# Depth effect in mirror reflections

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## Main Question or Discussion Point

A simple question but the answer is not apparent to me. I just know so little about optics.

When I look at an object, I gather an impression of depth because my eyes have to adjust to the object's distance from me. I assume this is due to the angle of the light from the object - light coming from close objects strikes my eye at a different angle from farther objects. Light is coming from different spatial points.

When I look at objects in a mirror, I see virtual objects that are rendered on the mirror's surface. And yet, I need to focus my eyes on these objects in ways that match the real objects (or so it seems, perhaps I am mistaken).

However, all of the virtual objects (the mirror images) lie on a flat plane. Light strikes the mirror and light is then emitted back at the exact same angle. However as far as I can see - naively - all light from that surface should obtain the same angles as if the (virtual) objects were all co-located on that plane. I should have no sense of depth for mirror images, but I do.

Why?

## Answers and Replies

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Well, isn't that a beautiful thing! And for the life of me, I couldn't see that. Thanks.

PeterO, Nik_2213 and Dale
I never realised the seemingly simple process of reflection was so complex. So now I am aware of specular and diffuse reflection. These are not different things at all, as I understand it, more properties of illuminated objects. I assume the reason for full-colour images in mirrors relates to the reflecting surface - it must be both uniformly smooth and totally neutral in regard to wavelengths reflected (ie all incident wavelengths are reflected)?

In effect, all surfaces are "mirrors" because we are really talking about the same phenomenon.

sophiecentaur
Gold Member
In effect, all surfaces are "mirrors" because we are really talking about the same phenomenon.
In fact there are basically two types of reflection (most objects actually exhibit both effects). The mirror or 'specular' type of reflection is off a perfectly flat surface. Diffuse reflection is what you get from rough surfaces and diffuse reflection does not produce a discernible image because the angles of incidence and reflection are different for each tiny part of the surface. Even the very best quality mirrors will have some dust or dirt on them which will always degrade an image a bit. You pays your money and you takes your pick of which you want.

Yes, but isn't the underlying effect simply that a surface either absorbs or reflects incident light? Any object that we can see is visible because in effect, a "reflection" of it impinges on our eyes. So the physical phenomenon is how objects absorb and reflect light, whether as a mirror or as a coloured object depends on the object's surface - its smoothness and albedo?

Actually no, I must be wrong in assuming this. A smooth black surface (low albedo) still reflects images. So something else is going on?

sophiecentaur
Gold Member
So something else is going on?
Yes. The geometry is "going on". So is absorption.

sophiecentaur
Gold Member
Any object that we can see is visible because in effect, a "reflection" of it impinges on our eyes.
We do not see it's reflection. We see the reflection of other sources of light being reflected in its surface. Of course, there are self luminous objects but that's another thing for later.

Sorry, you'll have to explain that to me. All there is, as far as our eyes are concerned, is light - EMR of particular wavelengths. We see an object because incident light is reflected back to us. A black object is black, as I understand it, because all incident light is absorbed. That is, we do not have any reflected light impinging on our retinas. Yet it appears my understanding is flawed - what is reflected back to us also depends on whether the surface is smooth or rough. A black object is only black when it is not smooth, that is, its colour depends on whether light incident on its surface is reflected either diffusely or specularly. Which means that rough surfaces absorb light more readily than smooth surfaces?

sophiecentaur
Gold Member
We see an object because incident light is reflected back to us.
I was taking my lead from your reaction to the diagram, posted by @A.T. but the basic mirror phenomenon is only part of how we construct our internal map of what we are looking at.

Your take on the topic isn't the whole story. 'Seeing' is something in our brains. An object with a mat surface is 'easy' because we can only identify the light reflected from its surface as coming from the surface. A mirror surface is pretty well invisible and we strive to identify 'something' so we 'see' the objects reflected in the mirror (perhaps the most obvious interpretation of what's in front of us). This happens even when the surfaces of the object are curved and at different angles. Two people can look at the same slightly shiny coloured object and one may 'see' a surface when the other 'sees' the reflections in the surface. The brain is all the time trying to make sense out of even paradoxical visions - especially when depth perception is involved.

Yes, but while I grasped the issue of depth perception in mirror images, now I am perplexed by the underlying physics. Why do rough surfaces appear solidly coloured and do not reflect images while glossy ones do the opposite? The mirror is a surface at the micro scale and the phenomenon is how light is emitted from that surface. Our sense of the reflected light having depth and appearing as an object is just how our visual system responds to the light (and as explained, the angles are what counts).

However, at the end of the day, we just have a surface emitting EMR at visible wavelengths. The colour of the surface depends on the absorptivity, I thought, yet now it seems the surface finish is more important.

