Question

A beam of light traveling horizontally is made of an unpolarized component with intensity I0 and a polarized component with intensity Ip. The plane of polarization of the polarized component is oriented at an angle u with respect to the vertical. Figure P33.59 is a graph of the total intensity Itotal after the light passes through a polarizer versus the angle a that the polarizer’s axis makes with respect to the vertical. (a) What is the orientation of the polarized component? (That is, what is u?) (b) What are the values of I0 and Ip?

Short answer

a) The orientation of polarized component is at angle$35°$ .

b)$l_o=10W/m^2$ and $l_p=20W/m^2$.

Step-by-step solution

About Malus’s Law

The unpolarized component of the light is given by $\frac{l_2}{2}$while the polarized component could be found from Malus’s law

$l=l_p{\cos }^2\left(a-\theta \right)$

Hence, the total intensity of light equals the summation of the components

$l_{total}=\frac{l_o}{2}+{\cos }^2\left(a-\theta \right)$

The orientation of the polarized component occurs at the maximum intensity. At $l_{max}$the term $\left(a-\theta \right)$= 1, which means$a=\theta$.At maximum intensity(25W/m2), the angle $a=35°$. Hence, the angle$\theta$will be

$a=\theta =35°$

The orientation of polarized component is at angle$35°$ .

Intensity of light.

At minimum intensity $l_{min}=5W/m^2$and$a=125°$ . Substitute these values with into equation to get $l_p$ .

$$\begin{gathered} l_{min}=\frac{l_o}{2}+l_p{\cos }^2\left(a-\theta \right) \\ 5W/m^2=\frac{l_o}{2}+l_p\cos \left(125°-35°\right) \\ 5W/m^2=\frac{l_o}{2} \\ {l}_o=10W/m^2 \end{gathered}$$

While at maximum values of intensity $l_{min}=25W/m^{2}a=35°$ and$\theta =35°$ Now, substitute these values with $l_o$to get $$\begin{gathered} l_{max}=\frac{l_0}{2}+l_p{\cos }^2\left(a-\theta \right) \\ 25W/m^2=\frac{10W/m^2}{2}+l_p{\cos }^2\left(35°-35°\right) \\ 25W/m^2=5+l_p \\ {l}_p=20W/m^2 \end{gathered}$$

$l_o=10W/m^2$and$l_p=20W/m^2$ .