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Hello Stan,

On Fri, Apr 8, 2011 at 3:28 PM, Stanislav Vitha <[log in to unmask]> wrote:
>
> My main question was not why I get degradation of resolution in the
> fluorescence channel (that I expected),  but rather why is the degradation
> worse for a lower-resolution objective  (20x/0.85 oil imm versus 100x/1.4 oil
> imm). The "image doubling" due to the shear of the objective DIC prism should
> be more noticeable with the high-resolution lens

I happened to have gone through this exercise when measuring the shear
in our DIC setup.
We tried to recap what we understood in the appendix of our paper
(http://www.mshalin.com/blog/wp-content/uploads/2010/09/mehtaao2010-samplefreecalibrationdic.pdf).

Here is an attempt at understanding the specific case that you described:

Nomarski prism is an angular polarizing beam-splitting devices. i.e.,
it splits the incoming light into two polarizations that travel at
different angles. The angular split caused by the Nomarski prism
(placed in the back focal plane) amounts to spatial shear in the
specimen plane. The relationship between angular shear (a) and spatial
shear (s) is rather simple: s=tan(a)*f. Where f is the focal length of
the objective. This fact can be seen easily from the geometry of fig.
11(d) in above paper. Now the image doubling is caused when the shear
is large with respect to the resolution. To judge the doubling, we
should normalize both sides of the above equation by lambda/NA. Then,
we have s/(lambda/NA) = tan(a) * f/(lambda/NA). Therefore, when
s/(lambda/NA) or s*NA is large (for given wavelength) we will have
image doubling.

In the Olympus setup there is only one DIC prism on objective side,
i.e., the angular shear is the same for all objectives. Therefore, the
objectives for which f*NA is large will give rise to larger s*NA and
hence image doubling.

The product of focal length and NA is large for 20X 0.85 than for
100X1.4. I compute the product of focal length and NA below.

Objective's magnification (M) = tube length (ft)/ objective's focal length (f).
Therefore, f=ft/M.
Therefore, f*NA = ft*(NA/M).
For, 20X 0.85: f*NA= 0.0425*ft.
For, 100x 1.4: f*NA= 0.014*ft.

So the 'image doubling'  for 20X 0.85 is almost 3 times more than  for 100x 1.4.

Now, it is interesting to see how this relates with Guy's intuitive observation.

NA of the objective = radius of the objective BFP (d)/ objective focal
length (f).
So, r=f*NA. As we have seen above,  the size of the BFP (ie., f*NA)
for 20X 0.85 is 3 times larger than for 100x1.4. So as Guy says, rays
collected by 20X 0.85 pass through larger (3 times larger) surface of
the Wollaston prism.

Best
Shalin

http://mshalin.com

Research Associate
Bioimaging Lab, Block-E3A, #7-10
Div of Bioengineering, NUS Singapore 117574
(M) +65-84330724