Suppose that a thin hollow glass slab with a refractive index of $1.5$ contains a liquid inside, which has a refractive index of $\mu$, and $\mu \neq 1.5$. If I now send a laser light at the medium and measure the angle of refraction, will it now be $ sin^{-1} (\frac{sin i}{\mu}) $ or $ sin^{-1} (\frac{sin i}{1.5})$? So will the refractive index of this medium become $\mu$ or does it remain $1.5$?
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$\begingroup$ Related : Refraction across two interfaces. $\endgroup$– FrobeniusCommented Jan 29, 2022 at 9:48
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2 Answers
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For an interface between two mediums $A$ and $B$ with absolute refractive index $\mu$ and angle from normal $i$, the relationship $\mu_{\rm A}\sin i_{\rm A} = \mu_{\rm B}\sin i_{\rm B} $ can be used.
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Can you please clarify what is meant by $i$ - is it the angle of incidence on the hollow glass? If so, according to Snell's law you may write ($\alpha$ is angle of refraction in the glass and $\beta$ is the angle of refraction in the water): $\sin i = 1.5 \sin \alpha = \mu \sin \beta$.
As you are interested in the angle of refraction in the medium, the right answer is the first one $\beta = \arcsin (\frac{\sin i}{\mu})$.
