91Ó°ÊÓ

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

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:

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

Where,

I = radiant intensity,

$\text{³¦´Ç²õ }\mathrm{θ}$= the angle between the direction of the incident light and the surface normal,

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,

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

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,

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

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 $\text{0.67}$.