0
$\begingroup$

I think I might have a solution to calculation the lateral displacement of light, however I'd like informed opinion on whether this is right. We first extend the incident line forwards. Now, we calculate the two components of the velocities of both the incident and refracted rays, and then, we use the thickness of the slab as well as the component of velocity of refracted ray along the normal at point of incidence to calculate time taken by the refracted ray to travel the length of the glass slab. Now, using the two components of velocity, we find the coordinates of the particle (here I take the ray to be a single photon travelling through air and glass) after the refracted ray hits the other side of the glass slab, taking the point of incidence to be origin. (No need to take the sign of the coordinate, they both lie on the same quadrant of the coordinate axis so we can ignore that). We first find the coordinates of the photon after refraction and upon hitting the other wall of the slab, and then the coordinates if the photon were not refracted, i.e. if the glass slab were absent. Now using distance formula, we can calculate the distance between the two points, thus, we get the value of lateral displacement.[An example upon which I tested my method.][1]

enter image description here

$\endgroup$
2
  • $\begingroup$ Please consider including diagrams/equations. $\endgroup$ Commented May 5 at 12:27
  • $\begingroup$ Done. I have added a few images of how I have successfully solved an example using my method. $\endgroup$ Commented May 5 at 13:24

0

Browse other questions tagged or ask your own question.