Chapter 24, Problem 82GP

(a)

To determine

The fraction by which the unpolarized incident light is transmitted by fours polarizers.

Answer to Problem 82GP

  $0.21{\text{I}}_0$

Explanation of Solution

Given data:

Polarizers are placed at ${\text{θ}}_2={\text{θ}}_3={\text{θ}}_4=30°$ .

Formula used:

  ${\text{I}}_1=\frac{{\text{I}}_0}{2}$

Where,

  $I_0$ is initial intensity.

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

  $\theta$ is an angle.

Calculation:

  ${\text{I}}_1=\frac{{\text{I}}_0}{2}$

  $\therefore {\text{I}}_2={\text{I}}_1{\text{cos}}^{\text{2}}{\text{θ}}_2=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}{\text{θ}}_{\text{2}}$

  $\therefore {\text{I}}_3={\text{I}}_2{\text{cos}}^{\text{2}}{\text{θ}}_3=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}{\text{θ}}_{\text{2}}{\text{cos}}^{\text{2}}{\text{θ}}_3$

  $\therefore {\text{I}}_4={\text{I}}_3{\text{cos}}^{\text{2}}{\text{θ}}_4=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}{\text{θ}}_{\text{2}}{\text{cos}}^{\text{2}}{\text{θ}}_3{\text{cos}}^{\text{2}}{\text{θ}}_4$

  $\therefore {\text{I}}_4=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}\text{30}°{\text{cos}}^{\text{2}}\text{30}°{\text{cos}}^{\text{2}}30°$

Hence,

  $\therefore {\text{I}}_4=0.21{\text{I}}_0$

Conclusion: The fraction by which the unpolarized incident light is transmitted by four polarizers will be $0.21{\text{I}}_0$ .

(b)

To determine

Whether transmitted light can be decreased by removing one polarizer.

Explanation of Solution

Given data:

If the second polarizer was removed, so, ${\text{θ}}_3\text{'}=60°$ .

Fourth polarizer is placed at ${\text{θ}}_4=30°$ .

Formula used:

  ${\text{I}}_1=\frac{{\text{I}}_0}{2}$

Where,

  $I_0$ is initial intensity.

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

  $\theta$ is an angle.

Calculation:

  ${\text{I}}_1=\frac{{\text{I}}_0}{2}$

  $\therefore {\text{I}}_3={\text{I}}_1{\text{cos}}^{\text{2}}{\text{θ}}_3\text{'}=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}{\text{θ}}_3$

  $\therefore {\text{I}}_4={\text{I}}_3{\text{cos}}^{\text{2}}{\text{θ}}_4=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}{\text{θ}}_3{\text{'cos}}^{\text{2}}{\text{θ}}_4$

  $\therefore {\text{I}}_4={\text{I}}_3{\text{cos}}^{\text{2}}{\text{θ}}_4=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}\text{60}°{\text{cos}}^{\text{2}}30°$

Hence,

  $\therefore {\text{I}}_4=0.094{\text{I}}_0$

Conclusion: The intensity of light is decreased when either the third or second polarizer will be removed.

(c)

To determine

Whether the intensity of light be extinguished if polarizers are removed.

Answer to Problem 82GP

No light is transmitted when the second and third polarizers are removed.

Explanation of Solution

Given data:

If second and third polarizer wereremoved, then ${\text{θ}}_4\text{'}=90°$ .

Calculation:

If the second or the third polarizers will be removed, then the two polarizers axes become perpendicular to each other. Hence, no light gets transmitted.

Conclusion: If, either the third or second polarizer will be remove then no lights will be transmitted.