An infinitely long cylinder of radius $R$ is made of an unusual exotic material with refractive index $-1$ (See figure). The cylinder is placed between two planes whose normals are along the $y$-direction. The center of the cylinder $O$ lies along the $y$-axis. $A$ narrow laser beam is directed along the $y$-direction from the lower plate. The laser source is at a horizontal distance $x$ from the diameter in the $y$-direction. Find the range of $x$ such that light emitted from the lower plane does not reach the upper plane.

(D) For a material with negative refractive index,Snell's law is given by,$-n = \frac{\sin \theta_i}{\sin \theta_r}$.
Given $n = -1$,we have $-(-1) = \frac{\sin \theta_i}{\sin \theta_r}$,which implies $\sin \theta_i = \sin \theta_r$,so $\theta_i = \theta_r$.
From the geometry,the ray enters at $B$ and exits at $C$. Due to the negative refractive index,the ray bends such that the total deviation is $4\theta_i$.
For the light not to reach the upper plate,the emergent ray must be directed downwards or sideways,meaning the deviation angle $4\theta_i$ must satisfy $\frac{\pi}{2} \leq 4\theta_i \leq \frac{3\pi}{2}$.
This simplifies to $\frac{\pi}{8} \leq \theta_i \leq \frac{3\pi}{8}$.
Using $\sin \theta_i = \frac{x}{R}$,and assuming small angles where $\sin \theta_i \approx \theta_i$,we get $\frac{\pi}{8} \leq \frac{x}{R} \leq \frac{3\pi}{8}$.
Thus,the range of $x$ is $\frac{\pi R}{8} \leq x \leq \frac{3\pi R}{8}$.