Hint: The deviation of the ray depends on the refractive index of the cylinder.

Step 1: Find the angle of deviation of the ray.

Since, the material is of refractive index -1, ${\mathrm{\theta }}_{\mathrm{r}}$ is negative and ${\mathrm{\theta }}_{\mathrm{r}}^{\text{'}}$ positive.
Now,                           $\left|{\mathrm{\theta }}_{\mathrm{i}}\right|=\left|{\mathrm{\theta }}_{\mathrm{r}}\right|=\left|{\mathrm{\theta }}_{\mathrm{r}}^{\text{'}}\right|$
The total deviation of the outcoming ray from the incoming ray is 4${\mathrm{\theta }}_{\mathrm{i}}$. Rays shall not reach the receiving plate if,
                                    $\frac{\mathrm{\pi }}{2}\le 4{\mathrm{\theta }}_{\mathrm{i}}\le \frac{3\mathrm{\pi }}{2}$                         [angles measured clockwise from the y-axis]

On solving,                    $\frac{\mathrm{\pi }}{8}\le {\mathrm{\theta }}_{\mathrm{i}}\le \frac{3\mathrm{\pi }}{8}$

                  

Step 2: Find the range of light.

Now,                              ${\mathrm{sin\theta }}_{\mathrm{i}}=\frac{\mathrm{\pi }}{\mathrm{R}}$
                                       $\frac{\mathrm{\pi }}{8}\le {\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{R}}\right)\le \frac{3\mathrm{\pi }}{8}$
or                                     $\frac{\mathrm{\pi }}{8}\le \frac{\mathrm{x}}{\mathrm{R}}\le \frac{3\mathrm{\pi }}{8}$
Thus, for light emitted from the source shall not reach the receiving plate if; $\frac{\mathrm{R\pi }}{8}\le \mathrm{x}\le \frac{3\mathrm{R\pi }}{8}$