Re: Why light wave can't see the details smaller than wavelength?


ola.andreas.hakansson@xxxxxxxxx wrote:
I don't see what Heisenberg's uncertaintly relation has to do with it?

The phrase "ligh wave can't see the details smaller than the
wavelength" is valid for any wavelengths including wavelengths of
several meters. I don't see how you can include Heisenberg for these
waves.

Andreas

http://personales.upv.es/andreas

Heisenberg's uncertainty relation is an alternate, equivalent way of
stating the diffraction limit.

Suppose an object of size D emits photons that are collected and imaged
by a lens. Let "theta" be the angle of the cone that is intercepted by
the lens.

Heisenberg's relation says:

[uncertainty in momentum] x [uncertainty in position] >= Planck's
constant "h"

Make the >= relation an = if we are at the diffraction limit.

A photon scattered from the object into the lens has momentum (h /
lambda) and uncertainty in momentum of
(h / lambda) x theta

The position uncertainty of the photon is the size of the object, D.

Put this all together:

[(h / lambda) x theta] x [D] = h

or

D x theta / lambda = 1

If D is made smaller than this condition, then the image will still
show a "D" that obeys this condition; smaller objects are not resolved.

This is the same as the usual diffraction-limit condition, neglected
numerical factors and assuming small theta = sin(theta) = tan(theta).
It is " valid for any wavelengths including wavelengths of several
meters."

Mark

p.s. I think it's pretty neat that the "h" terms cancel out, giving an
expression that is derivable from classical wave optics.

.

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