Then a slab is introduced in front of one of the slits. What will be the path difference offered by the slab? The equation must be, according to my school teacher:
$$\left({n/n'-1}\right)p=d\sin\phi$$
But my brother told me it should be:
$$(n-n')p=d\sin\phi$$ Where $n$ is the refractive index of the slab , $n'$ is the refractive index of the optical medium and $p$ is the thickness of the slab.
It was justified by the optical path of the light rays in the slab and the medium (which is why $n'$ is additionally in product in the second equation) .
But do we have not already considered optical path by writing the refractive index of the slab with respect to the medium in the first equation?
Two men dealing with the same concept has different equations.
I can't make up who is the correct one.
It would be a great help if anyone can write the correct equation for the path difference offered by the slab in this scenario.