Question

Light of original intensity I_0 passes through two ideal polarizing filters having their polarizing axes oriented as shown in the figure (Figure 1) . You want to adjust the angle \phi so that the intensity at point P is equal to I_0/10.0

A) if the original light is unpolarized, what should phi be?

B)If the original light is linearly polarized in the same direction as the polarizing axis of the first polarizer the light reaches, what should phi be?

Answer 1

Concepts and reason

The concept required to solve this problem is Malus law.

Initially use the Malus law to calculate for the angle between the light's polarization direction and the axis of the polarizer for each case.

Fundamentals

The law of Malus gives the intensity of emerging light pulse.

$I = {I_0}{\cos ^2}\theta$

Here, ${I_0}$ is the initial intensity, and $\theta$ is the angle between the light's initial polarization direction and the axis of the polarizer.

For an unpolarized light average value of ${\cos ^2}\theta$ is used that is $\frac{1}{2}$ , as there is no specific angle between the unpolarized light and polarizer.

(A)

Use law of malus for first incidence on the vertical polarizer.

Substitute $\frac{1}{2}$ for ${\cos ^2}\theta$ in the equation $I = {I_0}{\cos ^2}\theta$ .

$I = {I_0}\left( {\frac{1}{2}} \right)$

Rearrange the equation of law of malus and solve for angle.

$\begin{array}{c}\\I = {I_0}{\cos ^2}\phi \\\\\phi = {\cos ^{ - 1}}\left( {\sqrt {\frac{I}{{{I_0}}}} } \right)\\\end{array}$

Substitute $\phi$ for $\theta$ , $\frac{{{I_0}}}{2}$ for ${I_0}$ and $\frac{{{I_0}}}{{10}}$ for $I$ in the above equation $\theta = {\cos ^{ - 1}}\left( {\sqrt {\frac{I}{{{I_0}}}} } \right)$ and calculate the angle.

$\begin{array}{c}\\\phi = {\cos ^{ - 1}}\left( {\sqrt {\frac{{\left( {\frac{{{I_0}}}{{10}}} \right)}}{{\left( {\frac{{{I_0}}}{2}} \right)}}} } \right)\\\\ = {\cos ^{ - 1}}\left( {\sqrt {\frac{1}{5}} } \right)\\\\ = 63.43^\circ \\\end{array}$

(B)

Use law of malus for first incidence on the vertical polarizer.

Substitute $0$ for $\theta$ in the equation $I = {I_0}{\cos ^2}\theta$ .

$\begin{array}{c}\\I = {I_0}{\cos ^2}\left( 0 \right)\\\\ = {I_0}\\\end{array}$

Rearrange the equation of law of malus and solve for angle.

$\begin{array}{c}\\I = {I_0}{\cos ^2}\theta \\\\\theta = {\cos ^{ - 1}}\left( {\sqrt {\frac{I}{{{I_0}}}} } \right)\\\end{array}$

Substitute $\phi$ for $\theta$ , $\frac{{{I_0}}}{{10}}$ for $I$ in the above equation $\theta = {\cos ^{ - 1}}\left( {\sqrt {\frac{I}{{{I_0}}}} } \right)$ and calculate the angle.

$\begin{array}{c}\\\phi = {\cos ^{ - 1}}\left( {\sqrt {\frac{{\left( {\frac{{{I_0}}}{{10}}} \right)}}{{\left( {{I_0}} \right)}}} } \right)\\\\ = {\cos ^{ - 1}}\left( {\sqrt {\frac{1}{{10}}} } \right)\\\\ = 71.57^\circ \\\end{array}$

Ans: Part A

The angle between the vertical and polarizer is $\phi = 63.43^\circ$ .

Part B

The angle between the vertical and polarizer is $\phi = 71.57^\circ$ .