Understanding Reflection of Light on a Concave Mirror

[A.T.]

The model could be too easy and too hard for the OP, at the same time.

You can always make the hole bigger and a put a lens in it. But for starters the OP learn to distinguish between the real image formed by the mirror, and the real image in the eye/camera.

 

[ZapperZ]

View attachment 265704
In this image we can see that the person is standing between focus and the centre of curvature, so his real image is formed behind him (in the air)... so sir If a real image exists behind him, he might not expect to see it when he is looking toward in the front (in the mirror).

This is why I continue to think that you do not understand ray optics yet.

There is no real image anywhere in this scenario! This means that there isn't a real image behind him! Again, if you done a ray diagram and shown it to us, we can easily dispel this!

View attachment 265705
Then, how is he able to see his real and inverted image in the concave mirror.
Sir @jbriggs444 please help me...

He is not seeing a real image. His eyes cannot focus on light rays that is about to form a real image. I have said this numerous times! The light rays are DIVERGING, and the lens in his eyes is the one focusing the image onto his retina.

The image is also NOT behind him. It is behind the mirror.

Since you have refused to draw your own ray diagram, I did it.

Notice where the object is, where the VIRTUAL IMAGE is. There are no real image anywhere here. The virtual image is behind the mirror, not behind the object.

Since the light rays are DIVERGING after they have been reflected and when they go through the object, if the object is a person, then his eyes see diverging light rays, and the lens in his eyes will focus on those rays to cause a focused image on his retina. This enables him to "see" the virtual image at an apparent distance behind the mirror.

BTW, why do you automatically assume that everyone who answers your question on Physics Forums must be and can only be male? You have never seen women physicists or engineers or scientists before?

Zz.

 

[A.T.]

Since you have refused to draw your own ray diagram, I did it.

He is asking about the case where the object is between the curvature center and the focus:

Here you have a true image behind the object. But for what you see, only the true image on your retina matters.

 

[ZapperZ]

He is asking about the case where the object is between the curvature center and the focus:

https://www.physicsforums.com/attachments/265733

Here you have a true image behind the object. But for what you see, only the image on your retina manners.

Probably, but I needed him to show a ray diagram first, and I'm showing him the simplest case to establish some "baseline" here. I'm hoping to "goad" him to actually show a ray diagram by showing him mine.

Zz.

 

[A.T.]

He is asking about the case where the object is between the curvature center and the focus:

Here you have a true image behind the object. But for what you see, only the true image on your retina matters.

Looking at that diagram again: Since the real image on the retina is flipped, you should see yourself not flipped in this case. Assuming your eye can still focus these rays.

So the distance at which the image on your retina flips, is not the focus point but the center of curvature. The focus point is merely the distance at which the not flipped virtual image behind the mirror is replaced by a real image behind you.

 

[kuruman]

You can always make the hole bigger and a put a lens in it. But for starters the OP learn to distinguish between the real image formed by the mirror, and the real image in the eye/camera.

Perhaps it will be convincing to treat this question as a compound mirror and lens problem and put some math in it. Let $o_1$ = distance of man's face in post #19.
First we find the position of the real image produced by the mirror relative to the mirror: $\dfrac{1}{o}+\dfrac{1}{i_1}=\dfrac{1}{f_1}.$
The real image produced by the mirror is a virtual object for the lens at a negative distance $o'=-(i_1-o)$ from the lens.
This gives a second equation $\dfrac{1}{-(i_1-o)}+\dfrac{1}{i_2}=\dfrac{1}{f_2}$
Solving the system of two equations and two unknowns yields $i_1$ and $i_2$.
Judging from OP's photographs, $f_1\approx 60~\mathrm{cm};~~o\approx 90~\mathrm{cm}$. With a "commonly accepted value" from the web $f_2=2.4~\mathrm{cm}$, I got $i_2=+2.34~\mathrm{cm}$.

If we assume that $f_2$ is the distance from the eye lens to the retina, this value places the image in front of the retina which means that the image would be (slightly?) out of focus. The image will be interpreted as "flipped" because there is no flip in the second equation since the object is virtual. Furthermore, the answer for $i_2$, which I do not provide because I think it would be instructive for OP to derive it, is multiplied by an overall factor of $f_2$. This says that even if the eye muscles change the focal length of the lens, the image will still not be in focus. The reason why $i_2$ and $f_2$ are close is because $o$ and $f_1$ are close in value and much larger that $f_2$.

 

[A.T.]

The image will be interpreted as "flipped" ...

So where did I go wrong in post #35?

