Question

A beam of initially unpolarized light is sent through two polarizing sheets placed one on top of the other. What must be the angle between the polarizing directions of the sheets if the intensity of the transmitted light is to be one-third the incident intensity?

Short answer

The polarizing angle between the polarizing directions of the sheets is.$35°$

Step-by-step solution

Write the given data 

A beam of unpolarized light is sent through two polarizing sheets placed one on top of the other.

Determine the formulas for polarization 

If the incident light is un-polarized, then the intensity of the merging light using one-half rule is given by:

$\mathbf{I}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}{\mathbf{I}}_0$…… (i)

If the incident light is already polarized, then the intensity of the emerging light is cosine –squared of the intensity of incident light is as follows:

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }$ ……. (ii)

Here, $\theta$is the angle between polarization of the incident light and the polarization axis of the sheet.

Determine the polarizing angle 

Let $I_{0}andI$betheincident and transmitted intensity respectively.

Initially, the beam is unpolarized. When it passes through the first sheet, thus using equation (i), we can say that the transmitted intensity I is half the original intensity.$I_0$

$I_1=\frac{1}{2}{I}_0$

This then passes through the second sheet, for which, the intensity of the emergent light from sheet 2 due to incident intensity from sheet 1 is given using equation (ii) as:

$I_2=\frac{1}{2}{I}_0{\cos }^2\theta$ …… (iii)

It is given thattheintensity of the transmitted light is one third the original incident intensity. $I_0$That is given as:

$I_2=\frac{1}{3}{I}_0$ …… (iv)

Thus, the angle of polarizing between the two sheets can be given using equation (a) and (b) as follows:

$\begin{array}{l}\frac{1}{2}{I}_0{\cos }^2\theta =\frac{1}{3}{I}_0\\ {\cos }^2\theta =\frac{2}{3}\\ \cos \theta =\sqrt{\frac{2}{3}}\\ \theta =co{s}^{-1}\sqrt{\frac{2}{3}}\end{array}$

Solve further as:

$\begin{array}{l}\frac{1}{2}{I}_0{\cos }^2\theta =\frac{1}{3}{I}_0\\ {\cos }^2\theta =\frac{2}{3}\\ \cos \theta =\sqrt{\frac{2}{3}}\\ \theta =co{s}^{-1}\sqrt{\frac{2}{3}}\end{array}$

The angle between the polarizing directions of the sheets is.$35°$