Take a black object with a flat plane surface perpendicular to my line of gaze. If I have a light source somewhere above me, that surface will be black if it is a rough surface (ie matt). I thought this was because incident light is absorbed by the object and little is reflected back at visible wavelengths. Yet if that surface is highly uniform, perhaps precision ground and polished, visible wavelengths ARE reflected back. So well in fact that the surface will appear as coloured and not black. What IS affected in such cases seems to be luminance which says something to me about the bundling of light rays from the surface.

So what seems to count most, is the extent of specular "reflection" from a surface. Smoother objects reflect light more than rough objects, in particular ways. Which again, to my naive perspective, means that smoother objects have higher albedo? Yet I have no idea what phenomenon might drive that?

And yes, my lack of physics and optics knowledge causes me to confuse terms and concepts but hopefully you can see what I am getting at.

sophiecentaur
Gold Member
And yes, my lack of physics and optics knowledge causes me to confuse terms and concepts but hopefully you can see what I am getting at.
I think you are not wanting to include perception in this. The optics of the situation can be broken down into fairly simple ideas. The way light is absorbed, bent and reflected on its way from the lamp to the eye is not, in itself hard to appreciate. The problem is how the brain makes sense of things.

I agree about the brain part, and that was partly my original question which was answered to my satisfaction. But I am mystified by what is happening regarding smoothness of an object. Let me start with a small part of my question. Do you agree that an object with a rough surface, say roughened glass, which is one colour, say dark blue, when placed under a white light source is reflecting light at wavelengths around 450nm? I understand this to be because other wavelengths have been "absorbed" and primarily blue wavelengths are reflected? We say the object is blue because those wavelengths stimulate our eye and brain in that way, but physically there is no blue, just light at largely 450nm. Further, regardless of what other objects in the room are also reflecting the white light at various wavelengths, the blue glass absorbs all of that and only reflects back to us blue wavelengths. Hence the surface is a uniform blue.

If I am approximately right in saying that, I would go further to ask what would happen if that glass were highly polished and precision ground? I would say that the other objects in the room that are reflecting light at various wavelengths will now be reflected in the blue glass. That is, images will be visible on the blue glass. They will be coloured. In effect, for those areas on the glass where an image "appears" to my eye, the glass is reflecting wavelengths at other than 450nm.

So what light is absorbed and what is reflected depends upon the surface smoothness, not just the object's absorptivity. The smoother the surface, the more light is reflected as specular reflection.

What I am getting at then is that light is reflected from all surfaces (or not, for black surfaces) to create images in our brains. Mirror images are the same process, with the difference being the smoothness of the surface. A flat plane surface, polished to varying degrees, will present as varying degrees of reflected detail. For the matt surface, no images are seen other than the primary object (highly diffuse reflection). A somewhat polished surface will present as having perhaps fuzzy images of surrounding objects (mix of diffuse and specular reflection). A highly polished surface will present clear images of surrounding objects, such images having colour different from the surface colour of the plane surface (highly specular reflection).

So perception of an object (eg colour, or surface irregularities) depends both on the surface's absorptivity (wavelengths reflected) AND its smoothness (kind of reflectance).

Am I more or less on track?

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Drakkith
Staff Emeritus
Why do rough surfaces appear solidly coloured and do not reflect images while glossy ones do the opposite?
Specular versus diffuse reflection like you said. The difference between the two is that in diffuse reflection the object's surface is usually very rough, such that the incoming light gets scattered in many different directions.

The colour of the surface depends on the absorptivity, I thought, yet now it seems the surface finish is more important
The surface finish doesn't affect the color of the object in most cases.

Take a black object with a flat plane surface perpendicular to my line of gaze. If I have a light source somewhere above me, that surface will be black if it is a rough surface (ie matt). I thought this was because incident light is absorbed by the object and little is reflected back at visible wavelengths. Yet if that surface is highly uniform, perhaps precision ground and polished, visible wavelengths ARE reflected back. So well in fact that the surface will appear as coloured and not black. What IS affected in such cases seems to be luminance which says something to me about the bundling of light rays from the surface.
The color is determined by the difference in reflectivity between the different wavelengths when impinging on a diffuse surface. A highly specular surface that reflects mostly one color would look like a mirror that tints the reflected image a certain color.

