In a well-lit room, a light source shines light on the objects in the room, some of which reflects back off the objects. We can make the light source brighter by causing it to emit more photons, which means more photons are hitting the objects and also reflecting off the objects and thus they appear brighter as well.

Therefore it seems that if I place a mirror in the room that reflects the light coming off of an object back to it, the object should get brighter, because more photons are now hitting it than there were before. But when I physically do this, the object doesn't get brighter.

This holds true even if I place the mirror on a part of the room that was shadier than the object, to eliminate a possible effect that the object is simply getting less light from the direction of the mirror than it was before:

enter image description here enter image description here

(higher-res versions: one two)

The floor is equally bright on the area that the mirror is reflecting the floor's light back at it and the area where it isn't.

Shouldn't the floor be brighter with more photons on it than before?

  • 31
    $\begingroup$ They do. Human eyes with their superb compensation mechanisms are poor judges of brightness. $\endgroup$
    – DKNguyen
    Commented Apr 15, 2023 at 19:07
  • $\begingroup$ Also, cameras usually compensate settings (aperture, speed, iso) to get more or less "middle" exposure. $\endgroup$
    – Pablo H
    Commented Apr 17, 2023 at 15:00
  • $\begingroup$ Draw me a line of how the mirror in this position reflects light back at the same source it came from. Consider that two different parts of "the floor" are not the same, and that diffuse reflection leads to the scattering of light. $\endgroup$
    – Flater
    Commented Apr 17, 2023 at 23:38
  • $\begingroup$ I'll also note that you've chosen a terrible background to look for this effect. I feel like there is a very faint square-shaped light spot on the floor in the second photo, but the variable coloration of the flooring material makes it impossible to tell what's variability in the tile color versus variability in the lighting. $\endgroup$ Commented Apr 18, 2023 at 19:08
  • $\begingroup$ If you were shining sunlight onto a sunlit floor, you'd see an increase in brightness. But since you've placed the mirror out of direct sunlight, you'll see barely any effect, if at all. $\endgroup$
    – CJ Dennis
    Commented Apr 19, 2023 at 3:16

7 Answers 7


"Why don't objects get brighter when I reflect their light back at them?"

They do. It's just that:

  • To notice these phenomena in general, a trained eye is required, in part because we're so used to things being illuminated by bounced light that we don't even know what to look for, and in part because our brains play tricks on us; in addition, to have any chance to notice a difference in the kind of setup you're attempting requires very specific lighting conditions.
  • Bounced light is often less bright than the incoming light (you don't get total reflection), so the effect can be rather subtle.
  • If you just want to observe light bouncing off of a diffuse surface, a mirror might not be the best tool for the job.

For example, here's one thing that's a consequence of light bouncing around that you probably didn't think of, that can be observed in your images. For those points on the floor that are in shadow, the direct path to the light source is blocked. That is, no light is coming to them directly from the light source. Yet, the shadows are not pitch black - why is that? Why don't shadows appear as in the first image below?

enter image description hereenter image description here

It's because these areas are illuminated by the light that's bouncing off of the walls of your room (or, if you're outside, by the bouncing in the environment, and coming from the sky). The walls are diffuse reflectors (unlike a mirror), meaning that the light bounces off of them in all directions, basically randomly.

Let's look at another example - the back side of the mirror.

enter image description hereenter image description here

The casing is black, but it's also partially reflective (more like a mirror than like the walls). You can see the reflection of the floor in it - you're not seeing the dark color of the casing. The reason for this is because there's light bouncing off of the floor, and hitting the back side (and then bouncing off of the casing, and hitting your eye).

But also, take a close look at the floor itself, right next to the back side of the mirror. It's a little brighter than the floor further away. That's light bouncing off of the casing, and hitting the floor. In the image on the right, I tried to adjust the contrast so that the effect is better seen. Notice a rectangular band of light, with a dark strip in the middle - I'm guessing the dark strip is produced by some sort of writing or a label on the surface that's less reflective than the casing itself.

Now, I said that a mirror is a bad choice - if you want to see light bouncing of off a diffuse reflector. That's because, unlike the walls of the room that reflect incoming light in all directions from every point, a mirror reflects each light ray in one direction, and this direction depends on the angle of incidence (the angle at which the light is coming in with respect to the surface of the mirror).

You can certainly make objects brighter by bouncing light using a mirror. It looks like this:

enter image description here

You get a sharply delineated bright area.

