Question

A beam of partially polarized light can be considered to be a mixture of polarized and unpolarized light. Suppose we send such a beam through a polarizing filter and then rotate the filter through$\mathbf{\text{360°}}$while keeping it perpendicular to the beam. If the transmitted intensity varies by a factor of$\mathbf{\text{5.0}}$during the rotation, what fraction of the intensity of the original beam is associated with the beam’s polarized light?

Short answer

The fraction of intensity of the original beam associated with the beam’s polarized light is $\text{0.67}$.

Step-by-step solution

Given

Transmission intensity varies by factor $\text{5.0}$.

Determining the concept

Here, it is needed to consider both the polarized and non-polarized portion of light. The intensity of transmitted light can be calculated using the equation of the one-half rule and cosine-squared rule. By adding the intensity of the polarized portion and the intensity of the non-polarized portion, the total intensity can be found. Then taking the ratio of maximum and minimum intensity, solve for the fraction of polarized light.

The formula is as follows:

$$\begin{gathered} I=\frac{\text{1}}{\text{2}}{I}_0 \\ I=I_0{\text{cos}}^{\text{2}}\theta \end{gathered}$$

Where,

I = radiant intensity,

$\text{cos\hspace{0.17em}}\theta$= the angle between the direction of the incident light and the surface normal,

Determining the fraction of intensity of the original beam associated with the beam’s polarized light

Let us assume that ‘f’is the fraction of polarized light. So, (1-f) will be the fraction of unpolarized light. Hence, the intensity of polarized light will be,

$I_p=f{I}_0{\text{cos}}^{\text{2}}\theta$

Similarly, the intensity of an unpolarized portion of the light will be,

$I_u=\frac{\text{1}}{\text{2}}(\text{1}-\text{f})I_0$

So, the total intensity of transmitted light after passing through the filter will become,

$$\begin{gathered} I=f{I}_0{\text{cos}}^{\text{2}}\theta \text{+0.5}(\text{1}-f)I_0 \\ \text{0to1;} \end{gathered}$$

The filter is rotated through $\text{360°;}$the value of changes, therefore,

$\begin{array}{c}{I}_{max}=f{I}_0+\text{0.5}(\text{1}-f)I_0\\ I_{max}=\text{0.5}(\text{1}+f)I_0\frac{I_{max}}{I_{min}}=\frac{\text{1}+f}{\text{1}-f}\end{array}$

And minimum intensity is as follows:

$I_{min}=\text{0.5}(\text{1}-f)I_0$

Therefore, taking the ratio,

Butit is known that the transmitted intensity varies by a factor of 5.0.

It means,
$\frac{I_{max}}{I_{min}}=\text{5.0}$

So, the above equation will become,

$\text{5.0}=\frac{\text{1}+f}{\text{1}-f}$

Solving for f,

$f=\text{0.67}$

Hence, the fraction of intensity of the original beam associated with the beam’s polarized light is $\text{0.67}$.