0
$\begingroup$

A modification of the simplest case of Young’s double slit experiment is when the path length for one of the slits is changed.

I've been told that if a strip of material of thickness $ t $ and refractive index $ n $ is placed over one slit then it adds a path difference of $(n − 1)t$, which results in the fringes being shifted. However, I am not sure how $(n − 1)t$ is derived and why this gives the path difference?

$\endgroup$
  • $\begingroup$ It’s unclear what you’re asking. In the title it says block or cover the slit. In the follow up it looks like you’re talking about some transparent material with a refractive index blocking it. $\endgroup$ – Bill Alsept Mar 21 '18 at 21:55
0
$\begingroup$

Earlier that light had to travel the distance $t$ in vacuum with refractive index 1. Now the refractive index is $n$ so the path will be $n*t$ The additional path is $n*t - t$ which is $(n-1)*t$. So if the path difference between the two rays was some $x$, $(n-1)*t$ would be added to it.

$\endgroup$
0
$\begingroup$

Apply the definition of optical path.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.