91Ó°ÊÓ

Figure 33.15a with the light ray arrangement is given.

The ratio of final to initial intensity,$\frac{I_3}{I_0}=A{\cos }^n\mathrm{θ}$

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 cosine-square rule. Thus, using this concept of polarization and the given polarizing sheets, along with the equations for the intensity of light passing through the polarizing sheet, we can get the ratio of the final to the initial intensity of the polarized light.

Formulae:

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

$I=I_0{\cos }^2\mathrm{θ}$ (i)

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

For the first sheet, the incident light is polarized parallel to the x-axis.

As the first sheet is rotated through $60°$counterclockwise, the angle will be given as:

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

Thus, substituting this in equation (i), we can get the intensity value of emerging light from sheet 1 as: $I_1=I_0{\cos }^{2}30°$

Thus, using equation (i), we can get the intensity of the emerging light through sheet 2 as:

$I_2=I_1{\cos }^{2}0°$

Now, the third sheet has a polarization direction along the x-axis. So, its angle with the polarization direction of the incident light will be $30°$.

Thus, the emerging light through sheet 3 due to incident light from 1 is given using equation (i) as:

$\begin{array}{rcl}{I}_3& =& I_2{\cos }^{2}30°\\ & =& I_0{\cos }^{2}30°×{\cos }^{2}30°\\ & =& I_0{\cos }^{4}30°\end{array}$

So, the fraction of light intensity is,

$\frac{I_3}{I_0}=1{\cos }^{4}30°$

Comparing the above fractional value with the given equation,

$\frac{I_3}{I_0}=A{\cos }^n\mathrm{θ}$

We get $A=1,n=4,\mathrm{θ}=30°$.

Hence, the values of $A,n,\text{and}\mathrm{θ}$ if we rotate the polarizing direction of the first sheet $60°$ counterclockwise are $1,4\mathrm{and}30°$respectively.

The incident light is polarized parallel to the x-axis. So, when we rotate the polarization direction of the first sheet by clockwise, it will become parallel to the incident light and will allow the whole light to pass through. So, there will be no change in intensity.

Thus, the intensity of emerging light from sheet 1 is given by,

$I_1=I_0$

Now, the second sheet has a polarization direction through $60°$counterclockwise. So, the angle between the polarization of the incident light and the polarization axis of the sheet will be given as:

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

Thus, the intensity of the emerging light from sheet 2 due to incident light from sheet 1 is given using equation (i) as follows:

$\begin{array}{rcl}{I}_2& =& I_1{\cos }^{2}30°\\ & =& I_0{\cos }^{2}30°\end{array}$

Now, the third sheet has a polarization direction along the x-axis. So, its angle with the polarization direction of the incident light will be $30°$.

Thus, the intensity of the emerging light from sheet 3 due to incident light from sheet 2 will be given using equation (i) as follows:

$\begin{array}{rcl}{I}_3& =& I_2{\cos }^{2}30°\\ & =& I_0{\cos }^{2}30°×{\cos }^{2}30°\\ & =& I_0{\cos }^{4}30°\end{array}$

So, the fraction of light intensity is given by,

$\frac{I_3}{I_0}=1{\cos }^{4}30°$

Comparing the above condition with the given equation,

$\frac{I_3}{I_0}=A{\cos }^n\mathrm{θ}$

We get $A=1,n=4$and $\mathrm{θ}=30°$

Hence, the values of $A,n,\text{and}\mathrm{θ}$ if we rotate the polarizing direction of the first sheet $90°$clockwise $1,\text{4, 30}°$ , respectively