Chapter 24, Problem 81GP

(a)

To determine

The fraction of transmitted incident light intensity.

Answer to Problem 81GP

  $0.086{\text{I}}_0$ .

Explanation of Solution

Given:

Third polarizer are placed at ${\text{θ}}_2=62°$ .

Formula used:

It is known that: ${\text{I}}_1=\frac{{\text{I}}_0}{2}$

Where,

  $I_0$ is initial intensity.

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

  $\theta$ is an angle.

Calculation:

By three successive sheets,

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

  $\therefore {\text{I}}_2={\text{I}}_1{\text{cos}}^{\text{2}}{\text{θ}}_1=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}62°$

  $\therefore {\text{I}}_3={\text{I}}_2{\text{cos}}^{\text{2}}{\text{θ}}_2=\left(\frac{{\text{I}}_0}{2}\right){\text{cos}}^{\text{2}}62°{\cos }^{2}28°$

Hence,

  $\therefore {\text{I}}_3=0.086{\text{I}}_0$

(b)

To determine

Result when third Polarizer is placed in front of the other two Polarisers.

Answer to Problem 81GP

No light will be transmitted.

Explanation of Solution

Given:

Third polarizer are placed at ${\text{θ}}_3=62°$ .

Formula used:

It is known that: ${\text{I}}_1=\frac{{\text{I}}_0}{2}$

Where,

  $I_0$ is initial intensity.

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

  $\theta$ is an angle.

Calculation:

If the third polarizer placed in front of the other two polarisers, then the third polarizer will block the reflected light that come from other two polarisers.