One want to rotate the direction of polarization of a beam of polarized light throughby sending the beam through one or more polarizing sheets.

(a) What is the minimum number of sheets required?

(b) What is the minimum number of sheets required if the transmitted intensity is to be more than$\mathbf{\text{60\%}}$of the original intensity?

Use the concept of change in intensity of light by polarization. Consider the number of sheets required to get the required intensity.

The intensity of the polarized light-

$I=I_0{\cos }^2\theta$

Where,l is the intensity.

Polarized light cannot be rotated with a single sheet. If the direction of the incident light and the sheet iswith each other, then there is no transmission of polarized light. The rotation can be done with two sheets. Usethesheet with angle$\theta$,between 0 to${90}^0$,to the direction of polarization of incident light. Placethesecond sheet withwiththeoriginal direction of polarized light.Polarization is rotated again by${90}^0-\theta$. The intensity will not be zero if the first angle is specified between$0<\theta <{90}^0$.

Write the intensity as,

$I=I_0{\cos }^2\theta \times {\cos }^2({90}^0-\theta )$

$I=I_0{\cos }^2\theta \times {\sin }^2\theta$

Therefore, we need two sheets.

Find the number of sheets using the angle$\theta =\frac{{90}^0}{n}$relative to the direction of polarization of incident light. The polarizing direction of the successive sheets is rotated$\frac{{90}^0}{n}$ liketheprevious sheet. The direction of transmitted polarized light is${90}^0$withthedirection of incident light.

Write the intensity as,

$I=I_0{\cos }^{2n}\left(\frac{{90}^0}{n}\right)$

Using the value of$n=1,2,3\dots .$,

$\begin{array}{c}{I}_1=I_0{\cos }^2\left(\frac{{90}^0}{\left(1\right)}\right)\\ =0^0\end{array}$

$\begin{array}{c}{I}_2=I_0{\cos }^4\left(\frac{{90}^0}{\left(2\right)}\right)\\ =0.25I_0\end{array}$

$\begin{array}{c}{I}_3=I_0{\cos }^6\left(\frac{{90}^0}{\left(3\right)}\right)\\ =0.442I_0\end{array}$

$\begin{array}{c}{I}_4=I_0{\cos }^8\left(\frac{{90}^0}{\left(4\right)}\right)\\ =0.531I_0\end{array}$

$\begin{array}{c}{I}_5=I_0{\cos }^{10}\left(\frac{{90}^0}{\left(10\right)}\right)\\ =0.605I_0\end{array}$

Thus, $I>0.60I_0$with five sheets.

So minimum 5 sheets are required if the transmitted intensity is to be more than 60% of the original intensity.