 

[SHASHWAT PRATAP SING]

[QUOTE

Here you have a true image behind the object. But for what you see, only the true image on your retina matters.
[/QUOTE]

Thankyou A.T. From this diagram I understood that-

The Light Rays from the Man hit the concave mirror,Reflect back and converge in the retina itself due to which the Real image is formed in the Retina and The eyes assume it to be coming from behind the mirror due to which the person see his inverted and real image in the concave mirror...

Am I Correct Now...?

 

[SHASHWAT PRATAP SING]

BTW, why do you automatically assume that everyone who answers your question on Physics Forums must be and can only be male? You have never seen women physicists or engineers or scientists before?

I am really really sorry ZapperZ I didn't mean that,I am a new user so I didn't knew you are male or female
really sorry...

 

[SHASHWAT PRATAP SING]

Perhaps it will be convincing to treat this question as a compound mirror and lens problem and put some math in it. Let $o_1$ = distance of man's face in post #19.
First we find the position of the real image produced by the mirror relative to the mirror: $\dfrac{1}{o}+\dfrac{1}{i_1}=\dfrac{1}{f_1}.$
The real image produced by the mirror is a virtual object for the lens at a negative distance $o'=-(i_1-o)$ from the lens.
This gives a second equation $\dfrac{1}{-(i_1-o)}+\dfrac{1}{i_2}=\dfrac{1}{f_2}$
Solving the system of two equations and two unknowns yields $i_1$ and $i_2$.
Judging from OP's photographs, $f_1\approx 60~\mathrm{cm};~~o\approx 90~\mathrm{cm}$. With a "commonly accepted value" from the web $f_2=2.4~\mathrm{cm}$, I got $i_2=+2.34~\mathrm{cm}$.

If we assume that $f_2$ is the distance from the eye lens to the retina, this value places the image in front of the retina which means that the image would be (slightly?) out of focus. The image will be interpreted as "flipped" because there is no flip in the second equation since the object is virtual. Furthermore, the answer for $i_2$, which I do not provide because I think it would be instructive for OP to derive it, is multiplied by an overall factor of $f_2$. This says that even if the eye muscles change the focal length of the lens, the image will still not be in focus. The reason why $i_2$ and $f_2$ are close is because $o$ and $f_1$ are close in value and much larger that $f_2$.

Kuruman I am a High School Student so I didn't understand this, please check my #38 reply and tell- am I correct ? Please...

 

[ZapperZ]

[QUOTEView attachment 265749

Here you have a true image behind the object. But for what you see, only the true image on your retina matters.

Thankyou A.T. From this diagram I understood that-
View attachment 265746
The Light Rays from the Man hit the concave mirror,Reflect back and converge in the retina itself due to which the Real image is formed in the Retina and The eyes assume it to be coming from behind the mirror due to which the person see his inverted and real image in the concave mirror...View attachment 265748
Am I Correct Now...?
[/QUOTE]

I still don't understand this. How was the last picture captured? Was there a screen in the guy's head to capture what his eyes see? It looks like this was an image capture elsewhere of him!

This is the ray diagram which you should have drawn for this situation.

But here's the problem. A human eye cannot focus on converging rays. The person, being the object, will see converging light rays entering his eyes. Unless he wears corrective glasses to compensate for that, he won't see a focused image.

There will, however, be a focused image at the location shown, but this can only be seen on a screen.

The only way for an eye to see a focused imaged is if this is observed at a distance LARGER than where the focused image is. Stand far back enough, and you will start to be able to see a focused image of that person with your eyes. That is why I questioned how that second picture was taken! It certainly wasn't taken by that person in the picture, because I noticed no camera.

Zz.

 

[SHASHWAT PRATAP SING]

A human eye cannot focus on converging rays. The person, being the object, will see converging light rays entering his eyes. Unless he wears corrective glasses to compensate for that, he won't see a focused image.

if you say this then how the person is able to see himself in the concave mirror if the real image is formed behind him.

 

[ZapperZ]

if you say this then how the person is able to see himself in the concave mirror if the real image is formed behind him.
View attachment 265753

That is why I questioned how that image was captured! Was his head wired to a computer so that you get an image that formed on his retina? Think about it!

He's not holding a camera. Where is the camera that captured this picture?

Zz.

 

[A.T.]

A human eye cannot focus on converging rays.

So the seen image will get blurry, when you get past the focus point.

But you still should be able to see if it is flipped or not. And based on post #35 it seems like the seen image is not flipped, when you are between focus point and center of curvature.

 

[kuruman]

So where did I go wrong in post #35?

I don't think you went wrong; I did, forgetting that an inverted image on the retina is interpreted as erect by the brain.

