0
$\begingroup$

For a mechanical inspection model of an equipment, i am having several transparent acrylic glass rulers with a regular milimetric graduation pattern inscripted on each flat side. The back graduation lines can be looked from the front, refracted through the glass, and compared with the front graduation lines, as shown.

Lets say one of these plates, is located at the origin, oriented through arbitrary normal $\hat n$ and direction $\hat t$ unit vectors.

An observer is at the point $x$, measuring the distance between the graduation lines $\Delta d$.

For the 2D case, when $x=(1,0,0)$, $\hat n=(n_1,n_2,0)$ and $\hat t$ lie in the $xy$ plane, the solution is simply:

$$\Delta d=\sin(\theta)(1-\frac 1\eta)=\frac{| x \times \hat n|}{|x|} (1-\frac 1\eta)$$

where: $\theta=\arcsin(n_2)$ is the angle between $x$ and $\hat n$, and $\eta$ the refraction coefficient.

Which should be the solution for the 3D case?.

enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ Somethings seems to be wrong: according to the definitions $\sin\theta=n_2/n_1$ while $|x\times\hat{n}|/|x|=n_2$. $\endgroup$ Sep 2, 2017 at 8:12
  • $\begingroup$ Corrected, $\sin \theta=n_2$ actually. $\endgroup$
    – Brethlosze
    Sep 2, 2017 at 16:58

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.