Chapter 23, Problem 44E

To determine

The fraction of intensity transmitted for the given case.

Explanation of Solution

Write expression for intensity of the light after passing through first polarizer.

    $I_1=I_0{\cos }^2{\theta }_1$        (1)

Here, $I_0$ is the original intesity of the light and ${\theta }_1$ is the angle of first polarizer from the normal.

Write expression for intensity of light after passing through second polarizer.

    $I_2=I_1{\cos }^2{\theta }_2$

Here, ${\theta }_2$ is the angle from the first polarizer.

Write expression for intensity of light after passing through third polarizer.

    $I_3=I_2{\cos }^2{\theta }_3$

Here, ${\theta }_3$ is angle from second polarizer.

Substitute $I_1{\cos }^2{\theta }_2$ for $I_2$ in above expression.

    $I_3=I_1{\cos }^2{\theta }_2{\cos }^2{\theta }_3$        (2)

Write expression for fraction of intensity passed.

    $I_F=\frac{I_3}{I_0}$        (3)

Conclusion:

Substitute $0°$ for ${\theta }_1$ in equation (1).

    $\begin{array}{l}{I}_1=I_0{\cos }^{2}0°\\ I_1=I_0\end{array}$

Substitute $I_0$ for $I_1$, $30°$ for ${\theta }_2$ and $30°$ for ${\theta }_3$ in equation (2).

    $\begin{array}{l}{I}_3=I_0{\cos }^{2}30°{\cos }^{2}30°\\ I_3=0.56I_0\end{array}$

Substitute $0.56I_0$ for $I_3$ in equation (3).

    $\begin{array}{l}{I}_F=\frac{0.56I_0}{I_0}\\ I_F=0.56\end{array}$

Thus, the fraction of intensity passed is $0.56$.

(b)

To determine

The fraction of intensity transmitted for given case.

Explanation of Solution

Write expression for intensity of the light after passing through first polarizer.

    $I_1=I_0{\cos }^2{\theta }_1$        (1)

Here, $I_0$ is the original intesity of the light and ${\theta }_1$ is the angle of first polarizer from the normal.

Write expression for intensity of light after passing through second polarizer.

    $I_2=I_1{\cos }^2{\theta }_2$

Here, ${\theta }_2$ is the angle from the first polarizer.

Write expression for intensity of light after passing through third polarizer.

    $I_3=I_2{\cos }^2{\theta }_3$

Here, ${\theta }_3$ is angle from second polarizer.

Substitute $I_1{\cos }^2{\theta }_2$ for $I_2$ in above expression.

    $I_3=I_1{\cos }^2{\theta }_2{\cos }^2{\theta }_3$        (2)

Write expression for fraction of intensity passed.

    $I_F=\frac{I_3}{I_0}$        (3)

Conclusion:

Substitute $0°$ for ${\theta }_1$ in equation (1).

    $\begin{array}{l}{I}_1=I_0{\cos }^{2}0°\\ I_1=I_0\end{array}$

Substitute $I_0$ for $I_1$, $60°$ for ${\theta }_2$ and $30°$ for ${\theta }_3$ in equation (2).

    $\begin{array}{l}{I}_3=I_0{\cos }^{2}60°{\cos }^{2}30°\\ I_3=0.19I_0\end{array}$

Substitute $0.19I_0$ for $I_3$ in equation (3).

    $\begin{array}{l}{I}_F=\frac{0.19I_0}{I_0}\\ I_F=0.19\end{array}$

Thus, the fraction of intensity passed is $0.19$.