The floor is equally bright on the area that the mirror is reflecting the floor's light back at it

The problem is, the mirror is not reflecting the light coming from the floor back at it. When you place a mirror vertically on the floor like that, light that bounces off of the floor and hits the mirror is not going back to the floor, but is reflected towards the ceiling instead. Some if it hits your eye - which is why you can see the reflection. Think of a little billiard ball bouncing - if it came from the direction of the floor, it wouldn't come back to it with this setup. Some of the light coming from the walls and the ceiling is reflected by the mirror back on the floor, but this just compensates for the light that would have otherwise come from behind the mirror, that you are now blocking. The effect is too subtle to notice.

enter image description here

Also, trying to do it by making the mirror face the floor is not going to work well. You're trying to re-bounce the light coming off of your floor (a diffuse reflector) but you're not seeing much because the effect is so faint, it's hardly noticeable. Also, by directing the mirror at the floor and by bringing it closer and closer, you're actually blocking the incoming ambient light, so you might lessen the overall brightness.

enter image description here

Instead of a mirror, take a white sheet of paper, and place it against one of the walls, so that the surface of the paper faces the light. Try to do it relatively close to a bright light source. It's better if you can arrange things so that there's only one source of light in the room (so, turn off computer monitors, TV sets, etc.)

enter image description here

In the image above, notice that the darker area between the paper and the brightened area is not the shadow of the paper - that's just the brightness of the wall without the paper present. You can see the shadow way behind there.

Or, take something colored, and observe how the light it bounces takes on the color of the object (that's why the object appears to us to be colored that way in the first place - it only reflects certain wavelengths of light - only certain parts of the rainbow):

enter image description here

  • 4
    $\begingroup$ Best answer so far +1 $\endgroup$
    – justhalf
    Commented Apr 17, 2023 at 8:51
  • $\begingroup$ The image processing on the OP's photos is quite on point. Nicely done. $\endgroup$ Commented Apr 18, 2023 at 7:48
  • $\begingroup$ This answer is well done but I’m asking specifically of reflecting an object’s light back at it, not reflecting additional light from the light source onto the object (which is what the contrast-enhanced image and the mirror and papers at the end show). There is a thermodynamic argument that it’s not possible (see: what-if.xkcd.com/145), and I was wondering if that is valid and if so, why. The answer seems to be that it is possible just the effect isn’t very large? $\endgroup$
    – Cloudyman
    Commented Apr 19, 2023 at 7:06
  • $\begingroup$ @Cloudyman - the xkcd argument says that it is not possible to use a magnifying glass to make the spot in focus hotter than the surface of the source, not that light can't bounce back and forth multiple times. I guess you're right about my examples not showing exactly your scenario, but my point was, if it bounces once, no reason why it can't bounce again - light is light, doesn't matter if it's coming from the light source, or if it's reflected. 1/2 $\endgroup$ Commented Apr 22, 2023 at 16:35
  • 1
    $\begingroup$ You should be able to observe a second level bounce if you are in a completely dark room, and you take a narrow-beam LED flashlight (or maybe a bright laser pointer), have it hit a wall, and then use a mirror or a white piece of paper to bounce the light that has been reflected off of the wall back at the wall (i.e. the light source can be behind the mirror, so no direct light hits the mirror itself). Might be challenging to film though - if I succeed, I'll add it to the answer. 2/2 $\endgroup$ Commented Apr 22, 2023 at 16:35

Maybe. All that matters is the total flux of light from the solid angle subtended by the mirror from any point you are investigating, plus possibly your hand. It is possible that this is still lower than it was without the mirror, if ambient light is intense enough.

It is also possible that the total flux is actually slightly higher, but the direct illumination is so intense that the relative difference is too small to be observed. The response of the human eye is approximately logarithmic, as stated by the Weber-Fechner law. If the mirror subtends a solid angle of $1\ \mathrm{sr}$, which is about 16% of the upper hemisphere ($2\pi\ \mathrm{sr}$), and the total flux from that solid angle is 10% higher than it would be without the mirror, the resulting relative increase in illuminance 1.6 % would be probably difficult to see.

A very rough estimate

Imagine that the direct illuminance from the Sun is $100000\ \mathrm{lx}$ and from the rest of the upper hemisphere (sky) $10000\ \mathrm{lx}$. Since the surface is not in shadow, direct illumination is unchanged, but the mirror blocks some $1600\ \mathrm{lx}$ of ambient light, for a total of $108400$ lux with no reflections.

Let's say the surface is perfectly matte (or Lambertian), and reflects 50% of incident light, or $54200\ \mathrm{lx}$. Of this about 16% will reach the mirror. In fact it will be even less, since with a Lambertian surface most reflected light is concentrated around the normal, which is clearly not the case. So maybe $6000\ \mathrm{lx}$ of scattered reaches the mirror after one reflection. Even if the mirror is perfect, it can only reflect back that much.

Neglecting multiple reflections, the total illuminance is then $108400 + 6000 = 114400$ lux, or about $10^{5.0584}$, compared to $110000 \approx 10^{5.0414}$ lux without the mirror. This could be measured with specialized equipment or maybe with a consumer camera in manual mode, but is too small to see with human eyes, especially if it cannot be compared side-by-side.


