# How large is the smallest object that can be detected at a given wavelength?

+ 3 like - 0 dislike
769 views

What is the cross section of the smallest object that can be detected with say visible light ($\lambda$ ~380 - 750 nm) or X - band radar ($\lambda$ ~20 - 50 mm).

Does the object need to have one side larger then $\frac{\lambda}{2}$ or something like that to get a reflection?

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Theodor
detect how? for exemple for visible light: just visible againt the light or for example optical microscopy? that changes things

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Steve
@steve - Ok, say rather then detect, say reflect.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Theodor
I would say rather than detect or reflect, "affect," since this is what we usually mean by measure.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Jeremy
It is of the same order as the wavelength

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user user1708

+ 5 like - 0 dislike

Seems like in principle, there should be no minimum size. For example, a small hand-held radio (cell phone for those of you in the 21st century) can detect, generate, and scatter EM waves with wavelength much longer than the object itself.

For optical and X-ray scales, one would probably be limited by quantum or thermal effects, which may in principle be mitigated at very low temperatures. But even so, a single atom is MUCH MUCH smaller than 500nm, and can absorb and emit light at that wavelength.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Jeremy
answered Dec 21, 2010 by (505 points)
I think Jeremy is correct. An object many times smaller than the wavelength can still cause scattering or absorption, so it would create a detectable change to the incident beam of light, which is all we need for detection. The quantum efficiency (what fraction of photons are scattered) will be small, but as long as a few are, the particle is detectable. Think of the blue sky. The air molecules are only a few angstroms size, whereas light with wavelengths of a few thousand angstroms do suffer from Rayleigh scattering. And we see this as blue sky, as the blue is scattered more than the red.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Omega Centauri
is incorrect in saying that "there should be no minimum size." The question is not whether light can be scattered by an object, but whether light of a given wavelength can resolve an object of a certain size - at least that is how I interpret it.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user user346
"What is the cross section of the smallest object that can be detected"--That sure sounds more like scattering than resolving to me!

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Jeremy
Indeed, the question was very clearly about detection, not resolution.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Colin K
+ 1 like - 0 dislike

Detecting an object requires you to get enough of a signal from it that you can discriminate that signal from background noise. This is not necessarily dependent on the size of the object, although it can be.

For example, if you are trying to detect an object which has very low reflectivity at your given wavelength, you might get a stronger signal if the object is larger. Of course this also depends on the geometry of the object, and how much of the reflected light is reflected towards your detector. Ultimately the deciding factor in whether you can detect an object is the ratio of the signal strength to the noise in your detection system. Engineers typically call this the "Signal to noise ratio" or SNR.

The Rayleigh limit will describe the minimum distance between two point sources such that an imaging system can distinguish one point from the other, rather than seeing them as a single point. However, this has nothing to do with the fundamental requirement of detecting the presence of an object in the first place. It is quite possible to, for example, detect medium-wavelength infrared light (with a wavelength of 3 to 8 microns) emanating from a pinhole or the tip of an optical fiber which is only a couple microns across. In this case an imaging system would indeed see a spot of light with a radius given by:

$$\begin{array}{rl} \Delta l = 1.22 \frac{f\lambda}{D} &\mbox{for a circular aperture}\\ \Delta l = \frac{f\lambda}{D} &\mbox{for a square aperture} \end{array}$$

Simply by detecting this spot, you would have detected the presence of the object, although you would not be able to say anything about its size or shape, because it is below the resolution of your device.

A good way to think about this is to get away from the concept of an imaging system. Imagine you've got a lens focusing light onto a photodiode, instead of an imaging detector like the CCD in a camera. With a single photodiode, you can't determine the size or shape of anything. All you can do is detect a signal, or the lack of a signal. This is mostly independent of resolving power or the size of the object being detected.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Colin K
answered Dec 27, 2010 by (10 points)
+ 0 like - 0 dislike

The Rayleigh criterion determines the maximum minimum size of objects - in terms of the maximum minimum angular resolution $\theta$ - which can be resolved at a give wavelength $\lambda$ and with a lens of a given diameter $D$:

$$sin(\theta) = 1.22 \frac{\lambda}{D}$$

The spatial resolution is given by:

$$\Delta l = 1.22 \frac{f\lambda}{D}$$

where $f$ is the focal length. For visible light this gives a limit of $\sim$ 200 nm for the minimum resolvable distance.

Of course, this is all based on classical optics. With newer metamaterial based lenses one can do much better than the Rayleigh limit.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user user346
answered Dec 21, 2010 by (1,985 points)
Where is the magic number 1.22 come from? Is it like $\pi^2/8$ or something?

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user ja72
@jalexiou: "1.220 is approximately the first zero of the Bessel function of the first kind, of order one (i.e. J1), divided by π." from the wiki

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user hwlau
don't you mean "minimum size" and "minimum angular resolution"?

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Jeremy
@jeremy, thanks for the corrections !

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user user346
While the equations are correct, the question clearly talks about reflection and detection, not resolution. Which would change the answer completely.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Bruce Connor
Yes, but the question of what can be "detected" is far too broad to answer coherently. Also the OP clearly says: "Does the object need to have one side larger then $\lambda/2$ or something like that to get a reflection?" - sounds to me like he is speaking of something more than just detection.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user user346
Isn't that only true if you ignore near-filed effects? I recall that some years ago this limit was beaten by near-field microscopy.

This post imported from StackExchange Physics at 2014-04-01 16:23 (UCT), posted by SE-user Lagerbaer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.