4
$\begingroup$

The fraunhoffer treatment of circular apertures yields a diffraction pattern of circles, with the first minimum (dark ring) at an angular radius of $\theta$ where $\sin(\theta)=1.22\lambda/b$, where $b$ is the diameter of the circular aperture.

But this same result is extrapolated to lenses, to calculate its limit of resolution. But in the case of lenses, there is no opaque material which blocks of the light coming out from everywhere except the aperture.

In the absence of this opaque obstacle, how can diffraction take place, since according to the huygen's principle, the wavefront which would produce the secondary wavelets still extends everywhere and not just to a confined region, which is how the diffraction pattern arises, by confining secondary sources to a finite region or confining it from an "edge" till infinity?* In other words, Why does putting a circular lens in the path of plane wavefornt produce the same effect as allowing only a circular portion of that wavefront to propagate?

And even if there is diffraction, will the same formulae apply? will not the presence of light elsewhere change the case?

$\endgroup$
3
$\begingroup$

What does an aperture do? It "applies" Huygens principle to every point within the aperture, and ignores those outside the aperture because they are blocked.

There are a couple of things going on when you consider a lens. Let's make sure we understand them.

An aperture produces a diffraction pattern in the space of diffraction angles. Recall from the simple derivation: the diffracted rays from every point in the aperture are parallel to each other. (The diagrams accompanying the discussion often have to be fudged so that parallel lines appear to converge, although some authors are careful to include a lens as described below.) The diffracted intensity is a function of diffraction angle. The diffracted field from the screen comprises sets of parallel rays, each set corresponding to a particular interference condition (max, min, or in between). Again, the diffraction pattern is in "angle space". In order to see the pattern you need to be far enough away from the aperture that the rays from different angles and different points in the aperture no longer cross each other causing a confused pattern. You need your viewing screen to be "at infinity".

Now consider what a lens does. Any set of parallel rays entering a lens will be focused to a single point in the focal plane.

Consider placing a lens after an opaque aperture. Any parallel rays entering the lens will be focused to a single spot in the focal plane. Hence all the rays in each set of parallel rays from the aperture will focus onto the same spot in the focal plane. We've created the usual "Fraunhofer pattern at the focal plane". Allow the aperture and lens to approach each other, and you end up with an aperture containing a lens, producing the usual Fraunhofer pattern on the focal plane.

To finally answer your question: remove the aperture and leave the lens. The rays that hit the lens behave as before, forming the diffraction pattern at the focal plane. The rays that fall outside of the lens are also diffracted, according to Huygens principle. But these do not pass through a lens. So the diffraction pattern from these rays stays in "angle space". They exist, and in fact some of those rays overlap the rays from the lens. But we don't see them because they are spread out and weak, and if you are close to the lens their rays cross in a confused manner. Note, however, that those rays outside of the lens do form a legitimate diffraction pattern, which could be viewed "at infinity". The pattern would be a dark spot surrounded by light, the complement of the pattern for an aperture. Look up Babinet's Principle for more details.

$\endgroup$
2
$\begingroup$

There's a trick here: The lens behaves as the aperture because the lens also focusses the light which passes through it. If you don't have a lens but just an opaque sheet with a hole in it, the Fraunhofer pattern shows up at "infinity" distance.

The lens essentially performs a Fourier Transform, thus producing the Fraunhofer pattern at the focal length. All the light which does not pass through the lens is not focussed and at worst provides some background pedestal to the diffraction pattern you see.

It's interesting to create a pinhole aperture, which acts to focus the light without lens! If you play with the math a little, you'll see how that happens.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.