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I'm not an expert in lens design. I need to build a lens having fixed the focal point $f$, the lens diameter $D$, the maximum thickness $d$, the refractive index $n$ and the half-angle $\theta$ entering on the lens and the angle that I want at the exiting from the lens. With these, I want to find the radius of curvature $R_1,R_2$.

In some books of optics that I used in my course of physic and I found the Lensmaker equation (because I cannot use the thin lens approximation).

$\frac{1}{f} = (n-1)\left [ \frac{1}{R_1} - \frac{1}{R_2} + \frac{(n-1)d}{nR_1R_2} \right ]$

This equation don't allow me to play with the lens diameter, so looking around I found the relation between this and the numerical aperture NA from Fresnell equation but I'm not sure how can help me.

$n_1\sin\theta_{in} = n_2\sin\theta_{out}$

$NA = n\sin\theta = n\sin\left [ \arctan \left ( \frac{D}{2f} \right ) \right ]$

Can you help me figuring out these radius knowing all the previous variables and how take into account the whole lens diameter?

This is my geometry. The focal point is known since I have a diverging source and I want to have a planar wavefront after my lens.

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    $\begingroup$ Hi! You can put mathematical expressions in MathJax by wrapping them in $ symbols. To get greek letters you can just type the name preceded by a forward stroke eg \theta. $\endgroup$
    – jacob1729
    May 17, 2019 at 12:26
  • $\begingroup$ Didn't know. Thanks $\endgroup$
    – Shika93
    May 17, 2019 at 12:32
  • $\begingroup$ I might be misunderstanding the geometry, but aren't the radii of curvature independent of the lens-diameter? So you need to pick a lens diameter consistent with the angle you want to let in, which is given by the last equation in your post, and then given the focal length you want the lens-makers equation gives you the radii of curvature. (There will be lots of solutions, I'd probably pick $R_1=-R_2$ if you're doing this by hand) $\endgroup$
    – jacob1729
    May 17, 2019 at 12:36
  • $\begingroup$ The point is that I know where is the focal length (I have a divergent source hence the focal point) and I want that the divengency angle, if we want to call it like that, my $\theta_{in}$ fixed at 45°. I want a converging lens to get $\theta_{out}$ possibly 0°. The lens diameter is due to mechanical constrain that need to be fulfilled and the refraction index is something that I cannot change. $\endgroup$
    – Shika93
    May 17, 2019 at 12:51
  • $\begingroup$ I added a schema of the geometry. I need to converge, so I thought to start with a bi-convex lens and then we'll see. I guess the Lensmaker equation describes whatever shape changing the value of $R_1$ and $R_2$ (flat surface will have $R=\infty$ I suppose) $\endgroup$
    – Shika93
    May 17, 2019 at 13:02

1 Answer 1

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The focal length of a lens depends only on its thickness and radii of curvature, as appears in the lensmaker's formula you quote. Think of a single ray - as long as it enters the lens, it will be deflected according to the focal length you calculate. It simply doesn't care about the extent of your lens. A word of warning - this is true in the paraxial approximation, so if your lens is too big you no longer have a focal point. In this cases different rays will converge into different points, an effect known as spherical aberrations.

However, both the diameter and the focal length of a lens affect the amount of light it can collect from a source at its focal point, which is what the numerical aperture captures. This is of paramount importance in microscopy, where one wants to maximize the amount of light collected from a sample.

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  • $\begingroup$ In the third formula that I wrote, the $n$ is the refraction index of the lens or outside? Or it doesn't matter?. Because as you said, I guess that one of the "constraint" that I have is not actually a constraint but something that someone tried to calculate (wrong) and this explains a few things that I'm experiencing $\endgroup$
    – Shika93
    May 17, 2019 at 14:07
  • $\begingroup$ $n$ is the refractive index of the lens. What is the purpose of this lens if I may ask? $\endgroup$
    – eranreches
    May 17, 2019 at 14:17
  • $\begingroup$ Is a lens for a projector with a diameter D. The supplier gave to me the constrains that I said but I think that actually some of them (focus, input angle ecc) are computed, not actually constrains and I believe they used the approximation for thin lenses that is fancy from a physical point of view but impossible to build since we're talking of a big lens. I guess that given the Fresnell eq (fixing the output angle) I need to compute the input angle and then find the focus length with the third formula. At the end, the radius with Lensmaker eq. Am I right? $\endgroup$
    – Shika93
    May 17, 2019 at 15:01
  • $\begingroup$ You have the diameter and you don't need it anymore for any formula. You should just use the given focal length and invert the lensmaker formula. Note that you can simplify the equation by assuming a symmetric lens or a plano-convex lens. Those are typical types of lenses. $\endgroup$
    – eranreches
    May 17, 2019 at 15:50
  • $\begingroup$ The problem is when I want simulate this with a tool in geometric optics (i.e. Zemax) this tool asks me the radius of curvature and also the diameter of the lens. This is why I'm feeling that I'm missing something $\endgroup$
    – Shika93
    May 17, 2019 at 16:27

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