An infinitely long cylinder of radius R is made of an unusual exotic material with refractive index -1 (Fig). 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.
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We are given that, the refractive index of the material is -1, $\theta_\text{r}$, is negative and $\theta_\text{r}'$ is positive.
Now, $|\theta_\text{i}|=|\theta_\text{r}|=|\theta_\text{r}'|$
The total deviation of the out coming ray from the incoming ray is $4\theta_\text{i}$.
Rays shall not reach the receiving plate if $\frac{\pi}{2}\leq4\theta,\leq\frac{3\pi}{2}$ [angles measured clockwise from the y-axis]
Dividing all the terms by 4, we get
$\frac{\pi}{8}\leq\theta,\leq\frac{3\pi}{8}\ .....(1)$
Now, $\sin\theta_\text{i}=\frac{\text{x}}{\text{R}}$
$\Rightarrow\ \theta_\text{i}=\sin^{-1}\Big(\frac{\text{x}}{\text{R}}\Big)$
Substituting $\theta_\text{i}=\sin^{-1}\Big(\frac{\text{x}}{\text{R}}\Big)$ in (1), we get
$\frac{\pi}{8}\leq\sin^{-1}\frac{\text{x}}{\text{R}}\leq\frac{3\pi}{8}$
$\frac{\pi}{8}\leq\frac{\text{x}}{\text{R}}\leq\frac{3\pi}{8}$
$\Rightarrow\ \frac{\pi\text{R}}{8}\leq\text{x}\leq\frac{3\pi\text{R}}{8}$
Thus, the range of x such that light emitted from the lower plane does not reach the upper plane is
$\frac{\text{R}\pi}{8}\leq\text{x}\leq\frac{\text{R}3\pi}{8}.$
Now, $|\theta_\text{i}|=|\theta_\text{r}|=|\theta_\text{r}'|$
The total deviation of the out coming ray from the incoming ray is $4\theta_\text{i}$.
Rays shall not reach the receiving plate if $\frac{\pi}{2}\leq4\theta,\leq\frac{3\pi}{2}$ [angles measured clockwise from the y-axis]
Dividing all the terms by 4, we get
$\frac{\pi}{8}\leq\theta,\leq\frac{3\pi}{8}\ .....(1)$
Now, $\sin\theta_\text{i}=\frac{\text{x}}{\text{R}}$
$\Rightarrow\ \theta_\text{i}=\sin^{-1}\Big(\frac{\text{x}}{\text{R}}\Big)$
Substituting $\theta_\text{i}=\sin^{-1}\Big(\frac{\text{x}}{\text{R}}\Big)$ in (1), we get
$\frac{\pi}{8}\leq\sin^{-1}\frac{\text{x}}{\text{R}}\leq\frac{3\pi}{8}$
$\frac{\pi}{8}\leq\frac{\text{x}}{\text{R}}\leq\frac{3\pi}{8}$
$\Rightarrow\ \frac{\pi\text{R}}{8}\leq\text{x}\leq\frac{3\pi\text{R}}{8}$
Thus, the range of x such that light emitted from the lower plane does not reach the upper plane is
$\frac{\text{R}\pi}{8}\leq\text{x}\leq\frac{\text{R}3\pi}{8}.$
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