I know that for an object to be transparent, visible light must go through it undisturbed. In other words, if the light energy is sufficiently high to excite one of the electrons in the material, then it will be absorbed, and thus, the object will not be transparent. On the other hand, if the energy of the light is not sufficient to excite one of the electrons in the material, then it will pass through the material without being absorbed, and thus, the object will appear transparent.

My question is: For a non-transparent object like a brick, when the light is absorbed by an electron, it will eventually be re-emitted. When the light is re-emitted won't the object appear transparent since the light will have essentially gone through the object?

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    $\begingroup$ see physics.stackexchange.com/q/130459 $\endgroup$
    – njzk2
    Nov 15, 2015 at 21:32
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    $\begingroup$ From a biological point of view, this wouldn't make sense since then we wouldn't have evolved eyes if all objects were transparent and so we wouldn't even have the word "transparent". $\endgroup$ Nov 16, 2015 at 4:56
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    $\begingroup$ wot mate? This be circular logic. $\endgroup$ Nov 16, 2015 at 5:02

3 Answers 3


For an object to be transparent, the light must be emitted in the same direction with the same wavelength as initially. When light strikes a brick, some is reflected in other directions, and the rest is re-emitted in longer, non-visible wavelengths. That is why a brick is opaque to visible light.

Some materials we consider transparent, like glass, are opaque to other wavelengths of light. Most window glass these days, for example, is coated with infrared- and ultraviolet-reflective films to increase insulative capacity. You can see through these fine with your eyes, but an infrared-based night vision system would see them as opaque objects. Another example is that most materials are transparent to radio waves, which is why both radio broadcasts and radio telescopes are so successful.

  • $\begingroup$ Why would light have to be the same wavelength? Is information lost when its wavelength is changed? Also, you said that some of the light is reflected back, why is this? $\endgroup$ Nov 15, 2015 at 18:15
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    $\begingroup$ Yes, information is lost, namely the original light's colour. $\endgroup$
    – Graumagier
    Nov 15, 2015 at 22:41
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    $\begingroup$ I don't think this is totally correct. An object being transparent at certain wavelengths isn't because it re-emits photons in the same direction, it's because it never absorbed them to begin with. $\endgroup$
    – thanby
    Nov 16, 2015 at 5:42
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    $\begingroup$ @thanby, He doesn't say "re-emit", just "emit". So if the light is never absorbed, it's necessarily emitted in roughly the same direction as it entered. (Also, common parlance considers it transparent even if wavelength and angle are slightly distorted, as long as it still forms an appropriate image.) $\endgroup$
    – MichaelS
    Nov 16, 2015 at 9:59
  • $\begingroup$ window glass ... is coated with ... ultraviolet-reflective films to increase insulative capacity and also hopefully to keep birds from flying into it $\endgroup$ Nov 16, 2015 at 16:59

You say:

For a non-transparent object like a brick, when the light is absorbed by an electron it will eventually be re-emitted.

but this isn't true. In a solid the excited state can decay by transferring energy to lattice vibrations instead of emitting a photon. This means the energy of the incident photon is converted to heat and the photon is lost forever.

  • $\begingroup$ Yes, but for some objects the light should be re-emitted, no? $\endgroup$ Nov 15, 2015 at 18:16
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    $\begingroup$ Eventually this energy will still be emitted as photons due to thermal radiation ;) (although you'll not distinguish it from other thermal emission) $\endgroup$
    – Ruslan
    Nov 16, 2015 at 10:30
  • $\begingroup$ So are you saying that this is a non-equilibrium process in the sense that the principle of detailed balance does not apply? $\endgroup$
    – ProfRob
    Jun 19, 2016 at 7:21
  • $\begingroup$ @RobJeffries: no, obviously it's an equilbrium process, but the energy from the incident photon can leave the brick by various ways e.g. by conduction into the surrounding of the brick. So an incident photon does not mean there has to be a photon of the same wavelength leaving the brick. $\endgroup$ Jun 19, 2016 at 7:48
  • $\begingroup$ The principle of detailed balance says that in equilibrium, all rates are balanced by their reverse, for all processes. An object absorbing light at one wavelength and emitting it at another cannot be in equilibrium. Which may be the case. $\endgroup$
    – ProfRob
    Jun 19, 2016 at 9:51

Your problem is that light can do 3 things when hits a wall surfice: it can pass through (transparency), absorbed (black body) or re-radiated (reflection). You overlook the latter. That is, you overlook that there is no any limit on the direction that atom can reflect the photon. What is the chance to re-emit into the original direction? It is basically zero. Can you talk about the transparency if there is no chance to pass your photons through?

Actually, I once had the same misunderstanding/question in a physics lesson. I asked "Why do we see a gas, illuminated by distant stars in space, instead of the stars themselves? The gas should entail some delay, like transparent body, but it should re-emit the photon afterwards and let it go through." The teacher explained that the photons hit the gas atoms indeed but then they are emitted into all directions. That is why you do not see the empty space but start seeing it once cloud is placed there. Empty space does not radiate but cloud does.

Same thing happens with reflection. The inciting photons are reflected into all directions from the surfice but you cannot see them deep in the material. Since fraction of them is absorbed, they all are absorbed after some re-emissions-absorbtions in the depth of the matter. Here, by "absorbtion" I mean black-body absorbtion. The photons disappear accelerating the thermal motion of the atoms. Only those which are reflected back into the air immediately stay visible. Therefore the most objects you see are neither transparent nor black. They reflect the inciting light.

I also cannot cannot explain is why most reflections are mirror-like, which means that if they fall at angle $\alpha$, they are re-emitted at angle $-\alpha$, despite they are re-emitted in all directions arbitrarily, as I told you. Feynman tried to explain this popularly, but I failed.

So, anyway, the brick can either absorb the photons the black-body way but reflects a fair amount of photons from the surface before this happens. There is almost 0 chance for the light to pass the brick or any other object because of absorbtion and reflection into all directions. You need to assimilate the reflection process into your picture.

  • $\begingroup$ Look up "stimulated emission". $\endgroup$
    – ProfRob
    Jun 19, 2016 at 7:04

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