Why doesn't my reflection follow the inverse square law ?

[DaveC426913]

Which is an opposite function of the inverse square law ?

No. You are confusing two unrelated things.

A reflected image becomes more visual contracted the greater the distance the mirror is away ?

Yes, in the sense that the path of light has to travel a longer distance.

If you were standing under the Eiffel Tower, looking at a (very tall) mirror 1km away, the image of the Tower would appear to be 2km away, because the light rays travel 1km in each direction.
It would look exactly the same as if you were looking at the Eiffel Tower from 2km away without a mirror.


Draw a diagram like mine.

 

[Jerimiah]

No. You are confusing two unrelated things.


Yes, in the sense that the path of light has to travel a longer distance.

If you were standing under the Eiffel Tower, looking at a (very tall) mirror 1km away, the image of the Tower would appear to be 2km away, because the light rays travel 1km in each direction.
It would look exactly the same as if you were looking at the Eiffel Tower from 2km away without a mirror.
View attachment 4458

Draw a diagram like mine.

No. You are confusing two unrelated things.


Yes, in the sense that the path of light has to travel a longer distance.

If you were standing under the Eiffel Tower, looking at a (very tall) mirror 1km away, the image of the Tower would appear to be 2km away, because the light rays travel 1km in each direction.
It would look exactly the same as if you were looking at the Eiffel Tower from 2km away without a mirror.
View attachment 4458

Draw a diagram like mine.

If an objects reflection is equal and proportional to its own dimensions then why aren't you drawing parallel lines ?

 

[DaveC426913]

If an objects reflection is equal and proportional to its own dimensions then why aren't you drawing parallel lines ?

Here you go. Parallel lines showing the object and its reflection appear the same height - provided they both have the same path length.

 

[Jerimiah]

Here you go. Parallel lines showing the object and its reflection appear the same height - provided they both have the same path length.

View attachment 4464

But they don't appear the same size in the image in the mirror ! If we stood a mirror right next to the tower , that was the same size of the tower , the tower image in the mirror would be equal and proportional to the tower . If we then moved that mirror away following vector x, the image would visually contract , looking smaller . Why does the reflection do this when the towers reflection is parallel also ?

 

[DaveC426913]

But they don't appear the same size in the image in the mirror ! If we stood a mirror right next to the tower , that was the same size of the tower , the tower image in the mirror would be equal and proportional to the tower . If we then moved that mirror away following vector x, the image would visually contract , looking smaller . Why does the reflection do this when the towers reflection is parallel also ?

The angle the image subtends is proportional to total path length.
Measure the distance from you to the object.
Measure the distance from you to the mirror and then to the object.
If the path length is the same, then the image will subtend the same angle as the object, thus they will appear the same height and the same distance away.

Please, draw a diagram of what you think is happening, so we can figure out where your thinking is going awry. Simply drawing the diagram may correct your confusion.

How far have you gotten in school in geometry and/or optics?

 

[gmilam]

But they don't appear the same size in the image in the mirror ! If we stood a mirror right next to the tower , that was the same size of the tower , the tower image in the mirror would be equal and proportional to the tower . If we then moved that mirror away following vector x, the image would visually contract , looking smaller . Why does the reflection do this when the towers reflection is parallel also ?

That would only apply if your pupils were as big as the Eiffel Tower.

 

[DaveC426913]

That would only apply if your pupils were as big as the Eiffel Tower.

Oy. Don't get him started. I think his troubles are much more fundamental than that.

 

[Jerimiah]

The angle the image subtends is proportional to total path length.
Measure the distance from you to the object.
Measure the distance from you to the mirror and then to the object.
If the path length is the same, then the image will subtend the same angle as the object, thus they will appear the same height and the same distance away.

Please, draw a diagram of what you think is happening, so we can figure out where your thinking is going awry. Simply drawing the diagram may correct your confusion.

How far have you gotten in school in geometry and/or optics?

I didn't get very far in school and I don't know how to upload a picture , its asks for url ?

 

[Jerimiah]

That would only apply if your pupils were as big as the Eiffel Tower.

I already said the mirror was the size of the Eiffel Tower , so why does the parallel reflection downscale in visual size when the mirror is moved away ?
Why does it become narrow visioning ?

