Answer
$1.41\lt n_{prism}\lt1.88$
Work Step by Step
We know that when the reservoir is empty the light experience a total internal reflection toward the sensor.
And when there is liquid, some of the light reflects and the other part refracts outside the prism inside the liquid.
For total reflection,
$$\dfrac{n_2}{n_1}=\sin\theta_c$$
where $n_1$ is for the prism and $n_2$ is for the outside material.
Solving for $n_1$;
$$n_1=\dfrac{n_2}{\sin\theta_c}\tag1$$
The minimum value of the index of refraction of the prism is given when its hypotenuse is covered by air
$$n_{1,minimum}=\dfrac{n_{air}}{\sin\theta_c}$$
From the geometry of the given figure, we can see that the incident angle and the reflected angle are 45$^\circ$.
So,
$$n_{1,minimum}=\dfrac{1}{\sin45^\circ}=\bf 1.41$$
The maximum value of the index of refraction of the prism is given when its hypotenuse is covered by the liquid (water).
$$n_{1,maximum }=\dfrac{n_{water}}{\sin\theta_c}$$
From the geometry of the given figure, we can see that the incident angle and the reflected angle are 45$^\circ$.
So,
$$n_{1,maximum }=\dfrac{1.33}{\sin45^\circ}=\bf 1.88$$
Therefore, and from all the above,
$$\boxed{1.41\lt n_{prism}\lt1.88}$$
