91Ó°ÊÓ

The angle of the first sheet is,${\mathrm{θ}}_1=30°$.

The angle of the third sheet is,${\mathrm{θ}}_3=30°$.

Light can be polarized by passing it through a polarizing filter or other polarizing material. The intensity of an unpolarized light after passing through a polarized filter is given by the one-half rule or the cosine-square rule. Here, we need to use the concept of polarization using polarizing sheets along with the equations for the intensity of light passing through the polarizing sheet.

Formulae:

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

$\text{I}=0.5{\text{I}}_{\text{0}}$....................(1)

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

$I=I_0{\cos }^2\mathrm{θ}$.................(2)

Here, $\mathrm{θ}$is the angle between the polarization of the incident light and the polarization axis of the sheet.

As shown in the figure, for the first sheet, the incident light is unpolarized,

Thus, the intensity of the emergent light from sheet 1 is given using equation (1):

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

Now, the second sheet has a polarization direction along the x-axis. So, the angle of polarization will become

$\begin{array}{rcl}{\mathrm{θ}}_2& =& 90°-{\mathrm{θ}}_1\\ & =& 90°-30°\\ & =& 60°\\ & & \end{array}$

Using this angle in equation (2), we get the intensity of the emergent light from sheet 2 after passing sheet 1 as follows:

$\begin{array}{rcl}{I}_2& =& I_1{\cos }^2\left({60}^0\right)\\ & =& 0.25×\frac{1}{2}{I}_0\\ & =& 0.125I_0\\ & & \end{array}$

Now, the angle of the third sheet is with the y axis. So, the angle between the direction of polarization and the incident light for the third sheet will become

$\begin{array}{rcl}{\mathrm{θ}\text{'}}_3& =& 90°-{\mathrm{θ}}_3\\ & =& 90°-30°\\ & =& 60°\\ & & \end{array}$

Thus, the intensity of emergent light from sheet 3 after passing through 2 is given using equation (2) as follows:

$\begin{array}{rcl}{I}_3& =& I_2{\cos }^{2}60°\\ & =& 0.25×0.125I_0\\ & =& 0.03125I_0\\ & & \end{array}$

Thus, the fraction of light emerging from the system is given as follows:

$\begin{array}{rcl}\frac{I_3}{I_0}& =& 0.03125\\ & ≈& 0.032\\ & & \end{array}$

Hence, the value of the fraction is$0.032$.