Young's double slit experiment with two glass slabs

Question

Suppose I place two glass slabs with different thickness and refractive indices in a YDSE setup. I was told that if $$ (\mu_1-1)t_1=(\mu_2-1)t_2 $$ then there would be no change in the interference pattern, as compared to the one without any glass slab. Why isn't this true for $$\mu_1 t_1=\mu_2 t_2$$ i.e. equating their geometrical paths rather than the difference of geomterical and optical paths? Since the only difference caused presence of glass slab is that the ray travels a distance $\mu t$ rather than $t$.

Answer 1

For each path you started with $\dfrac t\lambda_{\rm air}$ wavelengths in a thickness $t$ of air.

Replacing the air with a medium of refractive index $\mu$ means that in a thickness $t$ there are now $\dfrac{t}{\lambda_{\rm air}/\mu} = \dfrac{\mu t}{\lambda_{\rm air}}$ wavelengths.

For there to be no change in the pattern the change in the number of waves must be the same for both paths, ie $\dfrac{\mu_1 t_1}{\lambda_{\rm air}} -\dfrac {t_1}{ \lambda_{\rm air}} = \dfrac{\mu_2 t_2}{\lambda_{\rm air}} -\dfrac {t_2}{ \lambda_{\rm air}}$ and the relationship follows.

Answer 2

That is because for the thinner slab the wave travels a distance t$_2$-t$_1$ trough vacuum. Equality of the phases requires $$\mu_1 t_1 +t_2-t_1 = \mu_2 t_2$$ which immediately gives the correct result.

Answer 3

For no change in interference pattern the sufficient condition is just that the central bright is observed at central line(as the other pattern would automatically follow and turn out to be the same)

Hence, we want path difference at the centre to be zero