Answer
The index of refraction of the liquid is $~~1.26$
Work Step by Step
We can write Snell's law:
$n_2~sin~\theta_2 = n_1~sin~\theta_1$
We can find $\theta_1$, the angle the incident ray makes with the vertical:
$tan~\theta_1 = \frac{L}{D}$
$\theta_1 = tan^{-1}~(\frac{L}{D})$
$\theta_1 = tan^{-1}~(\frac{1.10~m}{0.850~m})$
$\theta_1 = 52.3^{\circ}$
We can use Snell's law to find $n_1$:
$n_2~sin~\theta_2 = n_1~sin~\theta_1$
$n_1 = \frac{n_2~sin~\theta_2}{sin~\theta_1}$
$n_1 = \frac{1.00~sin~90^{\circ}}{sin~52.3^{\circ}}$
$n_1 = 1.26$
The index of refraction of the liquid is $~~1.26$.
