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I've thought I had a good understanding how resolution enhancement tricks works for projection lithography, until I tried to understand if it's possible to get sub-diffraction performance for focused laser beam:

Let's assume we have a laser with nice and shiny perfect aspherical focusing lens (with performance greatly exceeding diffraction limit for it's NA). This system moves around and draw stuff.

The question is - can we get enhanced resolution if we apply usual lithography trick of annular/quadruple illumination? Obviously, we won't reduce aberrations thanks to off-axis illumination, as there are none.

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It is possible to beat the diffraction limit. Check my answer here-->physics.stackexchange.com/questions/17881/… –  Antillar Maximus Dec 20 '11 at 15:30
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Yes, in some cases, for some applications, depending on your definition of "beating the diffraction limit."

In the normal sense of imaging resolution, you can't beat the diffraction limit with a normal free-space optical system. But in lithography, what you need isn't really the same as resolution in the imaging sense. Because you are using the light to etch a material to a certain depth, all you need is a sufficient contrast between the bright part of your focal spot and the rest of it. If that contrast is sufficient, then your etched pattern may only become deep over a region smaller than the typical spot size estimate given by $2.44 \lambda \mathcal{f}/\#$.

To illustrate, think of the focal spot due to a perfect lens system with a circular aperture. It takes the form (ignoring scale factors and such) of: $$ \left(J_1(r) \over r \right)^2$$

where $J_1$ is the Bessel function of the first kind. Optics people call this a "jinc" function, due to similarity to the sinc() function. It is squared to give intensity. An image of this spot looks like this: $$ \left(J_1(2\pi r) \over r \right)^2 $$ jinc suared

An annular aperture can be considered the difference of two circular apertures of slightly different size, so by superposition the field at the focus of a system with an annular aperture will be the difference of two jink functions with different sizes. Such a function looks like this: $$ \left(\frac{J_1(2\pi r)}{r} - \frac{J_1(2.1\pi r)}{r}\right)^2 $$ enter image description here

You can see that the spot is considerably more spread out in general, which would be a problem for an imaging system; but the central bright spot is significantly smaller. If the intensity is chosen properly, the outer rings will not affect the lithographic material, while the small central lobe will.

This is even more effective if the material is chosen such that it doesn't absorb light except for rare two-photon events. In this case the intensity dependance of the absorption goes like the square of the intensity, making the contrast between the central spot and the lobes even better. I'm not sure how common this is in industry yet.

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Thanks for great explanation, that makes sence now. There are chemically amplified resists commonly used nowadays which have this "square" sensitivity. –  BarsMonster Dec 21 '11 at 4:13
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