Chapter 31, Problem 59P

To determine

The optimal number of sheet.

Answer to Problem 59P

The optimal number of sheets is $11$ .

Explanation of Solution

Formula used:

The expression for intensity is given by,

  $I_{\text{N}+1}=I_0{\cos }^2\left(\frac{\pi }{2N}\right)$

Calculation:

The intensity for $2\%$ loss is calculated as,

  $\begin{array}{c}{I}_{\text{N}+1}=I_0\left(0.98\right){\cos }^2\left(\frac{\pi }{2N}\right)\\ \frac{I_{\text{N}+1}}{I_0}=\left(0.98\right){\cos }^2\left(\frac{\pi }{2N}\right)\end{array}$

For the ideal polarizer,

  $\frac{I_{\text{N}+1}}{I_0}={\cos }^2\left(\frac{\pi }{2N}\right)$

For the real polarizer,

  $\frac{I_{\text{N}+1}}{I_0}=\left(0.98\right){\cos }^2\left(\frac{\pi }{2N}\right)$

So, the percent transmission is given by,

  $x={\left(0.98\right)}^N$

The values of $\frac{I_{\text{N}+1}}{I_0}$ for different values of $N$ for ideal polarizer, real polarizer and percent transmission is shown in table below,

	For ideal polarizer	Percent transmission	Real polarizer
$N$	${\cos }^2\left(\frac{\pi }{2N}\right)$	${\left(0.98\right)}^N$	$\left(0.98\right){\cos }^2\left(\frac{\pi }{2N}\right)$
$1$	$0.000$	$0.980$	$0.000$
$2$	$0.250$	$0.960$	$0.240$
$3$	$0.422$	$0.941$	$0.397$
$4$	$0.531$	$0.922$	$0.490$
$5$	$0.605$	$0.904$	$0.547$
$6$	$0.660$	$0.886$	$0.584$
$7$	$0.701$	$0.868$	$0.608$
$8$	$0.733$	$0.851$	$0.624$
$9$	$0.759$	$0.834$	$0.633$
$10$	$0.781$	$0.817$	$0.638$
$17$	$0.848$	$0.739$	$0.626$
$18$	$0.857$	$0.724$	$0.620$
$19$	$0.865$	$0.709$	$0.613$
$20$	$0.872$	$0.695$	$0.606$
$21$	$0.878$	$0.681$	$0.598$
$22$	$0.884$	$0.668$	$0.590$

Table 1

The graph of intensity of transmitted light as a function of $N$ consider table 1 is shown below,

  

Figure 1

The inspection of the graph gives the optimal number of sheets $N$ is $11$ ,

Conclusion:

Therefore, the optimal number of sheets is $11$ .