JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Wave Optics

For light entering the liquid from the top, the steepest refracted ray inside corresponds to grazing incidence, where $\sin \theta =\frac{1}{\mu }$. Complete polarization on reflection at the liquid-glass interface would require that ray to strike at the Brewster angle, $\tan {\theta }_B=\frac{1.5}{\mu }$. For polarization to be impossible no matter how the beam enters, even this extreme ray must fall short of Brewster's condition, and the borderline case sets $\sin \theta =\frac{1}{\mu }$ consistent with that geometry. Solving $\frac{1/\mu }{\sqrt{1-1/{\mu }^2}}=\frac{1.5}{\mu }$ gives ${\mu }^2=\frac{9}{5}$, so the minimum value is $\mu =\frac{3}{\sqrt{5}}$.