Respuesta :
Answer:
Without the third polarizer inserted in the middle of the first and second polarizer the irradiance of light that would emerge from the second polarizer is
[tex]I_2=321.39 W/cm^2[/tex]
With the insertion of the third polarizer at the middle of the first and second polarizer the irradiance of light that would emerge from the second polarizer is
[tex]I_4 = I(\theta_d_1) = 337.3 W/cm^2[/tex]
Explanation:
From the question we are told that the
The angle between the transmission axis of first linear polarizer and the vertical axis is [tex]\theta_1 = 0^o[/tex]
The angle between the transmission axis of second linear polarizer and the vertical axis is [tex]\theta_2 = 50^o[/tex]
The angle between the transmission axis of third linear polarizer and the vertical axis is [tex]\theta_2 = 25^o[/tex]
The irradiance of the incident natural light is [tex]I = 1000 W/cm^2[/tex]
Generally a linear polarizer divides the irradiance of a natural light by 2
So For the first polarizer the irradiance of the natural light would become
[tex]I_1 = \frac{I}{2}[/tex]
Substituting values
[tex]I_1 = \frac{1000}{2}[/tex]
[tex]=500 \ W/cm^2[/tex]
Now looking at the question we can deduce that the angle between the transmission axis of the and second polarizer is
[tex]\theta_d = \theta _2 - \theta_1[/tex]
Substituting values
[tex]\theta_d = 50^o - 0^o[/tex]
[tex]=50^o[/tex]
According to Malus Law the irradiance of light that would come out from the second polarizer is obtained by this mathematical expression
[tex]I(\theta_d) = I_1 cos^2(\theta_d)[/tex]
Substituting values
[tex]I_2 = I(\theta_d) = 500 *cos (50)[/tex]
[tex]=321.39 W/cm^2[/tex]
When the third polarizer is inserted between the first and second polarizer, we have that
The angle between the first polarizer and the third polarizer is mathematically evaluated as
[tex]\theta_d_1 = \theta_3 - \theta_1[/tex]
Substituting the values
[tex]\theta_d_1 = 25^o -0^o[/tex]
[tex]= 25^o[/tex]
According to Malus Law the irradiance of light that would come out from the third polarizer is obtained by this mathematical expression
[tex]I(\theta_d_1) = I_1 cos^2(\theta_d_1)[/tex]
[tex]I(\theta_d_1) = I_1 cos^2 (25)[/tex]
Substituting values
[tex]= 500 cos^2(25)[/tex]
[tex]I_3 = I(\theta_d_1) = 410.69 W/cm^2[/tex]
The angle between the third polarizer and the second polarizer is mathematically evaluated as
[tex]\theta_d_2 = \theta_2 - \theta_3[/tex]
Substituting the values
[tex]\theta_d_2 = 50^o -25^o[/tex] [Note the third polarizer is placed at the
[tex]= 25^o[/tex] The middle of the first and second
polarizer
According to Malus Law the irradiance of light that would come out from the second polarizer is obtained by this mathematical expression
[tex]I(\theta_d_2) = I_3 cos^2(\theta_d_2)[/tex]
[tex]I(\theta_d_1) = I_3 cos^2 (25)[/tex]
Substituting values
[tex]= 410.69 cos^2(25)[/tex]
[tex]I_4 = I(\theta_d_1) = 337.3 W/cm^2[/tex]
