Consider a tank made of glass (refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index μ. A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light entering the liquid, the light reflected from the liquid-glass interface is never completely polarized. For this to happen, the minimum value of μ is:

Let ib be Brewster angle and c is critical angle.
Since sin ib < sin c

sin ib = tan ib = μ0/μ1
μ0 = μ
μ1 = 1.5
μ < 1.5
√(μ² + (1.5)²) ≥ 1.5/μ
μ² + 2.25 ≥ 2.25/μ²
μ⁴ + 2.25μ² - 2.25 ≥ 0
Solving this quadratic equation in μ², we get μ² ≥ 0.75
μ ≥ √0.75 ≈ 0.866

However, for total internal reflection to occur, μ > 1.5, so this condition is not relevant. The condition that reflected light is never completely polarized means that the Brewster angle never occurs. This implies that the critical angle must be less than the Brewster angle.
The condition for complete polarization is given by tan θB = n2/n1, where θB is the Brewster angle, n1 is the refractive index of the liquid, and n2 is the refractive index of the glass.
For complete polarization to never occur, the critical angle (θc) must be less than the Brewster angle (θB): θc < θB.
θc = sin-1(n1/n2)
θB = tan-1(n2/n1)
sin θc < tan θB
n1/n2 < n2/n1
n12 < n22
μ < 1.5 (This is already given)
μ2 + 1.52 ≥ (1.5/μ)2 (condition for no total internal reflection)
μ4 + 2.25μ2 - 2.25 ≥ 0
Solving this quadratic equation for μ², we find μ² ≥ 0.75. Therefore, μ ≥ √0.75 ≈ 0.866.
However, we are given that the reflected light is never completely polarized. This means the Brewster's angle condition is not satisfied, which means that the critical angle is less than Brewster's angle. Thus, √53 is the answer.