Published: 2015-02-12 | Category: [»] Optics.

Microscopy objectives are a key element composing the microscope as they determine both the magnification and the resolution of the system. However, good microscopy objectives are also expensive. For instance, the 4x Olympus Plan that I have used previously, and which offers a lateral resolution of 3 µm, is sold for a little less than $250US at [∞] ThorLabs. Objectives that have higher resolution power, such as the 100x Nikon Plan Fluorite with immersion oil offering details down to 0.25 µm, are clearly out of reach for the amateur with a price of about $2700US. As an order of magnitude, the resolution limit to visible microscopy is on the order of 0.2 µm so this last objective belongs to the best you will find.

In this post, I will show how middle-range microscopy objectives can be efficiently replaced by camera lenses at a fraction of the cost. By middle-range, I mean magnification ratio below 20x and resolution power of about 1 µm. For the tests, I have been using an old 50 mm Nikkor-S Auto f/1.4, a Nikkor AF-S 35 mm f/1.8 and a NAVTAR 6 mm f/1.4 in conjunction with a 200 mm, 2”, doublet achromat lens imaging a sample onto a monochrome camera with 5.2 µm pixels. To quantify the lateral resolution obtained, I have used a USAF-1951 resolution target. Don’t worry if you don’t understand every bits here, I will make it clearer in the following.

Microscopy objectives and camera lenses have a lot in common in how they are built. They are both composed of an assembly of several lenses whose roles are to compensate the various optical aberrations that you would get with a single spherical lens (that I often call “stock lenses” on this website). To make things easier, these assemblies can both be seen as a single, perfect, lens with an equivalent focal length and an equivalent aperture diameter. Microscope objectives and camera lenses are referenced by these two values or variants of these.

In the case of camera lenses, all the values are clearly labelled onto the lens. A schematic representation of camera lenses is given on Figure 1.

Figure 1 - Schematic of a camera lens.

Your camera lens may slightly differ from the one of Figure 1 but don’t worry, you can still use it. Basically, you will find a focus ring that will change the position of the lens inside the case (take a look at the back of your camera lens while you are turning the focus ring) and an aperture ring which controls an iris inside the lens. A large iris corresponds to a large aperture and a small iris corresponds to a small aperture. The value is coded as the f-number which is defined as the focal length divided by the aperture of the lens. All camera lenses allow reducing the aperture by a large amount but its highest opened value is fixed by the construction of the lens. This largest aperture is always given in the specification of the lens; for instance, a 50 mm f/1.4 such as used here has a focal length of 50 mm and a f-number of 1.4, leading to a maximum aperture of 50 mm / 1.4 = 36 mm. You may not have an aperture ring on your camera but don’t panic, just locate the small spring-loaded pin at the back of the lens as it controls the iris as well. If you have a lens with no aperture ring, you will have to fix the pin at the back of the lens with some rubber band.

Finally, the lens ends with a mount that is specific to the manufacturer. Here, I have represented a Nikkon F-mount with its bayonets on Figure 1.

Please note that the focal length of the camera lens should be taken from the back of the lens and not from its front. So in all your experiment, make sure that the back of the camera lens is pointing to the sample. This is a very common mistake when using camera lenses with short focal lengths.

Concerning microscopes now, most of them are built by placing the sample at the focal plane of the objective. The image is then formed at the focal plane of the second lens (the tube lens) which is here a 200 mm, 2”, achromat doublet. The focal length of the tube lens is manufacturer-dependent but is usually 180 or 200 mm. When using this microscopy setup configuration, the magnification is given by the ratio of the tube lens and the microscopy objective focal lengths. For example, an Olympus Plan 4x has a equivalent focal length of 45 mm which makes the magnification ratio 4x with a 180 mm tube lens and 4.5x with a 200 mm tube lens. You are free to use custom focal lengths for the tube lens at conditions that will be highlighted later.

From the diffraction theory, we expect that the ability of the optical system to resolve two points that lies on the same optical plane (an optical plane is an imaginary plane perpendicular to the optical axis of the system; or parallel to the plane of the lens if you prefer) will be a function of the wavelength of the light, the distance of the plane to the lens and the aperture of the lens. Smaller wavelengths (i.e.: blue or violet light) will show more details than longer ones (i.e.: red light), closer objects will show more details than farther ones and larger apertures will show more details than shallower ones. In the case of the microscope setup used (infinity corrected microscope) we know that the object distance on the optical axis is equal to the equivalent focal length such as represented on Figure 1. If you want to know more about the maths, have a check at books like Shamir’s Optical Systems and Processes (Spie Optical Engineering, 1999).

Figure 2 gives a schematic representation of the equivalent lens which is then written as a line with arrows at each ends. The optical axis is perpendicular to the lens and passes by the lens centre. Furthermore, the lens has an equivalent diameter and focal length as discussed previously.

Figure 2 - Equivalent lens of the microscopy objective when used in an infinity corrected microscope configuration.

