Question

In Fig.33-41, a beam of unpolarised light, with intensity ${\text{43\hspace{0.17em}W/m}}^{\text{2}}$, is sent into a system of two polarizing sheets with polarizing directions at angles ${\mathbf{\theta }}_1\mathbf{\text{=70°}}$and ${\mathbf{\theta }}_2\mathbf{\text{=90°}}$to theyaxis. What is the intensity of the light transmitted by the system?

Short answer

The intensity of the light transmitted by the system is${\text{19W/m}}^{\text{2}}$.

Step-by-step solution

Given data

The intensity of unpolarized light is$I_0=43{\text{\hspace{0.17em}W/m}}^{\text{2}}$.

The angle of the first polarizing sheet is${\theta }_1={70}^{\circ }$.

The angle of the second polarizing sheet is${\theta }_2={90}^{\circ }$.

Figure 33-41 is the two polarizing sheets.

Determining the concept

Use the concept of polarization of the wave. Using the equation of the one-half rule and cosine-squared rule, we can find the intensity of the transmitted light.Polarization is a property applying to transverse waves that specify the geometrical orientation of the oscillations. In a transverse wave, the direction of the oscillation is perpendicular to the direction of motion of the wave.

(a) Determining the intensity of the light transmitted by the system

Find the angle as,

$\begin{array}{c}\theta ={\theta }_2-{\theta }_1\\ ={90}^{\circ }-{70}^{\circ }\\ ={20}^{\circ }\end{array}$

The intensity of light from the first sheet is,

$I_1=\frac{1}{2}{I}_0$

The intensity of light from the second sheet is,

$I_2=\frac{1}{2}{I}_0{\cos }^2\theta$

Plugging the values,and we get,

$\begin{array}{l}{I}_2=\frac{1}{2}\left(43\right)\left({\cos }^2{20}^{\circ }\right)\\ I_2=18.99\\ I_2~19{\text{W/m}}^{\text{2}}\end{array}$

Therefore, the intensity of the light transmitted by the system is ${\text{19\hspace{0.17em}W/m}}^{\text{2}}$.