a series of isosceles triangles. The base angles of these triangles will be 10^\circ. The angle the ray of light makes with the radius is the angle of incidence. The sum of the angles in a triangle is 180^\circ. So, after the first reflection, the ray makes an angle of 180 - (2 \times 10) = 160^\circ at the center. The ray will continue to be reflected until the path exits the hole it entered. We can determine the number of reflections by finding how many times the 160^\circ angle at the center can be added to itself before it is a multiple of 360^\circ. In other words, we are looking for the least common multiple of 160 and 360. We can also find how many times the initial 10^\circ can be added until it is a multiple of 180. Since the angle of incidence is 10^\circ, and the sum of the angles in a triangle is 180^\circ, the angle between two successive reflection points is 180 - 2 \times 10 = 160^\circ. The light ray will continue to reflect until it returns to the starting point. The number of reflections will be the number of segments of 160^\circ that fit into a full circle of 360^\circ plus the one from the exit point, so 360 / 20 = 18 reflections. Let's look at it another way. The number of reflections inside the circular mirror is determined by the formula: N = \frac{180^\circ}{2 \times \alpha} - 1 where N is the number of reflections and \alpha is the angle of incidence. However, this formula only works if 180 is a multiple of 2 \times \alpha. In this case, 2 \times 10 = 20, and 180 / 20 = 9. So we can expect a total of 9 reflections. However, the laser beam is a ray of light that enters through a hole and exits from the same hole. The ray will form a star pattern with a number of points equal to 180^\circ / \alpha. So, the number of reflections will be the number of points in the star minus 2 (the entry and exit points are not reflections). The number of reflections will be equal to 180 / \text{incoming angle} - 1. N = 180 / 10 - 1 = 18 - 1 = 17 reflections. Let's verify this. The angle of incidence is 10^\circ. The angle of reflection is also 10^\circ. Each reflection changes the direction of the ray. The path of the ray forms a series of isosceles triangles with base angles of 10^\circ. The angle subtended at the center of the mirror by the chord connecting two consecutive reflection points is 180 - 2 \times 10 = 160^\circ. To return to the starting point, the total angle subtended at the center must be a multiple of 360^\circ. We need to find the number of reflections, N, such that N \times 160^\circ is a multiple of 360^\circ. 160 \times N = 360 \times k, where k is an integer. N = 360/160 \times k = 9/4 \times k. The smallest integer value for N is obtained when k=4, which gives N=9. This means the ray will complete a path of 9 reflections before returning to the initial point. However, the question states that the ray exits from the same hole. This means that the path has to return to the initial point, but it should not be reflected at the point where it exits. Therefore, the number of reflections will be 17. Let's look at it as a star with 180/10=18 points. The path of the ray of light forms a star shape with 18 points. The light ray enters the circle, is reflected 17 times, and then exits the circle at the 18th point, which is the same as the entry point. So the number of reflections is 17.
Post by: alancalverd on 31/08/2025 15:11:15
And if the sphere has nonzero thickness (as it must), the beam cannot entirely exit through the hole anyway.
Oh, the analytical power of a sceptical human brain!
Post by: hamdani yusuf on 01/09/2025 03:35:15
Half marks to ChatGPT, at least, for asking the critical question - 10 degrees to what?In most physics textbooks on optics related to reflection and refraction, incoming angle is measured from normal line.
And if the sphere has nonzero thickness (as it must), the beam cannot entirely exit through the hole anyway.
Oh, the analytical power of a sceptical human brain!
The laser beam can still enter through non-zero thickness mirror at non-zero angle as long as the hole also has non-zero diameter.
BTW, the first time I solved the problem almost 30 years ago, my reasoning was similar to Grok's.
Post by: alancalverd on 01/09/2025 15:26:15
If the shell has finite thickness, for the beam to exit the hole it must exit at the complement of the entry angle. The range of permitted angle depends on the ratio of shell thickness to diameter d, and as d/t decreases so does the range of permitted angle, whilst the number of required reflections has the interesting property of being 1 for θ = 0 but tending to infinity for any other angle.
Anyone who has ever dropped a plectrum inside an acoustic guitar will be familiar with the problem. My old teacher could execute a brilliant θ = 0 manoeuvre but us poor mortals can spend a miserable half hour just shaking it at random to get the plectrum out.