A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $i_C$ and the Brewster's angle of incidence is $i_B$, such that $sini_C/sini_B = = 1.28.$

The relative refractive index of the two media is

I know $tan i_B=\mu $(refractive index)

And in this case sin$i_C = \mu$

So putting these values in$sini_C/sini_B = = 1.28.$

I got $\sqrt{1+\mu^2}=1.28$

From this I got approx $\mu =0.6$

But answer is 0.8