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An infinitely long cylinder of radius $\mathrm{R}$ is made of an unusual exotic material with refractive index-1 (figure). The cylinder is placed between two planes whose normals are along $t$ $y$-direction. The centre of the cylinder $\mathrm{O}$ lies along the $\mathrm{y}$ axis. A narrow laser beam is directed along the $y$-direction from the lower plate. The Laser source is at a horizontal distance $\mathrm{x}$ from the diameter in the $\mathrm{y}$ direction. Find the range of $\mathrm{x}$ such that light emitted from the lower plane does not reach the upper plane.
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Verified Answer
As, the cylinder is made of material of refractive index $(-1)$ and is placed in air of $(\mu=1)$, so, when ray $\mathrm{AB}$ is incident at $\mathrm{B}$ to cylinder, $\theta_{\Gamma}$, is negative and $\theta_{\mathrm{t}}^{\prime}$ positive.
Now, $\quad\left|\theta_t\right|=\left|\theta_{\mathrm{r}}\right|=\left|\theta_{\mathrm{r}}^{\prime}\right|$
The total deviation of the outcoming ray from the incoming ray is $4 \theta_t$. Rays shall not reach the recieving plane if
$$
\frac{\pi}{2} \leq 4 \theta_{\mathrm{t}} \leq \frac{3 \pi}{2}
$$
So, angles measured clockwise from the $y$-axis.
On dividing by 4 to all sides
By solving, $\quad \frac{\pi}{8} \leq \theta_{\mathrm{t}} \leq \frac{3 \pi}{8}$
Now, $\quad \sin \theta_{\mathrm{t}}=\frac{\mathrm{x}}{\mathrm{R}}$
$$
\frac{\pi}{8} \leq \sin ^{-1} \frac{x}{R} \leq \frac{3 \pi}{8} \text { or, } \frac{\pi}{8} \leq \frac{x}{R} \leq \frac{3 \pi}{8}
$$
Hence, for light emitted from the source shall not reach the receiving plane. If $\frac{\mathrm{R} \pi}{8} \leq \mathrm{x} \leq \frac{\mathrm{R} 3 \pi}{8}$.
Now, $\quad\left|\theta_t\right|=\left|\theta_{\mathrm{r}}\right|=\left|\theta_{\mathrm{r}}^{\prime}\right|$
The total deviation of the outcoming ray from the incoming ray is $4 \theta_t$. Rays shall not reach the recieving plane if
$$
\frac{\pi}{2} \leq 4 \theta_{\mathrm{t}} \leq \frac{3 \pi}{2}
$$
So, angles measured clockwise from the $y$-axis.
On dividing by 4 to all sides
By solving, $\quad \frac{\pi}{8} \leq \theta_{\mathrm{t}} \leq \frac{3 \pi}{8}$
Now, $\quad \sin \theta_{\mathrm{t}}=\frac{\mathrm{x}}{\mathrm{R}}$
$$
\frac{\pi}{8} \leq \sin ^{-1} \frac{x}{R} \leq \frac{3 \pi}{8} \text { or, } \frac{\pi}{8} \leq \frac{x}{R} \leq \frac{3 \pi}{8}
$$
Hence, for light emitted from the source shall not reach the receiving plane. If $\frac{\mathrm{R} \pi}{8} \leq \mathrm{x} \leq \frac{\mathrm{R} 3 \pi}{8}$.
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