The brightness of a point on the floor is well approximated by the integrated/average brightness of everything that you could see if the pupil of your eye were located at that point. You can use that fact to get some intuition for the effect of the mirrors.

As you move your point of view around the floor, the solid angle subtended by the mirror varies, and the fraction of the mirror through which the floor is visible also varies, but they vary only gradually. There is no point on the floor from which a small motion causes a large change in the average visible brightness, except for light coming directly from the sun, which will be occluded completely by a small motion if the sun is close to an opaque object from that vantage point. Therefore, the only sudden variations in brightness that you should expect to see on the floor are the edges of shadows from direct sunlight. In your second image there may be an area of increased brightness, but with such a broad/fuzzy edge that it's practically impossible to see.

Also consider that the view blocked by the mirror is likely to be of something fairly bright, such as a building or sky, even if the mirror itself is in shadow. If the picture were taken indoors, you would most likely block a view of the wall or ceiling, which is likely to be about as bright as the floor.

  • $\begingroup$ A tiny correction: brightness should be approximated not by average brightness (zeroth moment of illuminance with respect to the cosine of the normal angle, $\oint_{2\pi} E \ \mathrm{d}\Omega$), but by flux (first moment, $\oint_{2\pi} E \cos \theta \ \mathrm{d}\Omega$). Rays further from the normal are dispersed over a proportionally larger area. For constant illuminance over the entire hemisphere this is just a factor of 2, but if you have a bright source near the horizon, the results will be very different. $\endgroup$ Commented Apr 17, 2023 at 8:34
  • $\begingroup$ @Martin'Kvík'Baláž My idea was that if you look normal to the surface then the flux through your pupil has the right angular factor, and the "integrated/average brightness" is the total flux through the pupil. Admittedly it doesn't work if you look around. It might be better to imagine a camera with a 180° FOV pointed normal to the surface. $\endgroup$
    – benrg
    Commented Apr 17, 2023 at 19:37

The answer is a mix of both physics and biology. The way our eye converts optical to electrical signals is due to the photoelectric effect. Sources which emit more photons/sec will look brighter than other which don't. However the human eye is designed by nature to capture logarithmically the brightness of a source meaning it hardly recognises the difference between 2 bight sources and is overexcited when it sees differences of 2 dim sources so the human eye isn't a good detector that's why you may be fooled.


Other answers have valid points, but I don't see where they mention that you just plain have the angles wrong for what you are trying to accomplish. In the first photo, a vertical mirror will clearly not reflect light back at the floor, but upwards instead. The second photo also looks like it would be reflecting the floor's image more horizontally or upwards rather than back at the floor. If you want to reflect the floor's light back at itself, the mirror would have to be parallel to the floor.

Martin's answer does a good job explaining the relative brightness and how it isn't linear. Think about a 100 watt bulb lighting an entire room. The wall might seem bright, but if the light bulb illuminates 200 square feet to that brightness, the mirror reflecting the light from 1/10th of a square foot of the wall isn't going to make much of a difference.

And you might be thinking of the mirror as emitting a rectangular beam of light. After all, when you use it to reflect the light from the sun that is what it looks like, you see a bright rectangular patch. But that is only because it is reflecting a small bright source. The mirror is actually reflecting the sky around the bright area of the sun's reflection, but it is so dim in comparison and spread out that it isn't noticeable. In your example there would be a small general increase in brightness if the angles were correct, but it would be spread out and hard to differentiate.


They do.

Your experiment is not set up right and fails to detect it. Specifically, you have two problems:

  • You expect the brightness increase to be easily detectable by eye. You need a precise measuring device, because in many cases (such as flat mirrors) the brightness will increase by a small fraction. A camera will do but you will need to put it in manual mode, with no auto-balance, auto-exposure, etc.
  • Your control changes too many variables. You compare floor with mirror-on-floor, but the mirror itself casts a shadow regardless of reflection. What you need to do is cover the mirror with a black sheet and compare to that.

If you do both of the above, you should be able to observe that some spots on the floor are darker when the mirror is covered and brighter when it's not.

Actually, you don't even need to do the experiment. You can simply observe that a room with white walls is brighter inside than a room with black walls (or black curtains hanging on the walls without obstructing the windows).


Despite the many (and some highly-upvoted) answers to the contrary, I believe the answer is actually that they can't, because if they did they would violate Conservation of Etendue and by extension the 2nd Law of Thermodynamics.

From the Wikipedia:

The etendue of a given bundle of light is conserved: etendue can be increased, but not decreased in any optical system. This means that any system that concentrates light from some source onto a smaller area must always increase the solid angle of incidence (that is, the area of the sky that the source subtends). For example, a magnifying glass can increase the intensity of sunlight onto a small spot, but does so because, viewed from the spot that the light is concentrated onto, the apparent size of the sun is increased proportional to the concentration.

