Answer:
The intensity of the light transmitted through the third filter is [tex]I_3 = \frac{I_o}{8}[/tex]
Explanation:
From the question we are told
The intensity of the unpolarised light [tex]I_o[/tex]
The angle between the first and second polarizer is [tex]\theta _1 = 45^o[/tex]
The angle between the first and third polarizer is [tex]\theta _2 = 90^o[/tex]
Generally the intensity of light emerging from the first polarizer is mathematically represented as
[tex]I_1 = \frac{I_o}{2}[/tex]
According to Malus law the intensity of light emerging from the second polarizer is mathematically represented as
[tex]I_2 = I_1 cos^2 (\theta_1)[/tex]
Substituting for [tex]I_1[/tex] and [tex]\theta _1[/tex]
[tex]I_2 = \frac{I_o}{2} cos^2 (45)[/tex]
[tex]I_2 = \frac{I_o}{4 }[/tex]
According to Malus law the intensity of light emerging from the third polarizer is mathematically represented as
[tex]I_3 = I_2 cos ^2 (\theta_2 - \theta_1)[/tex]
Substituting for [tex]I_2[/tex] and [tex]\theta _1 \ and \ \theta _2[/tex]
[tex]I_3 = \frac{I_o}{4} cos ^2 (90 - 45)[/tex]
[tex]I_3 = \frac{I_o}{8}[/tex]
