Chapter 22, Problem 61P

(a)

To determine

What is the intensity of light exiting the last polarizer?

Answer to Problem 61P

The intensity of light exiting the last polarizer is $0.0938I_0$.

Explanation of Solution

The light is un-polarized at first. When an un-polarized light is passed through a polarizer, its intensity becomes half of the initial.

Write the equation to find the intensity of light from first polarizer.

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

Here, $I_1$ is the intensity of light from first polarizer, $I_0$ is the intensity of un-polarized light

Write the equation to find the intensity of light coming from second polarizer.

$\begin{array}{c}{I}_2=I_1{\cos }^2\theta \\ =\frac{1}{2}{I}_0{\cos }^2\theta \end{array}$ (II)

Here, $I_2$ is the intensity of light coming from second polarizer, $I_1$ is the intensity of light from first polarizer

Write the equation to find the intensity of light coming from third polarizer.

$I_3=I_2{\cos }^2\theta$ (III)

Here, $I_2$ is the intensity of light coming from second polarizer, $I_3$ is the intensity from third polarizer

Conclusion

Substitute $60°$ for $\theta$ in equation (II) to get $I_2$

$\begin{array}{c}{I}_2=\frac{1}{2}{I}_0{\cos }^{2}60°\\ =0.125I_0\end{array}$

Substitute $0.125I_0$ for $I_2$ , $30°$ for $\theta$ in equation (III) to get $I_3$

$\begin{array}{c}{I}_3=0.1250I_0{\cos }^{2}30°\\ =0.0938I_0\end{array}$

Therefore, The intensity of light exiting the last polarizer is $0.0938I_0$.

(b)

To determine

What is the intensity of light exiting the last polarizer if the light is incident vertically from left?

Answer to Problem 61P

The intensity of light exiting the last polarizer is $0.188I_0$.

Explanation of Solution

Here the light has the same polarization as the first ideal polarizer; there will be no change in intensity for light that passes the first polarizer.

$I_1=I_0$ (IV)

Write the equation to find the intensity of light coming from second polarizer.

$\begin{array}{c}{I}_2=I_1{\cos }^2\theta \\ =I_0{\cos }^2\theta \end{array}$ (V)

Write the equation to find the intensity of light coming from third polarizer.

$I_3=I_2{\cos }^2\theta$ (VI)

Conclusion:

Substitute $60°$ for $\theta$ in equation (IV) to get $I_2$

$\begin{array}{c}{I}_2=I_0{\cos }^{2}60°\\ =0.25I_0\end{array}$

Substitute $0.25I_0$ for $I_2$ , $30°$ for $\theta$ in equation (VI) to get $I_3$

$\begin{array}{c}{I}_3=0.250I_0{\cos }^{2}30°\\ =0.188I_0\end{array}$

Therefore, The intensity of light exiting the last polarizer is $0.188I_0$.

(c)

To determine

Can one polarizer be removed from this series of filters so that light incident from the left is not transmitted at all if un-polarized light is incident as in part a? If so, which one should be removed and answer same questions for vertically polarized incident light as in part b.

Answer to Problem 61P

Yes, polarizer can be removed. In both cases remove the middle polarizer.

Explanation of Solution

In both the cases in part a and part b the light is vertically polarized. The angle of polarization between the first and last polarizers is $90°$, there are three polarizers altogether. Therefore if the middle polarizer is removed then the light will not be transmitted. So we can remove the middle polarizer.

Conclusion:

Therefore, yes, polarizer can be removed. In both cases remove the middle polarizer.

(d)

To determine

Which polarizer should be removed to maximize the amount of light transmitted in part a?

Answer to Problem 61P

In case a first polarizer should be removed and in case b last polarizer should be removed.

Explanation of Solution

Consider case a.

Write the equation to find the intensity if the first polarizer is removed.

$I=\frac{1}{2}{I}_0{\cos }^2\theta$ (VII)

Write the equation to find the intensity if the last polarizer is removed.

$I=\frac{1}{2}{I}_0{\cos }^2\theta$ (VIII)

Consider part b.

Write the equation to find the intensity if the first polarizer is removed.

$I=\frac{1}{2}{I}_0{\cos }^2{\theta }_1{\cos }^2{\theta }_2$ (IX)

Here, $I$ is the intensity if first polarizer is removed, $I_0$ is the un-polarized intensity, ${\theta }_1$ is the angle of polarization between first polarizer, ${\theta }_2$ is the angle of polarization for last polarizer

Write the equation to find the intensity if the last polarizer is removed.

$I=\frac{1}{2}{I}_0{\cos }^2\theta$ (X)

Here, $I$ is the intensity if first polarizer is removed, $I_0$ is the un-polarized intensity, $\theta$ is the angle of polarization

Conclusion:

Substitute $30°$ for $\theta$ in equation (VII) to get $I$.

$\begin{array}{c}I=\frac{1}{2}{I}_0{\cos }^{2}30°\\ =0.375I_0\end{array}$

Substitute $60°$ for $\theta$ in equation (VIII) to get $I$

$\begin{array}{c}I=\frac{1}{2}{I}_0{\cos }^{2}60°\\ =0.125I_o\end{array}$

Here for maximum intensity first polarizer should be removed.

Substitute $60°$ for ${\theta }_1$ , $30°$ for ${\theta }_2$ in equation (IX) to get $I$.

$\begin{array}{c}I=\frac{1}{2}{I}_0{\cos }^{2}60°{\cos }^{2}30°\\ =0.188I_0\end{array}$

Substitute $60°$ for $\theta$ in equation (X) to get $I$

$\begin{array}{c}I=\frac{1}{2}{I}_0{\cos }^{2}60°\\ =0.250I_o\end{array}$

Here for maximum intensity last polarizer shall be removed.