The quantity controlling the resolution of the system is the numerical aperture (NA) which is related to the maximum cone of light caught by the lens from an emitter that would be located at the position of the object:

following the notations of Figure 2.

The transversal resolution of the objective is then given by the Rayleigh criterion:

where ∆x is the smallest distance that can be resolved on the object plane and λ is the wavelength of the light used.

When buying a microscopy objective, the manufacturer will tell you the numerical aperture directly. For instance, the 4x Olympus Plan discussed previously has a numerical aperture of 0.1 which lead to ~3 µm of lateral resolution when using blue (475 nm) light. In the case of camera objectives, we have to use the f-number to compute the numerical aperture. Analysis of Figure 2 allows us to write a formula to transform the f-number, f/#, into a numerical aperture:

So, using blue light, a f/1.4 lens has a maximum transversal resolution of about 0.86 µm, a f/1.8 lens 1.08 µm, a f/2.8 lens 1.65 µm and so on. This was verified experimentally by manually closing the aperture of the Nikkor-S 50 mm to various f-stops using the aperture ring and repeating the experiment with a blue LED source (475 nm), a green one (525 nm) and a red one (625 nm). The effective resolution (at the camera sensor) was estimated for each case using the 2D Fourier spectrum of the image. The sample used is a resolution test target US Air Force 1951 which is a known standard for resolution measurements. Example images along with their Fourier spectrums are given for f/5.6 and f/16 as an illustration on Figure 3 (under blue light).

Figure 3 - Resolution at two different apertures.

Although it is relatively difficult to estimate the resolution on the image directly, a look-up on the Fourier spectrum gives an accurate measurement of the highest spatial frequencies found in the image and therefore gives the transversal resolution of the system once the magnification ratio is known (4x here).

The complete results were summarized in Figure 4. As expected, smaller wavelengths gave sharper images than the longer ones and lower f-numbers gave better resolution too. However, I was not able to check the values below f/4 because the transversal resolution became smaller than the size of a pixel (the figure is given relative to the sample and so the size of a pixel was divided by the magnification ratio, i.e.: 5.2 µm/4).

Still, the Rayleigh criterion was verified for the other points except that the const factor was found to be 0.26 instead of 0.61. This can be explained by how the 0.61 factor was derived from the maths when compared to how we analyze resolution and it has little consequences here.

Figure 4 - Testing the resolution at various apertures and wavelengths

As a conclusion, the key quantity to select a camera lens as microscopy objective is its f-number. But because life is hard, low f-number lenses are larger, more difficult to manufacture and then, obviously more expensive. For example, the Nikkor AF-S 35 mm f/1.8 cost about $200US and the same lens but with f/1.4 goes above $1800US! Fortunately, there are good other camera lenses such as the Nikkor AF-S 50 mm f/1.4 and the Sigma 30 mm f/1.4 (untested) at about $400US. If you are lucky, you can even find a Nikkor-S Auto 50 mm f/1.4 for a little more than $100US on e-bay! But the real surprise comes here with the C-mount camera lens NAVTAR 6 mm f/1.4 that I have bought last year on [∞] ThorLabs for less than $150US and for which I have measured an effective transversal resolution of 0.89 µm (0.86 µm expected from the Rayleigh criterion) in the experiments done here with the resolution target.

You may ask: “why should I buy the NAVTAR 6 mm which is a bit more expensive than the Nikkor-S 50 mm knowing they have the same numerical aperture and hence the same transversal resolution?”

Well, this is true but you should remember that the magnification is given by the ratio of the objective and the tube lens focal lengths. Also, we are limited in the recording resolution by the size of the sensor pixels which is 5.2 µm here (it may be different for your camera sensor so remember to adapt the computations). To prevent aliasing from happening, we would like that the transversal resolution of the system span over at least 2 pixels. You can do more than two but you will reduce your field of view for no more details (this is called empty magnification). The ideal magnification ratio is then a function of the transversal resolution and the size of the pixel elements:

and so

with psize the size of one pixel on the camera sensor and ∆x the transversal resolution.

If we apply the formula on our f/1.4 lenses, we get an ideal magnification ratio of 12.1 when using blue (475 nm) light and 5.2 µm pixels. With the 6 mm lens, this requires a tube lens of about 75 mm (75/6 = 12.5). On the other hand, with the 50 mm lens, we would have to use a 600 mm tube lens which is not realistic. To be more exact, it is not realistic but also not even wanted because at such distance the numerical aperture of the tube lens will decrease and we will start to impair the resolution due to this second lens. Mathematically speaking, we would like that the numerical aperture of the tube lens should be higher than the numerical aperture of the first lens divided by the magnification ratio. This is usually enforced with 2” lenses up to 200 mm.

As a conclusion, we should then favour small focal length over larger ones for a given f-number.

For this last reason, I would recommend the NAVTAR lens because it matches all the requirements of our system: small f-number and small focal length. So if you have some time, go check what lenses you can find for C-mount cameras, they definitively worth spending some money on them!

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