# Reflection by a mirror of wavelength type dimensions

Suppose there is a perfectly reflecting mirror, which is somehow made to the size of wavelength is visible light. Will there be any reflection now, or will it act in some other way?

I thought there might be two diffraction patterns generated. One on the screen away from the source where the mirror acts an obstacle to the incoming light, and the other on the screen towards the source where mirror acts as a new source of light. But i was still confused about the reflection thing. Will we be able to see two rays with angle of incidence equal to angle of reflection?

P.S. I am a highschool student, so am not well versed with some advanced physics that might play some role here. But I know the concepts of interference and diffraction.

Any help will be appreciated!!

$$\let\l=\lambda$$ I'm sorry you received no answers till now. Yet your question is far from trivial. I hope what follows may remedy, at lest partly.

It might help to begin with a "big" mirror and then think of reducing its size. Until mirror is much larger than wavelength it will behave as geometrical optics says. Assume you send on the mirror a wide beam of parallel rays. You'll see a reflected beam the same size as mirror is and beyond the mirror a shadow, also sized and shaped like the mirror.

If mirror is still larger than wavelength, but not very much, diffraction effects will begin to be felt, both in reflected beam and in the shadow. You know that diffraction will involve a region of angular size $$a/\l$$, where $$a$$ is mirror's size and $$\l$$ is wavelength. Diffraction will cause an alternation of light and dark along the borders of reflected beam and of geometric shadow regions. Note that if you put a screen at a distance $$D$$ of the order $$a^2/\l$$ the diffraction zone will cover the full reflection and shadow regions - diffraction is dominant if $$D \ge a^2/\l$$.

If mirror's size is comparable with $$\l$$ then $$a^2/\l\simeq a$$ and geometric optics is no longer applicable, even as a rough approximation. Actually the diffraction region is so large that there is no sharp distinction between backward reflection and forward shadow: you may pass from one to the other in a continuous way.

This pattern is still more apparent if the mirror is smaller than wavelength. In that case physicists speak of "scattering" rather than of diffraction.

Could all this be calculated? In principle it could, but it's a formidable task. As far as I can understand the only treatable case is the one of normal incidence on a mirror of simple form, e.g. circular. I've never seen such calculation but I'd bet it was undertaken around end of 19th century, when wave propagation was one of the main topics of theoretical research. I have some idea of how the calculation could be done - not easy but feasible. Unfortunately I have no time to engage in it. But if you need some further clarification don't hesitate to ask.

• Thanks a lot for this amazing answer, but can you please explain this sentence : Actually the diffraction region is so large that there is no sharp distinction between backward reflection and forward shadow: you may pass from one to the other in a continuous way. Does it mean that we won't see any boundary like thing due to the mirror? And what will the pattern look like? – Eagle Feb 21 at 14:26
• I found this relating to the difference between diffraction and scattering. In the image of the first answer there, it says something about the surface being rough on length scales comparable to the wavelength of light for scattering. But in this case, the size of mirror itself is comparable to the wavelength. So why does scattering take place? – Eagle Feb 21 at 14:32
• @Natasha I owe you two answers. First. Does it mean that we won't see any boundary like thing due to the mirror? I was thinking of what you see far from the mirror. And what will the pattern look like? Don't know. A computation is needed. I told you what I could basing myself on my general knowledge of wave phenomena. – Elio Fabri Feb 23 at 16:13
• @Natasha I found this relating to the difference between diffraction and scattering. I've seen. You're right, I'd forgotten that "scattering" has two quite different meanings, the first in common speaking, the other more technical, used mainly (not only) in QM and particle physics. An important chapter in every Theoretical Physics course brings the title "Theory of scattering." I was thinking of this, not of light scattering on solid rough surfaces, which is an entirely different phenomenon. A professional bias, I suppose. – Elio Fabri Feb 23 at 16:13