Chapter 24, Problem 59P

To determine

The percentage reduction in intensity.

Answer to Problem 59P

  $44.5\%$ .

Explanation of Solution

Given data:

Polarizers are oriented at ${\text{θ}}_2=38°$ .

Light polarized by $19°$ .

Formula used:

It is known that: ${\text{I}}_1={\text{I}}_0{\text{cos}}^{\text{2}}{\text{θ}}_1$

Where,

  $I_0$ is initial intensity.

  ${\text{I}}_1$ is intensity after passing the first Polarizer.

  $\theta$ is an angle.

Calculation:

By two successive sheets,

  ${\text{I}}_1={\text{I}}_0{\text{cos}}^{\text{2}}{\text{θ}}_1$

  $\therefore {\text{I}}_2={\text{I}}_1{\text{cos}}^{\text{2}}{\text{θ}}_2={\text{I}}_0{\text{cos}}^{\text{2}}{\text{θ}}_1{\cos }^2{\text{θ}}_{\text{2}}$

Hence,

  ${\text{I}}_2={\text{I}}_0{\text{cos}}^{\text{2}}\left(19°\right){\cos }^{2}38°=0.555{\text{I}}_0$

The percentage of reduction will be,

  $\therefore \frac{{\text{I}}_0-\text{I}}{{\text{I}}_0}\times 100=\frac{1-.555}{1}\times 100=44.5\%$

Conclusion:

The percentage reduction in intensity is $44.5\%$ .