What is the critical angle for the interface between water and sodium chloride?

We will try to solve this according to Snell's Law:
n1*sin(theta1) = n2*sin(theta2)this means that:n1 > is the refractive index of 1st medium theta1 > is the incident angle inside 1st medium n2 > refractive index of 2nd medium theta2 > refracted angle inside 2nd medium 
In critical angle, refracted angle inside the 2nd medium is 90 degreesthe sine of 90 is = 1. The equation will then become: n1*sin(theta1) = n2 
Let us now solve:sin(theta1) = n2/n1 theta1 = arcsin(n2/n1) Water is 1.33 fused quartz is 1.46so theta1 = arcsin(n2/n1) = arcsin(1.33/1.46) = 65.64 degrees.