91Ó°ÊÓ

We need to calculate the minimum possible value of N for light inside a cylinder of (a) water, (b) polystyrene plastic, and (c) cubic zirconia.

Figure shows a regular polygon with interior angle of $2{\mathrm{θ}}_c$where ${\mathrm{θ}}_c$is the critical angle of the medium-air boundary.

The interior angle must be greater or equal to $2{\mathrm{θ}}_c$to undergo multiple internal reflections.

localid="1648572916833" $$\begin{gathered} {\mathrm{θ}}_c={\sin }^{-1}\left(\frac{n_1}{n_2}\right) \\ {\mathrm{θ}}_c={\sin }^{-1}\left(\frac{1.00}{1.33}\right) \\ {\mathrm{θ}}_c=48.75° \\ Thereforeinteriorangle2{\mathrm{θ}}_c=2×48.75°=97.5° \end{gathered}$$

Since a regular pentagon is having an interior angle of 108^o, The minimum value for light inside the cylinder is 5.

$$\begin{gathered} {\mathrm{θ}}_c={\sin }^{-1}\left(\frac{n_1}{n_3}\right) \\ {\mathrm{θ}}_c={\sin }^{-1}\left(\frac{1.00}{1.59}\right) \\ {\mathrm{θ}}_c=38.97° \\ Thereforeinteriorangle2{\mathrm{θ}}_c=2×38.97°=77.9° \end{gathered}$$
Since a regular quadrilateral is having an interior angle of 90^o, The minimum value for light inside the cylinder is 4.

$$\begin{gathered} {\mathrm{θ}}_c={\sin }^{-1}\left(\frac{n_1}{n_4}\right) \\ {\mathrm{θ}}_c={\sin }^{-1}\left(\frac{1.00}{2.18}\right) \\ {\mathrm{θ}}_c=27.3° \\ Therefore,interiorangle2{\mathrm{θ}}_c=2×27.3°=54.6° \end{gathered}$$

Since a regular triangle is having an interior angle of 60^o, The minimum value for light inside the cylinder is 3.