Question

In Fig. 33-40, initially unpolarized light is sent into a system of three polarizing sheets whose polarizing directions make angles of ${\mathbf{\theta }}_1\mathbf{\text{=40°}}$, ${\mathbf{\theta }}_2\mathbf{\text{=20°}}$, and${\mathbf{\theta }}_2\mathbf{\text{=40°}}$with the direction of theyaxis. What percentage of the light’s initial intensity is transmitted by the system? (Hint: Be careful with the angles.)

Short answer

The percentage of the light’s initial intensity is transmitted by the system is$3.1\%I_0$

Step-by-step solution

Given

${\theta }_1={40}^0;{\theta }_2={20}^0;{\theta }_3={40}^0$

Determining the concept

Find the intensities transmitted through each sheet by using the formulas of Malus law. Then comparing the final intensity with the initial intensity, find the percentage.Malus’ law states that the intensity of plane-polarized light that passes through an analyzer varies as the square of the cosine of the angle between the plane of the polarizer and the transmission axes of the analyzer.

 Step 3: (a) Determining thepercentage of the light’s initial intensity is transmitted by the system

Intensity transmitted through the first sheet:

$I_1=\frac{I_0}{2};$

Intensity transmitted through the second sheet:

$$\begin{gathered} \theta ={\theta }_1+{\theta }_2=40+20={60}^0 \\ {I}_2=I_1(\theta )=\frac{I_0}{2}\left(60\right) \\ {I}_2=\frac{I_0}{8} \end{gathered}$$

Intensity of light transmitted by the third sheet:

$$\begin{gathered} \theta ={\theta }_2+{\theta }_3=20+40={60}^0 \\ {I}_3=I_2\left(\theta \right)=\frac{I_0}{8}\left(60\right) \\ {I}_3=\frac{I_0}{32} \\ {I}_3=0.031I_0 \end{gathered}$$

Hence,

$I_3=3.1\%I_0$

Therefore, the percentage of the light’s initial intensity is transmitted by the system is.

$3.1\%I_0$