You are experimentally determining the critical angle for a novel transparent solid whose index of refraction is less than that of water $(n=1.33)$ but greater than that of air $(n=1.00)$. Your experimental method involves underwater work in which you observe whether there is any transmitted ray at the water/solid interface as you vary the angle of incidence on the solid surface. When your device reaches an incident angle of 53 degrees you find that the transmitted ray inside the solid vanishes. What is the index of refraction for this new solid?
A) 0.57
B) 1.06
C) 1.16
D) 0.53

Question

Quizplus · Accepted Answer

The critical angle ($\theta_c$) can be found using Snell's Law, $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, where $n_1$ and $n_2$ are the indices of refraction of the two media, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively. At the critical angle, $\theta_2 = 90^\circ$, so $\sin(\theta_2) = 1$, and the equation simplifies to $n_1 \sin(\theta_c) = n_2$. Given that $n_1 = 1.33$ (for water) and $\theta_c = 53^\circ$, we can solve for $n_2$ (the index of refraction of the solid) as follows: $1.33 \sin(53^\circ) = n_2$. Calculating this gives $n_2 \approx 1.06$, making choice B correct.

You Are Experimentally Determining the Critical Angle for a Novel (n=1.33)(n=1.33)(n=1.33)

You Are Experimentally Determining the Critical Angle for a Novel $(n=1.33)$