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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.

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Your teacher is right. We can expand the expression to $$ np=n' d \sin\theta + n'p$$

For the 0th order, the rays from both slits must have the same optical path lengths. From simple geometry of the double slit experiment we know that the extra path one of the rays travels is $d\sin\theta$, which is the first term on the right. Without the slab, from there on, the optical path lengths are equal. But with the slab, along the thickness of it, one of the rays has an optical path of $np$, while the other only has $n'p$, which are the term on the left, and the second term on the right, respectively. From the point after the slab to the screen, the paths are identical again.

Sketch without slab from researchgate: Sketch

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  • $\begingroup$ According to you sir, one of the rays has an optical path of np.But if the slab is placed in a medium of absolute refractive index n', the refractive index of slab must be taken relative and it must be n/n'. Then the optical path of the ray must be (n/n')p accordingly. Am I right? Please correct me if I'm wrong. $\endgroup$ Jul 30, 2019 at 13:49
  • $\begingroup$ The answer contains a whole argument how to arrive at the formula for $n/n'$. I'm not sure what your question is. $\endgroup$
    – noah
    Jul 30, 2019 at 14:02

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