Question

A light beam can use reflections to form a closed, N-sided polygon inside a solid, transparent cylinder if N is sufficiently large. What is the minimum possible value of N for light inside a cylinder of (a) water, (b) polystyrene plastic, and (c) cubic zirconia? Assume the cylinder is surrounded by air.

Short answer

a) Minimum possible value of N in case of water is 5

b) Minimum possible value of N in case of polystyrene plastic is 4

c) Minimum possible value of N in case of cubic zirconia is 3

Step-by-step solution

Step 1. Given information is :Cylinder is surrounded by airRefractive index of air, n1 =1.00Refractive index of water, n2 =1.33Refractive index of polystyrene plastic, n3 =1.59Refractive index of cubic zirconia, n4 =2.18

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.

Step 2. a) Water-Air boundary Using the concept of critical angle.

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

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

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

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

Step 3. Part b) Polystyrene-Air boundary

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

Step 4. Part c) Polystyrene-Plastic boundary 

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

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