In my high school, I am studying geometrical optics. I studied why smooth surfaces like mirrors form an image of an object but rough surfaces like a wall don't (due to regular reflection and irregular reflection), but it answered my question by considering particle nature of light. Can anyone answer the question why mirrors form an image of an object, while a wall doesn't, considering the wave nature of light?
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1$\begingroup$ Do you mean it answered your question considering light as rays (not particles) as in geometrical optics but would like an explanation considering the wave nature of light? $\endgroup$– Not_EinsteinCommented May 8 at 17:14
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$\begingroup$ The ray (geometric optics) model for light is perfectly good for describing the formation of images due to reflection of mirrors. One can derive the law of reflection (and more generally reflection and refraction-i.e., Snell's law-at the interface between two materials) from wave optics, which you'll find in any optics or EM book, but the derivation is likely beyond the scope of a high school class. $\endgroup$– marchCommented May 8 at 17:23
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$\begingroup$ I tried to search it on the internet the most logical answer was that, when a light wave strikes a surface it vibrates the atoms of the surface which in result radiates a wave of same frequency. Is it correct or not? $\endgroup$– Himanshu NirwamCommented May 8 at 17:31
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$\begingroup$ Yes, and it vibrates the atoms due to the Lorentz force that it exerts on the nuclei and the electrons in the atoms. $\endgroup$– GhosterCommented May 8 at 17:36
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$\begingroup$ This might help, though it isn't really an answer to your question - If we repeatedly divide a colorful solid in half, at what point will the color disappear? $\endgroup$– mmesser314Commented May 8 at 18:30
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4 Answers
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Both previous answers still rely on ray optics for their arguments while OP clearly wants to know how to explain things via wave description.
I will attempt at giving an answer based on waves.
Imagine a water tank, or the sea, does not matter, with very calm waters and a nice long wave approaching the shore. Now consider two cases: 1: the shoreline is a nice even concrete wall and 2: the shoreline is full of randomly located boulders of different sizes and shapes.
In case 1, the incoming water wave will be nicely reflected back while in case 2 the wave will be completely scattered: part of it will be lost in the crevices between the rocks, and some others will create wavelets at random directions and there is no discernible long and well-behaved wave any more after "reflection".
That is basically similar to what a mirror vs a rough wall does. The smooth wall (water)/mirror (light) preserves a lot of the characteristics needed for the considered wave: phase and amplitude, while the boulders (water)/rough wall (light) divide the incoming wave into a lot of random phase and amplitude wavelets.
EDIT: I realized that I also need to point out that a very important thing is that the k-vectors of the wavelets (in a sense their direction) are also a very important part for preserving the information needed to form an image.
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$\begingroup$ This is a good analogy but it does not explain internal mechanics. $\endgroup$ Commented May 11 at 15:22
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$\begingroup$ @CapoMestre Nor should it. It's a highschool student that clearly thinks that ray optics means "particle" behaviour. I will not go into medium polarization and plasmonics for a question that is more about waves in a broad sense rather than about light. $\endgroup$ Commented May 15 at 8:23
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$\begingroup$ I would not try to impose what answer should be given and what not. It is like deciding for the student. Here, different answers should be provided and it should be up to the student to decide what works for him/her better. And I personally think it would be great if you went into plasmonics, etc. At least, I know for myself that if I were a high-school student I would rather prefer to hear real stuff and thus to feel inside a gradient "force" to learn more. $\endgroup$ Commented May 15 at 11:18
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When light hits a mirror, it follows a specific law known as the law of reflection. This law states that the angle of incidence (the angle between the incoming light ray and the normal, which is an imaginary line perpendicular to the surface of the mirror) is equal to the angle of reflection (the angle between the reflected light ray and the normal). This means that light bounces off the mirror in a predictable manner, preserving the direction of the light rays.
On the other hand, when light hits a wall, it can interact in multiple ways. Some of the light may be absorbed by the wall, some may be transmitted through the wall, and some may be reflected. However, the reflection from a wall is not as orderly or predictable as it is with a mirror. The surface of a wall is rough and irregular at a microscopic level, causing light rays to scatter in different directions rather than reflecting uniformly like they do on a smooth mirror surface. As a result, walls don't form clear images like mirrors do because the reflected light is dispersed in various directions instead of being reflected at equal angles.
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PERFECT MIRROR
One can think of the reflection as constructive/destructive interference of waves (see Figure 1). In order to reach a detector (e.g., human eye), light waves from a point light source can in principle travel all these trajectories shown in Figure 1 (there is always a non-zero probability for that to happen). Then, one just has to sum up all these waves together in the immediate proximity of the detector. Considering only parts of trajectories that have same length from the detector (i.e., between black markers and the detector in Figure 1), the phase of light waves at the detector for trajectories that are away from reddish ones accumulates much faster than the phase of light waves along the reddish trajectories. As a result, the phase at the detector is nearly random for the grey trajectories, and nearly the same for the reddish trajectories. Adding those waves together results in the reddish waves interfering constructively, thus maximizing probability for light to travel along the reddest paths to the detector. It turns out that the reddest trajectories are also the trajectories of the minimum length and shortest travelling time, which is obvious. More on a similar explanation is here: Feynman lecture
Figure 1. Infographics explaining the law of reflection using wave terminology. Red lines (the ones marked with numbers from -3 to 3) represent light trajectories where light interferes constructively. Shades of red represent gradual dephasing of light beams at the detector (human eye). Grey lines represent trajectories where light interferes destructively due to nearly random phases (at the detector) in the adjacent trajectories, thus no light is observed travelling along those trajectories. In comparison of waves along all 29 trajectories (bottom part of the figure), only paths a constant distance away from the detector were considered (these paths towards the detector start from black dots).
ROUGH SURFACE
As for the case of rough surface, one can view it as, for example, consisting of many tiny mirrors oriented at random angles to the light source and detector. Then, one has to do the same summing-waves procedure for every such tiny mirror. As a result, all minimum-distance rays from different mirrors will reach the detector with nearly random phases resulting in destructive interference and no clear reflection as seen by the human eye (detector).
EDIT: This picture in the case of rough surface assumes that the mirror patches are as large as the wavelength of light or larger. If these mirror-patches are much smaller, then reflection at each of them would happen into various directions (a.k.a. spherical wave) due to uncertainty principle (see, e.g., Wikipedia or Hyperphysics). In this case, one would need to perform the mentioned above summing-waves procedure but taking into account that a tiny mirror patch now actually reflects in all directions. Mind blowing!
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$\begingroup$ If I may, I do not think that a QED approach is the correct approach for teaching wave phenomena. We have perfectly valid models and equations that do not need the mental gymnastics of for example source-> point -14 ->detector. Also because this will confuse any highschool student learning the laws of reflection and refraction. $\endgroup$ Commented May 15 at 8:31
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$\begingroup$ I am not sure it is incorrect either. Maybe you want to expand why you think it is definitely incorrect. Again, this platform implies that answers should be provided here, and "best" ones chosen by those who ask. Potential confusion you mention is not necessarily a bad thing as it may serve as a motivation to learn/clarify/understand more. $\endgroup$ Commented May 15 at 11:23
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A perfect mirror conserves the direction of the light rays. Therefore reflected light can be imaged by a lens. A rough surface on the other hand scatters the rays in all directions. It can the no longer be imaged. However a rough surface is a good projection screen. If already ray optics can explain a mirror, then wave optics is not required.
