ACB is a right-angled glass prism with refractive index 1.5. Angle A, B, C are 60$^{\circ}$,30 $^{\circ}$and 90$^{\circ}$ respectively. A thin layer of liquid is on the AB. For a ray of light which is incident normally on AC to ;be totally reflected at AB, the refractive index of the liquid on AB should be

Given, $\mu$=refractive index of liquid.

From figure it is itself clear that , angle of incidence, $i=60^{o}$

Now, for internal reflection , angle of refraction $r=90^{o}$

Now, $\dfrac{\sin i}{\sin r}=\dfrac{\mu}{\mu_1}$

$\implies \sin 60^{o}=\dfrac{\mu}{1.5}$

$\implies \mu=\dfrac{\sqrt{3}}{2}\times 1.5$

$\implies \mu=1.3$

Hence, answer is option-(C)