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## Optics – Why is there a maximum focus distance when using an extension tube?

The objective of the objective is to create an image from the outside by projecting it onto the surface of a film or digital image sensor. To accomplish this, light from objects traverses the transparent lens. The shape (Figure) of the lens and the density of the lens material cause these imaging beams to change direction of travel. We can draw a trail of these exiting rays. When we do this, a trace reveals that this revised path takes the form of a cone of light. This action is called refraction and means to bend inwards.

The distance from the lens to the tip of this light cone can be measured. You will find that light rays from distant objects are closer to the lens than light rays from objects closer to the lens. In other words, the length of the cone of the imaging rays is a variable based on the distance of the object from the lens. The focal length of a lens is a lens-to-tip measurement when the lens images a faraway object (objects with the infinity symbol ∞). A measurement of the lens-to-vertex distance that is performed when the lens observes nearby objects is referred to as a "back focus".

The key to answering your question is the fact that a lens has only a limited refraction (bending inwards). When we focus the camera, we set the distance lens on the sensitized surface to make the vertex of the cone of image-forming rays kiss only the surface. If the distance is right, the object is in focus. If this point becomes too short or too long due to focus adjustment, the resulting image is out of focus.
As we focus, we again change the distance between the lens and the sensor. The focus mechanism of most cameras limits the movement of the lens to the sensor. This is because most camera lenses are optimized for distant subjects. A macro lens design is optimized for close-up and slightly degraded when you want to edit distant subjects.

Standard camera lenses typically keep forward motion at a distance of about ½ meter (20 inches). If we want to shoot with a standard lens very close, we need to install tubes / rings or bellows or resort to additional close-ups.

We usually use extension tubes when working closely together. The tube only increases the distance between the lens and the sensor. In other words, they extend the distance lens to the vertex, and this compensates for the now extended distance of the rear focus. When a tube is inserted, the lens is adjusted to the distance of the subject. Since the lens has only a limited refractive power, the now revised focus range is limited to close-ups. We can only work with narrow distances of objects.

## Macro Converter – Maximum focus distance with extension tube

I use the Nikon D850 together with the Sigma 50mm / 1.4 type and the
MEIKE MK-N-AF1-A Extension Tube (s).

In the first shots, I found that the combination has not only minimal focus removal (which is good), but also maximum focus removal (which is confusing).

I did not measure it, but with the 20-mil extension tube, I can only focus on objects that are only 1.5 to 2.5 inches from the lens (both manual and auto-focus).

I tried different aperture settings from 1: 1.4 to 1: 8, but the maximum focus distance has not changed.

Why is focusing on more distant objects a problem?

OR:

Why does the use of an extension tube result in maximum focus removal?

## Differential Equations – Non-Newtonian Pulse Equation in a Rectangular Tube

I try to solve the momentum equation for a non-Newtonian fluid in a rectangular-section tube with Mathematica (numerically).

Here are the assumptions:

Thus, the flow runs only in the z-direction (axial direction) and varies only in the x and y direction. The sides of the rectangle are 2a and 2b in the x and y directions, respectively.

The Z-momentum equation is:

The boundary conditions are:

The non-Newtonian model I use is called the Carreau model, where fluid viscosity is a function of the scalar shear rate.

All parameters in the above model are constants, with the exception of the gamma point, the scalar shear rate, which is evaluated as follows:

So I'm using NDSolve and NDSolveValue, but I was not lucky (error: PDE spatial order should not exceed two).

I paste the notebook I've put together into this link (https://community.wolfram.com/groups/-/m/t/1832082). Can you please look at it and let me know how I can handle it?

Important: Note that in the notebook, due to the symmetry (0 <= x <= a and 0 <= y <= b), I actually try to solve the equation for only 1/4 of the geometry.

To edit: The dUz / dx and dUz / dy should actually be partial derivatives in my description above.

Many thanks

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