Sunday, 10 November 2013

The size of the little things - Part II

I've been thinking about the amplification factor during these last days, and I thought about test this parameter in some of my lenses. I would like to share here my results.


I've tested three of my lenses:
  • Tamron 90mm Macro f/2.8 SP Di VC USD.
  • Tamron AF 70-300mm F/4-5.6 Di LD Macro 1:2 (at 300mm).
  • Nikon Nikkor 18-105mm f/3.5-5.6G AF-S DX ED VR (at 105mm)

The first one is the lens that I'm usually using for macrophotography. The second one is the lens that I used for my first macrophotographies. The last one is a normal multi-use lens.

The method used is the method described in the last post. I've taken pictures of a rule and I've measured the size of the photographed object. With these data, I compare the real size of the object with the size of the projected image and I calculate the scale factor.

Firstly, the qualitative analysis: in the graph we can view the three behaviours. The first one (TAMRON 90mm, blue) starts in a scale factor 1:1 (real macro) and it is increasing when we use larger extension tubes. It is the best lens for macrophotography: its scale factor is bigger than others in the graph.

The second one (TAMRON 70-300mm, red) starts in 0.5:1 (or 1:2, it's the same). It's increasing in a very slow way. Later we will know that It's due to the focal. With 68mm of extension tubes (the maximum that my gear lets) we have a scale factor of 1:1 (real macro), but it's very dark and unstable.

The last one (NIKON 18-105mm, yellow) starts very down (in 1:5), but its increasing rate is high. There aren't values for high lengths of the extension  tubes. It's due to the focus distance for these configurations was too much short for be possible.

Could we explain these behaviours? We are going to try it with some maths (it could have several mistakes). Firstly, some visual schemes.



The first scheme is the ray trace of a normal plane lens. The second one is the ray trace of a system displaced some distance (the extension tube). We are going to use some relations of triangles for this.



And now, let me use some geometry.



We can view that the scale ratio and the length of the extension tube have a linear relationship. The increasing rate is related with the inverse of the focal distance, so lenses with longer focal distances are less sensitive to the extension tubes.

We can use the previous data to calculate the focal length of the used lenses. Using a linear regression, the linear coefficients are:

  • Tamron 90mm: A ~ 0,0142 mm⁻¹ => f ~ 70mm.
  • Tamron 70-300mm: A ~ 0,0064 mm⁻¹ => f ~ 155mm.
  • Nikon 18-105mm: A ~ 0,0142 mm⁻¹ => f ~ 70mm.

The results are very very bad... I think maybe could be due to the crop factor. If we multiply the obtained focal by the crop factor, we have:

  • Tamron 90mm: f ~ 105mm.
  • Tamron 70-300mm (at 300mm): f ~ 233mm.
  • Nikon 18-105mm (at 105mm): f ~ 104mm.

The result is nearer to the real, but too much far yet (except in the last case). We have used geometric optic for plane lenses, and my TAMRON 70-300 is not a plane lens.

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Conclusions

In this post we've measured the native scale factor of three of my lenses:

  • Tamron 90mm: m = (103058 ± 12) x 10⁻⁵ (~ 1:1).
  • Tamron AF 70-300mm: m = (50880 ± 5) x 10⁻⁵ (~ 0.5:1 = 1:2).
  • Nikon Nikkor 18-105mm: m = (21878 ± 4) x 10⁻⁵ (~ 0.2:1 = 1:5).

We've also tried to measure the focal lengths:

  • Tamron 90mm: f = (105 ± 2) mm (real value = 90mm).
  • Tamron AF 70-300mm: f = (233 ± 224) mm (real value = 300mm).
  • Nikon Nikkor 18-105mm: f = (104 ± 45) mm (real value = 105mm).

In the first case, the focal length is near to the real, but it's not include in the uncertainty interval. In the second case, the real focal length is included in the uncertainty interval, but the central value is too much far to the real. The uncertainty is very high. In the last case the result is very near to the real value and it's included in the uncertainty interval. The best result is obviously found in the third case.

Lens Extension
Of the tube
(mm)
m Δm Amplification f
(mm)
Δf
(mm)
Tamron 90mm
Macro f/2.8 SP Di VC USD
0 mm 1.03058 1.2E-4 1.0 : 1 105 mm 2 mm
12 mm 1.22789 1.6E-4 1.2 : 1
20 mm 1.32527 1.6E-4 1.3 : 1
32 mm 1.53828 1.7E-4 1.5 : 1
36 mm 1.55641 1.8E-4 1.6 : 1
48 mm 1.75037 2.0E-4 1.8 : 1
56 mm 1.80805 2.2E-4 1.8 : 1
68 mm 2.01903 2.4E-4 2.0 : 1
Tamron AF 70-300mm
F/4-5.6 Di LD Macro 1:2
(at 300mm)
0 mm 0.50880 5.3E-5 0.5 : 1 233 mm 224 mm
12 mm 0.56031 6.0E-5 0.6 : 1
20 mm 0.61463 6.8E-5 0.6 : 1
32 mm 0.72090 8.0E-5 0.7 : 1
36 mm 0.71962 8.0E-5 0.7 : 1
48 mm 0.80129 9.6E-5 0.8 : 1
56 mm 0.81834 9.6E-5 0.8 : 1
68 mm 0.96258 1.0E-4 1.0 : 1
Nikon Nikkor 18-105mm
f/3.5-5.6G
AF-S DX ED VR
(at 105mm)
0 mm 0.21878 4.0E-5 0.2 : 1 104 mm 45 mm
12 mm 0.40754 4.8E-5 0.4 : 1
20 mm 0.50739 6.0E-5 0.5 : 1
32 mm 0.68594 8.0E-5 0.7 : 1
36 mm 0.73980 9.6E-5 0.7 : 1

The results of this little experiment are not perfect at all. What factors have influenced in these deviations? I've thought about three of them:

  • The length of the extension tubes: I think it is the most important deviation factor, because, for example, the combination 20mm + 12mm (32mm) is very similar to use a single tube of 36mm (sometimes it's equal). I think an important length error in the combination of several tubes exists.
 
  • The "plane lens" approximation: my geometrical calculus is based in the plane lens hypothesis. All my lens is so thin that it is occupying a single plane. However, the length of my TAMRON 70-300 is several decimetres. My previous calculus is only a approximate estimation.
 
  • The lens distortion: my method is based in the measure of distances in the photographed image. If this image is distorted, the measure is also distorted, so you have an additional error source.
 
 
Looking to the above graph, It seems that the better option for macrophotography is my TAMRON 90mm. It's a very good lens, with a great quality, and It offers the best scale factor of all my gear (from 1:1 to 2:1).
 
I hope that my little analysis be useful and interesting.

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