 

[SHASHWAT PRATAP SING]

So the seen image will get blurry, when you get past the focus point.

But you still should be able to see if is flipped or not. And based on post #35 it seems like the seen image is not flipped, when you are between focus point and center of curvature.

What does this mean I am confused ZaperZ is saying something else and A.T. something else I am confused...

 

[SHASHWAT PRATAP SING]

A.T. please read my #38 reply and tell am I correct...

 

[A.T.]

What does this mean I am confused ZaperZ is saying something else and A.T. something else I am confused...

It would help, if you would answer the questions in post #43.

Just like ZaperZ I'm skeptical that, the flipped image was taken from between the center of curvature and the focus point. How where these points determined?

 

[SHASHWAT PRATAP SING]

It would help, if you would answer the questions in post #43.

Just like ZaperZ I'm skeptical that, the flipped image was taken from between the center of curvature and the focus point. How where these points determined?

No, it was a video explaing the rules of image formation in the concave mirror from there I took the screen shot (when the person approaches the centre of curvature and focus).
I think The camera was somewhat placed behind the person at a far distance and to show this image formation it zoomed,

Please check out this video- Please...

 

[hutchphd]

I think The camera was somewhat placed behind the person at a far distance and to show this image formation it zoomed,

The man is describing what the camera (you) are seeing not what he is seeing. He cannot see the image when it is behind his head. The camera is well beyond the center of curvature. Not a great video.

 

[ZapperZ]

I think The camera was somewhat placed behind the person at a far distance and to show this image formation it zoomed,

And that is my point in my previous post. The image was made by a camera way beyond the location of the focused image! This is where the rays are diverging once more and a focused image can be viewed with a human eye, or a camera in this case.

The entire time, you seem to make the point that the person in the image, i.e. the object, is the one making the observation. This is clearly wrong!

It is why knowing what's what is important. The camera that made that images here are NOT from real images, i.e. the light rays are diverging. They are from virtual images of that person (object).

Zz.

 

[hutchphd]

It is why knowing what's what is important. The camera that made that images here are NOT from real images, i.e. the light rays are diverging. They are from virtual images of that person (object).

Sorry but this statement makes no sense to me.
When the object (guy) is outside the focal length the image is real.

 

[ZapperZ]

Sorry but this statement makes no sense to me.
When the object (guy) is outside the focal length the image is real.

The focused image is real AT THAT POINT. If you go BEYOND that distance, it is no longer focused. You cannot capture it anymore on a screen.

Zz.

 

[ZapperZ]

Sigh.. I'm making the same sloppy notation that I'm telling the OP not to do.

1. our eyes can only focus on diverging rays. So once the eye moves further away from the focused image location, you can now see the image with the human eye. But you cannot see a focus image on a screen.

2. I should not have said real or virtual, because this is not correct.

3. Inside the length of the focused image, you cannot see the image with your eyes, because the rays are converging.

Are we OK now?

Zz.

 

[hutchphd]

The camera is focused on that image plane. If you put a screen at the image plane the picture in the camera would be unchanged (of course you couldn't interfere with the object...this is easier with a lens!)

 

[hutchphd]

I think we are OK if you agree with the above.! I can see why the OP was confused by the video though.

 

[ZapperZ]

The camera is focused on that image plane. If you put a screen at the image plane the picture in the camera would be unchanged (of course you couldn't interfere with the object...this is easier with a lens!)

If you put a screen farther than where the focused image is, you will not get a focused image.

If you try to view the image with your eyes at a distance father than the location of the focused image, you will see the object.

If you try to view the image at a distance smaller than where the focused image location is, you will not see a focused image with your eyes.

Is there anything contradictory to what you just said here?

Zz.

 

[hutchphd]

Nope. I think I said the screen needs to be put in the image plane (which is determined by object distance). So we are copacetic...hope the OP is.

 

[A.T.]

The man is describing what the camera (you) are seeing not what he is seeing. ... The camera is well beyond the center of curvature.

Would it be correct to say:
- For a distant camera the image flip happens when the object is at the mirror's focal point.
- For a camera at the object the image flip happens when the object is at the mirror's center of curvature.

Assume a camera that can focus converging rays or a pinhole camera to avoid the issue if the image is sharp.

 

[SHASHWAT PRATAP SING]

hutchphd by saying this-
The man is describing what the camera (you) are seeing not what he is seeing. He cannot see the image when it is behind his head. The camera is well beyond the center of curvature.
you have confused me even more.

Does this mean the real image what we are seeing and what the man is seeing is different... please reply..
A.T. AND ZAPPERZ please help me.