Is it possible to calculate the length of a tube if the 'near' diameter and 'far' diameter are known? When looking down and through a length of tube, the 'far' diameter is less than than that diameter that is closest to the viewer's eye.
If, for example, the tube's internal diameter is measured at 80mm and (again, as an example) the internal diameter of the tube that is farthest away (the other end of the tube!) is observed to be 40mm, assuming that the tube does not taper, is it possible to calculate the length of the tube without actually shoving a measuring tape either into or outside the tube?  
 A: Clarifying Wolphram jonny's drawing / equations a little bit...:

You can write the following equations:
$$\tan\theta_1 = \frac{d}{2D}\\
\tan\theta_2 = \frac{d}{2(L+D)}$$
For small angles ($d<<L$), the observed angle $\theta_1 = 2\theta_2$ when $L=D$. So if I correctly interpret your statement "the internal end of the tube that is farthest away is observed to be 40 mm" to mean "the apparent angle is half", then the length of the tube is equal to the distance from your eye to the near end of the tube. If the small angle approximation doesn't quite hold, the above equations can still be solved (quite likely, "observed" means the ratio of the tangents, rather than the ratio of the angles, is 2 - so it would still be easy. But I can't know exactly what your statement means, as the aperture is said to be 80 mm so I don't know how it can be observed to be 40 mm...). 
A: 
If you know the distance from your eye to the closest end of the tube you can, otherwise you can't. The reason if that you need that distance to calculate  the angle of aperture and thus the distance to the far end. Because the ratio of perceived size depends on this distance. Hope the drawing helps (My drawings are becoming worse with time, I apologize)
(the right equations has to be $d=\theta_2 R$)
So you have $d/r=\theta_1$ and $d/R=\theta_2$
The ratio of diameters will be 1/2, so you have $\frac{ \theta_2}{\theta_1}=.5=\frac{ d}{r}+\frac{ d}{R}$, and from here you obtain R (and the length of the tube, $L=R-r$)
