Answer:
The fraction of the intensity of the original light that got through the last polarizer is [tex]\frac{I_3}{I_o} = 0.0955[/tex]
Explanation:
From the question we are told that
The angle between polarizer 2 and polarizer 1 is [tex]\theta_1 = 40.7^o[/tex]
The angle between polarizer 3 and polarizer 2 is [tex]\theta_ 2 = 54.8^o[/tex]
The angle between polarizer 3 and polarizer 1 is [tex]\theta _3 = 95.5^o[/tex]
The intensity of light emerging from the first polarizer can be obtained using the one half rule as follows
[tex]I_1 = \frac{I_o}{2}[/tex]
Here [tex]I_1[/tex] is the intensity of light emerging from the polarizer 1
[tex]I_o[/tex] is the intensity of the unpolarized light
The intensity [tex](I_2)[/tex]of light emerging from the second polarizer is obtained using the cosine-squared rule the intensity of light incidenting on the second polarizer is already polarized by the polarizer 1
So the intensity is mathematically represented as
[tex]I_2 = I_1 cos^2 \theta_1[/tex]
substituting for [tex]I_1[/tex]
[tex]I_2 = \frac{I_o}{2} cos^2 \theta_1[/tex]
Substituting values
[tex]I_2 = \frac{I_0}{2} cos^2 (40.7)[/tex]
The intensity [tex](I_3)[/tex] emerging from polarizer 3 is obtained using the cosine-squared rule as follows
[tex]I_3 = I_2 cos^2 \theta_2[/tex]
substituting for [tex]I_2 \ and \ \theta_2[/tex]
[tex]I_3 = \frac{I_o}{2} cos^2 (40.7) \ cos^2 (54.8)[/tex]
[tex]I_3 = \frac{I_o}{2} \ 0.5748 * 0.3323[/tex]
[tex]I_3 =(0.0955) I_o[/tex]
[tex]\frac{I_3}{I_o} = 0.0955[/tex]
