A planar structure of length $\mathrm{L}$ and width $\mathrm{W}$ is made of two different optical media of refractive indices ${\mathrm{n}}_1=1.5$ and ${\mathrm{n}}_2=1.44$ as shown in figure. If $\mathrm{L}>>\mathrm{W},$ a ray entering from end $\mathrm{A}\mathrm{B}$ will emerge from end $\mathrm{C}\mathrm{D}$ only if the total internal reflection condition is met inside the structure. For $\mathrm{L}=9.6\mathrm{ }\mathrm{m},$ if the incident angle $\mathrm{\theta }$ is varied, the maximum time taken by a ray to exit the plane $\mathrm{C}\mathrm{D}$ is $\mathrm{t}\times {10}^{-9}\mathrm{s},$ where $\mathrm{t}$ is ____.
[Speed of light $\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}]$

Solution

For maximum time taken by a rat to exit, it must travel maximum distance.
So, for this, it should be incident at minimum possible angle i.e., ${\theta }_C$ at $n_1-n_2$ boundary.
Applying Snells' law at $n_1-n_2$ boundary
$\Rightarrow 1.5\mathrm{s}\mathrm{i}\mathrm{n}{\theta }_C=1.44\mathrm{s}\mathrm{i}\mathrm{n}90°$
$\Rightarrow \mathrm{s}\mathrm{i}\mathrm{n}{\theta }_C=\frac{1.44}{1.50}=\frac{24}{25}$
$\therefore \mathrm{s}\mathrm{i}\mathrm{n}{\theta }_C=\frac{x}{d}=\frac{24}{25}$
$\Rightarrow d=\frac{25x}{24}$
$\therefore$ total length travel by light $=\frac{25}{25}\times 9.6=10m$ $\left(\mathrm{a}\mathrm{s},x_{\mathrm{m}\mathrm{a}\mathrm{x}}=L=9.6m\right)$
$\therefore t=\frac{S}{\left(\frac{C}{n_1}\right)}=\frac{10}{\frac{3\times {10}^n}{1.5}}\left(t=\frac{\text{distance\hspace{0.17em}travelled}}{\text{speed\hspace{0.17em}of\hspace{0.17em}light\hspace{0.17em}in\hspace{0.17em}medium}}\right)$
$=\frac{1}{2}\times {10}^{-7}=5\times {10}^{-9}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{s}$
$t=50ns$
$t=50\times {10}^{-9}$
$\therefore \mathrm{A}\mathrm{n}\mathrm{s}\mathrm{w}\mathrm{e}\mathrm{r}=50$

Hence, the correct answer is 50.