Question

Two polarizing sheets, one directly above the other, transmit p% of the initially unpolarized light that is perpendicularly incident on the top sheet. What is the angle between the polarizing directions of the two sheets?

Short answer

The polarizing angle between the polarizing directions of the sheets is.${\text{cos}}^{-1}\sqrt{\frac{p}{50}}$

Step-by-step solution

 Step 1: Given data

A beam of unpolarized light is sent through two polarizing sheets placed one on top of the other, transmitting p% of the initially unpolarized light.

 Step 2: Understanding the concept of polarization

Light can be polarized by passing it through a polarizing filter or other polarizing material. By using the relation between original intensity and transmitted intensity, we can find the polarizing angle between the polarizing directions of the sheets.

Formulae:

If the incident light is un-polarized, then the intensity of the merging light using one-half rule is given by,$I=0.5I_0\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

If the incident light is already polarized, then the intensity of the emerging light is cosine –squared of the intensity of incident light,$I=I_0{\cos }^2\theta \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Here,$\theta$is the angle between polarization of the incident light and the polarization axis of the sheet.

Calculation of the polarizing angle

Let $I_{0}andI$be the incident and transmitted intensity respectively.

Initially the beam is unpolarized. When it passed through first sheet using equation (1), the transmitted intensityI is half the original intensity. $I_0$ That is given as:

$I=\frac{1}{2}{I}_0$

This is then passed throughthesecond sheet, for which the emergent intensity from sheet 2 from sheet 1 is given using equation (2) as follows:

$\begin{array}{c}{I}_2=I_1{\cos }^2\theta \\ =\frac{1}{2}{I}_0{\cos }^2\theta \cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

It is given thatthe sheets transmit p% of the initially unpolarized light. Thus, the given emergent intensity from sheet 2 is given as follows:

$\begin{array}{c}I=p\%\text{of}{I}_0\\ =\frac{p}{100}{I}_0\cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)\end{array}$

Using equations (3) and (4), we can get the polarizing angle between the two sheets as follows:

$\begin{array}{c}\frac{1}{2}{I}_0{\cos }^2\theta =\frac{p}{100}{I}_0\\ {\cos }^2\theta =\frac{p}{50}\\ \cos \theta =\sqrt{\frac{p}{50}}\\ \theta ={\text{cos}}^{-1}\sqrt{\frac{p}{50}}\end{array}$

Hence, the angle between the polarizing directions of the sheets is.${\text{cos}}^{-1}\sqrt{\frac{p}{50}}$