 

[DaveC426913]

I already said the mirror was the size of the Eiffel Tower , so why does the parallel reflection downscale in visual size when the mirror is moved away ?
Why does it become narrow visioning ?

Your eyes do not see parallel lines. They only see rays that converge on the eye.

In the above image, you are the magenta figure. You cannot see the green rays; you can only see the red rays -the only rays that converge on your eyes.

 

[DaveC426913]

I didn't get very far in school and I don't know how to upload a picture , its asks for url ?

Does your upload dialogue not look like this?

 

[Jerimiah]

Does your uplaod dialogue not look like this?
View attachment 4467

No , it justs says image URL then an insert slot .

 

[Jerimiah]

Your eyes do not see parallel lines. They only see rays that converge on the eye.

In the above image, you are the magenta figure. You cannot see the green rays; you can only see the red rays -the only rays that converge on your eyes.

The thing is though a mirror should detect the parallel light rays and reflect a proportional image .

How can a reflection plane narrow ?

Seems spooky

If xy=xy of object and mirror , then the reflection detected by the mirror should always be xy ?

 

[DaveC426913]

The thing is though a mirror should detect the parallel light rays and reflect a proportional image .

It does reflect parallel rays. But those parallel rays are perpendicular to the mirror (green, in diagram), and don't converge on your eye, so you can't see them, which means those rays can't be part of the image you see.

It also reflects oblique rays, which are the only ones that matter to us.

In my pic, the only ray leaving the top of the tower that can reach your eye via the mirror is the top red one.

It starts off at the top (324m) and can only reach your eye if it reflects off the mirror at half height (163m) before entering your eye at 2m.

 

[Jerimiah]

It does reflect parallel rays. But those parallel rays are perpendicular to the mirror (green, in diagram), and don't converge on your eye, so you can't see them, which means those rays can't be part of the image you see.

It also reflects oblique rays, which are the only ones that matter to us.

In my pic, the only ray leaving the top of the tower that can reach your eye via the mirror is the top red one.

It starts off at the top (324m) and can only reach your eye if it reflects off the mirror at half height (163m) before entering your eye at 2m.

What do your eyes have to do with the reality of the mirror ? The visual size decrease can be measured physically , the mirror should detect the parallel rays without any sort of contraction . It is as if reflections are ''diamond'' formation like ?

 

[DaveC426913]

What do your eyes have to do with the reality of the mirror ?

Your eyes are what tell you how large and how far away an image is. Usually the image is the thing that's of interest about mirrors. But the term "image" has no meaning unless there is someone (or some thing) to detect it.

What the mirror does to light rays you're not looking at is a separate question that can't be examined by your eyes. You need a diagram with geometry to analyze it.

Which one are you interested in?

 

[Jerimiah]

Your eyes are what tell you how large and how far away an image is. Usually that's the only thing that's of interest about mirrors.

The term "image" has no meaning unless there is someone (or some thing) to detect it.

What the mirror does to light rays you're not looking at is a separate question that can't be examined by your eyes. You need a diagram with geometry to analyze it.

Which one are you interested in?

I am interested in the density function of the reflection .

 

[DaveC426913]

I am interested in the density function of the reflection .

"Density function"?

My question was: are you interested in parallel rays in a mirror, or are you interested in images seen in a mirror?

Given ideal circumstances, a reflection in a mirror could fool you into thinking you are actually looking at reality rather than a reflection (that is, after all, the entire purpose of a mirror. If it produced distortions, it would cease to serve its purpose as a mirror).

So, light that is reflected off a mirror behaves (for most purposes) just like light that that is not reflected off a mirror. And images seen in a mirror look (ideally) like the real thing (except the whole 'reversed' thing).

The inverse square law is not affected by a mirror. You are over-thinking this.

 

[DaveC426913]

Here is a more detailed diagram. Is there anything on here you disagree with or don't understand?

 

[origin]

Here is a more detailed diagram. Is there anything on here you disagree with or don't understand?

View attachment 4470

Dave, very nice job. I hope he gets it with this, I can't think of a better way to showing him where his confusion is coming from.