Chapter 16, Problem 10STP

To determine

The light intensity emerging from the second polarizing filter.

Answer to Problem 10STP

  $\frac{I_0}{4}$

Explanation of Solution

Given:

Initial intensity of the nonpolarized light is $I_0$ .

Formula used:

Malus’s Law: Light’s intensity out of a polarizing filter $\left(I_2\right)$ can be written in terms of the light entering into it of intensity $\left(I_1\right)$ as,

  $I_2=I_1{\cos }^2\theta$

Where $\theta$ is the angle between the polarizing axes of the two filters.

Calculation:

Now, the light coming out of the first polarizing filter will be,

  $I_1=\frac{I_0}{2}$

Now, the light of above intensity enters the second polarizing filter. Then, the intensity of the light emerging from the second polarizing filter will be,

  $\begin{array}{c}{I}_2=I_1{\cos }^2\theta \\ =\frac{I_0}{2}{\cos }^{2}45°\\ =\frac{I_0}{4}\end{array}$

Conclusion:

Hence, the intensity of the light emerging from the second polarizing filter is $\frac{I_0}{4}$ .