So what light is absorbed and what is reflected depends upon the surface smoothness, not just the object's absorptivity. The smoother the surface, the more light is reflected as specular reflection.
It's not the amount of light that gets reflected that makes an object a mirror, it's the degree of disruption the surface induces in the incoming cone of light. A mirror reflects the cone of light with very little disruption, while a diffuse surface reflects that cone of light in all directions. Shine a flashlight in a mirror in a dark room and look how the mirror reflects the cone of light coming from the flashlight versus shining it on a diffuse wall.

jbriggs444
Homework Helper
2019 Award
What I am getting at then is that light is reflected from all surfaces (or not, for black surfaces) to create images in our brains.
It might be worthwhile considering a person in a well lit room looking out through a dirty window onto a city street.

The person can choose to look through the window, seeing a car passing under a street lamp. An image of the street lamp and the pattern of textures and car edges is focused on the retina. The visual cortex and the brain do edge detection and convey the impression of a moving car. There is feedback to the eye itself, manipulating the lens to optimize edge sharpness so that the car remains "in focus". Spurious image elements that are not in focus or are not properly duplicated in both eyes are ignored.

The person can focus in tighter on the reflected image of his own face in the glass. It is the same thing all over again. The image of the face is sharp. The visual cortex and brain do edge detection and convey the picture of a face. Feedback to the eye keeps the facial image in focus. Spurious image elements (e.g. street lamp and car) are ignored.

The person can focus in more tightly yet and see streaks and flecks of dust on the window pane. It is the same thing all over again. The streak edges and dust motes are sharp. The visual cortex presents the image and works with the lens to retain a crisply focused image. Spurious image elements (e.g. a reflected face or a street lamp or car) are ignored.

Or you can take a final step and focus on the tip of your own nose. It is hard to see because the two eyes disagree about what is there and tend to suppress it. But close one eye and there it is -- plain as the nose on your face.

All of this is subject to conscious control. One can exercise that control and choose which version of the scene to look at. With a little practice is is even possible to watch carefully and see the automatic focus and eye alignment steps at work.

With even more practice, and a suitable set-up, it is possible to exert manual control and de-couple binocular vision alignment from the auto-focus mechanism in one's eyes. (i.e to go cross-eyed at will without losing focus). By default, the two are tied tightly together. As the eyes cross to track a nearby object, the lenses automatically focus in tighter, expecting to see an object at the predicted distance. Staring at pegboard and aliasing the columns between the two eyes allows one to decouple those responses.

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sophiecentaur
Gold Member
@Graeme M The most successful way to deal with Science is usually to deal with one bit at a time and, once you have got all the individual parts, put them together. I think you are trying to do the job all at once, which is definitely the hard way through.

A.T.
So perception of an object (eg colour, or surface irregularities) depends both on the surface's absorptivity (wavelengths reflected) AND its smoothness (kind of reflectance).
Note that it's not just about smooth vs. rough, but also about how exactly the roughness is structured:
https://en.wikipedia.org/wiki/Structural_coloration

Sophiecentaur, I suppose so, but this seems a very small part of the overall matter of things. Thank you (and the other commenters) for the very good comments that have helped me grasp how mirrors work. I am still hung up on the question of how opaque objects reflect light. If you can bear with me a little longer.

Here is Wikipedia:
• Light arriving at an opaque surface is either reflected "specularly" (that is, in the manner of a mirror), scattered (that is, reflected with diffuse scattering), or absorbed – or some combination of these.
• Opaque objects that do not reflect specularly (which tend to have rough surfaces) have their color determined by which wavelengths of light they scatter strongly (with the light that is not scattered being absorbed). If objects scatter all wavelengths with roughly equal strength, they appear white. If they absorb all wavelengths, they appear black.
• Opaque objects that specularly reflect light of different wavelengths with different efficiencies look like mirrors tinted with colors determined by those differences. An object that reflects some fraction of impinging light and absorbs the rest may look black but also be faintly reflective; examples are black objects coated with layers of enamel or lacquer.
As mentioned earlier, when I read this I take it to mean that any surface will either absorb some or all wavelengths present, or reflect some or all wavelengths. As I understand it, this is due to the photons exciting electrons at either resonant or non-resonant frequencies. Those wavelengths of light that cause electrons in the object's surface to resonate will be converted to thermal energy and are not re-emitted as light. Those that do not cause resonance are re-emitted as light.

When an opaque surface is rough enough, the re-emitted light is emitted in diffuse form, that is at angles determined by the various angles of incidence on the many planes of the surface at the micro-scale.

When an opaque surface is smooth enough however, the re-emitted light is emitted in specular form, that is at the same angle as the angle of incidence. This is because the surface has far fewer planes at the micro-scale, perhaps just one.

But what I am tripping up on is what this means for surface reflectance.

