91影视

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

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

$\mathbf{I}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}{\mathbf{I}}_0$鈥︹�� (i)

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

$\mathit{I}\mathbf{=}{\mathit{I}}_{\mathbf{0}}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{胃}$ 鈥︹��. (ii)

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

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:

$I_2=\frac{1}{2}{I}_0{\cos }^2胃$ 鈥︹�� (iii)

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

$I_2=\frac{1}{3}{I}_0$ 鈥︹�� (iv)

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

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

Solve further as:

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

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