Understanding Reflection of Light on a Concave Mirror

[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.

 

[SHASHWAT PRATAP SING]

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!

Ok, Does this mean the person seeing in the concave mirror will see something different than us...

t 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).

What Does this mean ?
When the object (guy) is outside the focal length the image is real.

 

[hutchphd]

I am saying that if the camera were in the man's eye socket we would not see an image for the part of the trip between cc and focal point. There will be an image formed that is inaccessable to camera view.

 

[SHASHWAT PRATAP SING]

I am saying that if the camera were in the man's eye socket we would not see an image for the part of the trip between cc and focal point. There will be an image formed that is inaccessable to camera view.

I mean that the Real image what we are seeing-

And what the person is seeing in the concave are they both different-

Will this person(Object) see something different than us ?
Please tell this...

 

[A.T.]

I am saying that if the camera were in the man's eye socket we would not see an image for the part of the trip between cc and focal point.

Not an image at all for the entire range between cc and focal point? Or just not see a sharp image? Assuming a camera that can focus converging rays on the sensor (or a pinhole camera), would it not see an image like shown in the diagram?

 

[SHASHWAT PRATAP SING]

Not an image at all for the entire range between cc and focal point? Or just not see a sharp image? Assuming a camera that can focus converging rays on the sensor (or a pinhole camera), would it not see an image like shown in the diagram?

A.T. please reply me at my post-#63

 

[SHASHWAT PRATAP SING]

I am saying that if the camera were in the man's eye socket we would not see an image for the part of the trip between cc and focal point. There will be an image formed that is inaccessable to camera view.

hutchphd please reply me at my post-#63

 

[SHASHWAT PRATAP SING]

You really need to work on your description. Say exactly what you mean. Say exactly what each picture shows.

Try to avoid phrases like "image must be forming". If you are standing still the image is either there or it is not. It is not "forming". Try to avoid phrases like "real image in the mirror". The real image is not in the mirror. It is behind your head.

You have provided a number of pictures. Let us go through them slowly. Let us identify carefully what they show.

The first picture is of someone standing nose to mirror. There are two locations marked on the floor. One is marked "C". One assumes that this is at the center of curvature of the concave mirror. The other is marked "F". One assumes that this is at the focal length of the concave mirror.

This impression is reinforced by the fact that the distance from the mirror to C is approximately double the distance from the mirror to F.

But you are not asking about the situation in the first picture. So that picture is mostly wasted.The second picture is of someone standing between the focal point F and the center of curvature C. We see a blue shape drawn in space behind the person. One assumes that this is the place where the [inverted] real image of the person would form.

The person stands in the way of some of the reflected light that might otherwise go into forming this real image. As a result, the real image may not be viewable from all angles. However, this detail is irrelevant to the question that we are trying to address. Let us ignore that situation and proceed.The third picture is the same as the first but without the "C" and "F" labels drawn in. We can ignore it.The fourth picture is the same as the second. But instead of "C" and "F" labels and an image drawn behind the person, we have an arrow in front of him pointing toward the mirror.

This would appear to be an attempt to depict the person's line of sight toward the mirror.The final picture appears to have been taken from a little to one side as the man stands in the position depicted in the second and fourth pictures. That is, it was taken while the man is standing between the C and the F. It shows an inverted image.

This appears to be an attempt to show what the person himself sees. I may be mistaken, but that might be the camera's reflection in the person's right hand.If I understand your concern, it is that if we are looking away from a real image, we might not expect to be able to see that image. Certainly, our intuitive expectations learned from everyday life tend to support this. If person is standing behind me, I cannot see them when I am looking toward the front. If an image exists behind me, I might not expect to see it when I am looking toward the front.

This expectation is incorrect. The photographic evidence you have provided demonstrates that it is incorrect.

The light rays that hit your eyes are light rays that would have converged at the real image behind your head. But they did not converge there. Your eyes were in the way. The lens in your eyes (or in your camera) caused them to converge on your retina (or on the photodetector array in your camera) instead.

You "see" the image because the pattern of illumination on the retina (or on the photodetector array) corresponds to the shape of the person.

Human eyes are not designed to be able to clearly image converging light rays. It requires that they relax to a focal length longer than the distance from lens to retina. A focal length equal to the retinal distance would allow one to clearly focus on objects "at infinity". A shorter focal length allows one to focus on objects closer in. A longer focal length allows one to focus on images corresponding to convergent light rays -- that is real images behind the head.

There is no evolutionary call for eyeball lenses that can focus out past infinity. And little technical call for cell phone lenses that can do the same. The cell phone image of the reflected inverted man is not sharp. But with a narrow aperture, one can push the limits.

As jbriggs444 said, I think this is what happening in this case

 

[hutchphd]

Not an image at all for the entire range between cc and focal point

When I say image I mean that light from each point in the the object plane will be found at only one point in the image plane. This is a geometric optics construction (no diffraction limits) and assumes a perfect lens/mirror. With this (correct) definition for an image the statement stands. If you want to talk about fuzzy images and depth of field and circles of confusion that is fine, but it is a different and largely distinct conversation.

 

[SHASHWAT PRATAP SING]

The light rays that hit your eyes are light rays that would have converged at the real image behind your head. But they did not converge there. Your eyes were in the way. The lens in your eyes (or in your camera) caused them to converge on your retina (or on the photodetector array in your camera) instead.

You "see" the image because the pattern of illumination on the retina (or on the photodetector array) corresponds to the shape of the person.

As jbriggs444 says I think this is what happening in this case-

Please tell

 

[A.T.]

When I say image I mean that light from each point in the the object plane will be found at only one point in the image plane.

Replace the eye in my diagram with a camera that has a flat sensor at the back and a lens that creates an image in the plane of the sensor. Why couldn't this camera record a sharp picture?