As shown below, etendue is conserved as light travels through free space and at refractions or reflections. It is then also conserved as light travels through optical systems where it undergoes perfect reflections or refractions. However, if light was to hit, say, a diffuser, its solid angle would increase, increasing the etendue. Etendue can then remain constant or it can increase as light propagates through an optic, but it cannot decrease. This is a direct result of the fact that entropy must be constant or increasing.

Conservation of etendue can be derived in different contexts, such as from optical first principles, from Hamiltonian optics or from the second law of thermodynamics.

The floor acts as a diffuser in this case (or maybe more accurately, we can see it due to diffuse reflections which I understand to mean that it increases the etendue).

Also see Can you use a magnifying glass and moonlight to light a fire? and the related follow-up physics SE question.

To summarize my understanding of it, basically you cannot take light from a light source and focus it to make some other spot brighter than the light source. i.e. the sun is ~6000K so you can magnify the sun to burn paper, but the moon is only ~400K and you can't magnify this to burn paper. You can easily verify this by looking at a wall through a magnifying glass -- even though you can get more of the wall to focus onto your pupil, the brightness of the wall remains constant.

Basically if we take the light from the floor and redirect it onto a completely unlit surface, that surface could only become as bright as the floor, no matter how many reflections or magnifications are done etc.

But that's onto an unlit surface -- couldn't we reflect the light back onto itself to make a spot brighter?

Proof by contradiction: let's say we could. The light from the floor goes into some optic system that redirects the light back to the floor and makes some portion of it brighter. Now we could introduce another optic system that magnifies this brighter spot (e.g. like magnifying the sun (that's small relative to the sky) with a magnifying glass) and projects it onto an unlit surface, to make that surface brighter than the floor initially was. This violates conservation of etendue, therefore it's impossible.

From a thermodynamic point of view, it would mean we could focus an object's thermal radiation such that to make another object hotter than it. Or even more blatantly, to make an object able to heat itself up forever with its own thermal radiation! Also impossible -- therefore we can't do it.

More straightforwardly, it's the brightness theorem (from the Wikipedia again):

A consequence of the conservation of etendue is the brightness theorem, which states that no optical system can increase the brightness of the light emitted from a source to a higher value than the brightness of the surface of that source (where "brightness" is defined as the optical power emitted per unit solid angle per unit emitting or receiving area).

In this case the floor becomes the light source. This brightness cannot be increased -- if we reflected the light back onto the object and it became brighter (even a tiny bit), it would violate the brightness theorem.

To address the other answers in turn:

@Filip Milovanović: Despite being well-written and providing examples of reflections making objects brighter, the examples all focus on light from the initial light source being redirected to make a spot brighter than it otherwise would be -- but the question was about the light from the object itself reflected back at it.

@Martin 'Kvík' Baláž: It's a relevant calculation showing the effect would be small, but as above if we could do it in reality then it seems we'd violate some laws of physics. I would want to see a measurement of this in reality with said specialized equipment.

@benrg @appliedSciences: It is compelling that the effect would be hard to see, but, as per above, it seems impossible.

@Jason Goemaat: It's a relevant point that the angles may indeed be wrong, but as above it seems no combination of angles, mirrors, lenses (be they concave or convex), etc., would work.

@gomennathan: The only point not already covered is that a room with white walls is brighter than a room with black walls -- that is true but it can be explained by the white walls reflecting more light/becoming a brighter light source than a black wall, which can then in turn make further objects brighter than they would be -- but, not brighter than the walls themselves! In other words it also doesn't answer the question of reflecting the light from the wall back to itself to make the wall brighter.

There is perhaps an important caveat -- the light from the light sources that reflects due to specular reflection would continue to have the same etendue as the light from the light source, as opposed to the increased etendue from the diffuse reflections. This specular component therefore should be able to be reflected to increase the brightness of the floor. But in this case, it isn't the light from the floor per se making the floor brighter, rather it's a more efficient use of the light from the light source (that wasn't initially diffusely reflected) making the floor brighter.

I wasn't aware of these distinctions when initially writing the question, but I see that the spirit of the question was such as to be asking why light from an object that has only diffuse reflections, can't be reflected to make that object brighter -- or equivalently, light from a light source directly itself. @Martin 'Kvík' Baláž's answer, for example, does make this assumption of a perfectly diffuse (i.e. Lambertian surface).

To conclude, it seems like you ought to be able to do this, but you can’t trick nature and you can’t get free energy …

I am open to this being wrong, but I would want to see an example of this actually happening in reality in an observable way (using specialized equipment if needed), along with an explanation of why the above reasoning doesn't follow, to be convinced!


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