Question

In Fig. 33-42, unpolarized light is sent into a system of three polarizing sheets, which transmitsthe initial light intensity. The polarizing directions of the first and third sheets are at angles${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}{\mathbf{0}}^{\mathbf{°}}$and${\mathit{\theta }}_{\mathbf{3}}\mathbf{=}\mathbf{90}\mathbf{°}$.What are the

(a) smaller and

(b) larger possible values of angle${\mathit{\theta }}_{\mathbf{2}}\mathbf{(}\mathbf{<}\mathbf{90}\mathbf{°}\mathbf{)}$for the polarizing direction of sheet 2?

Short answer

1. ${\text{19.6}}^{\text{°}}$
   
2. ${\text{70.4}}^{\text{°}}$
   

Step-by-step solution

Step 1: Given data

The intensity factor is $\text{0.05}$

${\theta }_1=0$

${\theta }_2=90°$

Determining the concept

Here usetheequation of the one-half rule and cosine squared rule, which givestherelation between transmitted intensity and the incident intensity. Using these equations, the intensity transmitted through the system can be found, and then equating it withthegiven value of the fraction, the smaller and larger angles forthepolarizing sheet 2 can be determined.

The formula is as follows:

$I=I_0{\cos }^2\theta$

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

Where,

I = radiant intensity,

cosθ = the angle θ between the direction of the incident light and the surface normal,

(a) Determining the smaller value of θ2

The intensity of light after passing through the second sheet can be expressed as,

$I_2=I_1{\cos }^2{\theta }_2$

Here $I_1=\frac{I_0}{\text{2}}$

$I_2=\frac{I_0}{2}{\cos }^2{\theta }_2$

The intensity of light after passing through the third sheet is as follows:

$\begin{array}{l}{I}_3=I_2{\cos }^2(90-{\theta }_2)\\ I_3=\frac{I_0}{2}{\cos }^2{\theta }_2{\cos }^2(90-{\theta }_2)\\ I_3=\frac{I_0}{2}{\cos }^2{\theta }_2{\sin }^2{\theta }_2\\ I_3=\frac{I_0}{2}\times \frac{1}{4}\times {\left(\sin \left(2{\theta }_2\right)\right)}^2\end{array}$ (1)

Here it is given,

$I_3=\text{0.0500}{I}_0$

Now plug the value in equation 1, and we get,

$\begin{array}{l}0.0500I_0=\frac{I_0}{2}\times \frac{1}{4}\times {\left(\sin \left(2{\theta }_2\right)\right)}^2\\ 0.4={\left(\sin \left(2{\theta }_2\right)\right)}^2\\ \sin \left(2{\theta }_2\right)=0.6324\end{array}$

Solving further as,

$\begin{array}{c}\left(2{\theta }_2\right)={\sin }^{-1}\left(0.6324\right)\\ 2{\theta }_2=39.34\\ {\theta }_2={19.6}^{°}\end{array}$

Hence, the smaller value of ${\theta }_2$is${19.6}^{°}$

(b) Determining the larger value of θ2

The larger angle can be determined using the property of inverse trigonometric function and the concept of polarization.The same polarization fraction for the angle can be determined,

$\begin{array}{l}{\theta }_2\text{'}={90}^{°}-{19.6}^{°}\\ {\theta }_2\text{'}={70.4}^{°}\end{array}$

So larger angle is ${\text{70.4}}^{°}$

Hence,the larger value of ${\theta }_2\text{'}$is${\text{70.4}}^{°}$