Chapter 24, Problem 84GP

To determine

The number of polarizesr.

Answer to Problem 84GP

  $24$

Explanation of Solution

Given data:

Each polarizer is placed at $\text{θ}=10°$ .

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:

Let the initial intensity is $I_0$ ,

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

Each polarizer placed at an angle $\theta$ to the preceding one will reduce the intensity,

  $\therefore {\text{I}}_{\text{i}+\text{1}}={\text{I}}_{\text{i}}{\text{cos}}^{\text{2}}\text{θ}$

Therefore, for n polarizers,

  $\therefore {\text{I}}_{\text{f}}=\frac{{\text{I}}_0}{2}{\left({\text{cos}}^{\text{2}}\text{θ}\right)}^{\text{n-1}}$

Hence, the number of polarizers $\text{n}$ required to reduce the intensity $\frac{1}{4}$ times of its original value:

  $\therefore \frac{{\text{I}}_0}{4}=\frac{{\text{I}}_0}{2}{\left({\cos }^{2}10°\right)}^{\text{n}-1}$

  $\therefore \frac{1}{2}={\left({\cos }^{2}10°\right)}^{\text{2n}-2}$

  $\therefore -\ln 2=\left(2\text{n}-1\right)\ln \left(\cos 10°\right)$

  $\therefore \text{n}=23.6\approx 24$

Conclusion: The required number of polarizers is 24 to drop below the initial intensity to $\frac{1}{4}$ times of its original value.