a student shines a laser through a semicircular transparent material of unknown index of refraction as…
a student shines a laser through a semicircular transparent material of unknown index of refraction as shown. he slowly increases the angle of incidence until the angle shown on the diagram when no light refracts into the air as shown. this is called total internal reflection! what is the index of refraction of the material?
Answer
Explanation:
Step1: Recall Snell's law for total internal reflection
For total - internal reflection, the critical angle $\theta_c$ is related to the refractive indices of the two media by $n_1\sin\theta_c=n_2\sin90^{\circ}$, where $n_1$ is the refractive index of the material (unknown), $n_2$ is the refractive index of air ($n_2 = 1$), and $\theta_c$ is the critical angle.
Step2: Identify the critical angle
From the diagram, the critical angle $\theta_c=36.0^{\circ}$.
Step3: Solve for the refractive index of the material
Substitute $n_2 = 1$ and $\theta_c = 36.0^{\circ}$ into the total - internal reflection formula $n_1\sin\theta_c=n_2\sin90^{\circ}$. Since $\sin90^{\circ}=1$, we have $n_1=\frac{n_2}{\sin\theta_c}$. Plugging in the values, $n_1=\frac{1}{\sin(36.0^{\circ})}$. We know that $\sin(36.0^{\circ})\approx0.5878$, so $n_1=\frac{1}{0.5878}\approx1.70$.
Answer:
$1.70$