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Imagine a mirror house i.e. completely made of mirror with no traces of whatsoever light absorbent.

Now, you introduce a light beam into that room and observe somehow through a hole.

Would the room be

  1. complete dark - since nothing is actually absorbing it!! or would it be
  2. very bright - since light is getting reflected from every place possible?

What I want to compare the above situation is with a white washed room.

EDIT:::Clarifications

This question is the result of a debate with my friend, who says the room would be dark and me who says otherwise.

Now being said that let's move to specifics...let's assume that the room is a cuboid and the light source is say a light bulb i.e. an isotropic source like a typical room at ceiling. and the observer with infinitesimally small view port looking from one of the wall presumably neither the ceiling nor floor.

I hope this suffices.

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    $\begingroup$ I think you might be interested in my answer to this question: How are we able to view an object in a room with bulb?. It's about specular reflections, but it talks about the appearance of a room entirely made of ideal mirrors (other than the viewer and the light source). My viewer is large and absorbent, but as AdamRedwine notes you can't really have an infinitesimal viewport, and the viewer always absorbs the light it uses to form an image. $\endgroup$
    – Kevin Reid
    Commented Nov 1, 2011 at 15:06

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If the walls are perfectly reflecting then unless your viewpoint is perfectly normal to a single perfectly flat wall then all rays from your eye will ultimately intersect with the bulb and so the room should appear as bright as the bulb filament.

It's just an integrating sphere - used to make a uniform bright field to test optics.

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    $\begingroup$ ""eye will ultimately intersect with the bulb and so the room should appear as bright as the bulb filament."" Violates 1st law. $\endgroup$
    – Georg
    Commented Nov 1, 2011 at 17:13
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    $\begingroup$ This is correct, it is the same as Olber's paradox for the images of the original light source, and it doesn't violate 1st law, because you assume that the candle is on constantly, filling up the room with thermal radiation until it is full. $\endgroup$
    – Ron Maimon
    Commented Nov 1, 2011 at 19:02
  • $\begingroup$ You can't reach a higher surface brightness than the filament true. But is that the same as any view into the box intercepts the filament? $\endgroup$
    – Martin Beckett
    Commented Nov 1, 2011 at 19:04
  • $\begingroup$ It's the same in the geometric optics limit, and long times compared to the length of the room in units where c is 1. The images are at integer positions, and all lines come arbitrarily close except for measure zero. $\endgroup$
    – Ron Maimon
    Commented Nov 1, 2011 at 22:22
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You use the verb "to be," which is rather deceptive in this situation (and in questions of optics generally). The room itself would not "be dark" or "be light," it would be a collection of various particles, some of which would be photons in the visible spectrum. It really only makes sense to ask how the room would appear to an observer looking through the viewport.

Even if you had completely reflective surfaces, the way the room appeared would depend significantly on the orentientation of the mirrors, the orientation of the viewing hole, the focus of the light, and perhaps most importantly, the refractive and diffusive properties of the mirrors. Ultimately, the appearance of the room would be a result of the light that falls upon the viewport. While there are an infinite variety of possible arrangements, the two extremes roughly align with the two possibilities you suggest.

For the room to look completely dark, no light paths fall on the viewport. A laser perpendicular to two parallel mirrors with the line of sight also parallel to the mirrors would produce this effect.

For the room to look completely bright, all (or however much you require to meet that definition) light paths fall on the viewport. A room in the shape of a truncated paraboloid with the viewport at the focal point would produce this effect.

Edit

I decided to move these up to my answer to avoid a prolonged comment conversation.

Your edits do help, but still do not provide enough specificity. If the viewport is coincident with the wall (that is with no collimating effect), it would be possible to see the lightbulb so it would certainly not be totally dark. Additionally, assuming that the mirrors are perfectly reflective with no effective refraction (no beveling of the mirror surface, etc.) the room would appear mostly bright. However, the exact nature of the appearance still depends on the orientation of the viewport and light source. The most "average" appearance would be a line of bulbs of decreasing size.

It is certainly not the case, however, that "light is getting reflected from every place possible." Stand in between two parallel mirrors and see what happens. Now imagine that the image does not get successivly weaker (due to imperfect reflectivity). Even though there are a large number of reflections (not infinite but a lot), they do not take up all of the mirror's surface. The situation would slightly change with mirrors on all sides, but there would still be dark spots... again depending on the arrangement.

Sadly, however, my response even includes unstated assumptions; I have, for example, assumed a macroscopic viewport. It is, I think, unwise to use words like "infinitesimal" when talking about optics. Optical phenomena are inherently macroscopic. It just doesn't make sense to talk about an infinitesimal viewport and an isotropic light. These notions are helpful for field theories like electromagnetism, but do not apply well to the aggregate behavior of photons.

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  • $\begingroup$ please see the edited notes... $\endgroup$
    – Vineet Menon
    Commented Nov 1, 2011 at 12:33
  • $\begingroup$ true that infinitesimal viewport may be a farfetched idea, but what's the problem with an isotropic source? A typical incandescent lamp is isotropic, I believe!! $\endgroup$
    – Vineet Menon
    Commented Nov 1, 2011 at 18:43
  • $\begingroup$ okay, okay, I'll grant you the isotropic light. :) $\endgroup$
    – AdamRedwine
    Commented Nov 1, 2011 at 19:12
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It would be bright. Really bright.

If the light was on for an hour, then the number of photons in the room would be an hours worth, still banging about, which would be an incredible number.

Light would be in all directions, so it does not matter where the hole is. It could be that this requires a light source of macroscopic size, but that is in the question. A pinpoint of light would still fill the room randomly, due to wavelength effects.

Opening up a small hole, would of course not change any of that, but the hole would have to be pretty small! A 1 meter room gets 3e8 reflections per second, so opening a hole that is one part in a billion of the surface area would only let the light accumulate for a few seconds.

In the end, the amount of light coming out of the hole would be exactly equal to the total luminosity of the bulb. So if the hole was 1e-6 the size of the filament, the hole would be a million times brighter per cm**2 than the light.

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