It seems to me that the initial condition of what light can be reflected or absorbed is determined by the composition of the object, not its surface smoothness. An opaque object that reflects blue light wavelengths absorbs other wavelengths - that is, green, red, yellow, etc wavelengths are absorbed and converted to thermal energy. I take this to mean that any object, regardless of surface smoothness, will absorb and reflect according to composition. A smooth blue cube is blue, as is a rough blue cube (bear in mind that I really mean something like that the EMR field sampled at a given point in space by a measuring device pointed at the cube will peak at blue wavelengths for the visible spectrum).

However, as per our discussion, if our cube has a smooth enough surface it will reflect wavelengths other than blue. That is how we can have mirrors and how any smooth enough surface can reflect images of objects that shine on the surface being observed. In that case, the EMR field will be sampled with different peaks for the visible spectrum according to where on the cube's face we point our measuring device.

My best guess at explaining this is that not all non-blue wavelengths are fully absorbed. Other wavelengths are partly absorbed and partly re-emitted according to the cube's composition (ie arrangements of atoms). Thus for a rough surface (ie the case of diffuse reflection) re-emitted non-blue wavelengths are at lower intensity than blue wavelengths (due to part absorption). I guess this might mean they are either scattered more by the rough surface or are scattered at the same "density" as blue wavelengths but affect the measuring device less. In the case of our eyes, the blue wavelengths are intense enough to activate cone cells, but the other wavelengths present are not. For a smooth surface however, the scattering is much less to non-existent and so the non-blue wavelengths achieve sufficient intensity to affect the measuring device noticeably. In the case of our eyes, to activate cone cells. This is why a uniformly blue (or black, for example) surface will not appear uniformly blue or black when it is very smooth.

I think if I boil it all down, where I may have been going wrong is to believe that when sources say light is absorbed or reflected, I took that as an absolute statement. I suspect it is not.

Have I grasped what's going on? If I am way out, I'll leave it there.

A.T. Structural colouration! That's brilliant.

sophiecentaur
Gold Member
@Graeme M It strikes me that you want several complete chapters of an optics text book to be condensed into just one long post on PF. You really can't hope to achieve this. Each of the mechanisms that affect light as it travels from one medium into another need some significant attention on its own. On top of the straight optics, there is the psycho visual process that goes on in our eyes and brains.
I took that as an absolute statement.
You can't afford to do that very often in Science.

Drakkith
Staff Emeritus
No real surface has 100% absorption or reflection. The Moon's surface is blacker than asphalt in many places, but it still reflects enough light to be the brightest object in the night sky. And all real mirrors will absorb some portion of the incoming light.

Regarding the example of the blue object, you
Do you agree that an object with a rough surface, say roughened glass, which is one colour, say dark blue, when placed under a white light source is reflecting light at wavelengths around 450nm?
This may be one source of your confusion. An object that we call blue usually reflects all visible wavelengths, more or less. If you shine white light on it and look at the spectrum of reflected light you may see an increased reflectivity over some wide range and not at all at a specific, single wavelength. Shine a red laser pointer on objects of different colors. You'll be very lucky to find an objects that does not reflect the some red light back, no matter what is the object's color.
The same thing is true when the object is polished to make a mirror. A "blue" mirror will show images of objects in all different colors. Same for a "black" mirror or any other color.

You should not confuse the notion of "black body" as a model of a perfect absorbent object with zero reflectance with an actual object that we call "black". They are not the same thing. Shine you laser pointer (of any color) on "black" objects and you will see that they are "black" but not "black bodies". I can see my face in natural colors as reflected in my black shiny cell phone display frame.

Thanks again for the various comments. I think maybe I am making it all sound harder than it is, my question about surfaces is really very simple conceptually though I understand there is much complexity behind that.

As I mentioned, conceptually I had in mind that objects are perceived as being a particular colour (eg blue) because they reflect more light at those wavelengths while other wavelengths are "absorbed". Most sources I read say that. Then, I thought of reflections (ie virtual images on a surface) as some different thing. Now I realise that each is an aspect of the same phenomenon.

All that happens is that incident light is reflected back by objects. The smoother an object's surface, the more the angles of reflected light approach the same angle of incidence. Rough objects reflect light that is dominated by wavelengths that correspond to the object's "colour". This is due to diffuse reflection. Smooth objects reflect light that is dominated by whatever dominant colour the incident light is, with some attenuation caused by the surface colour of the smooth object. This is due to specular reflection.

In both cases, the same phenomenon is occurring - light from all local sources hits a surface and is reflected to some degree. Generally speaking, the smoothness simply acts to reduce the homogeneity of the light field reflected from the object. The smoother an object, the less it is a uniform colour, as perceived by us. Conversely, the rougher an object the more uniform is the perceived colour.