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Imagine you have two identical perfect linear polarizers and a source of natural light. Place them one behind the other and position their transmission axes at 0 and 50 degrees respectively. Now insert between them a third linear polarizer with its transmission axis at 25 degrees. If 1000 W/cm2 of natural light is incident, how much will emerge with and without the middle polarizer in place?

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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


I_2=321.39 W/cm^2

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


I_4 = I(\theta_d_1) = 337.3 W/cm^2

Step-by-step explanation:

From the question we are told that the

The angle between the transmission axis of first linear polarizer and the vertical axis is
\theta_1 = 0^o

The angle between the transmission axis of second linear polarizer and the vertical axis is
\theta_2 = 50^o

The angle between the transmission axis of third linear polarizer and the vertical axis is
\theta_2 = 25^o

The irradiance of the incident natural light is
I = 1000 W/cm^2

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


I_1 = (I)/(2)

Substituting values


I_1 = (1000)/(2)


=500 \ W/cm^2

Now looking at the question we can deduce that the angle between the transmission axis of the and second polarizer is


\theta_d = \theta _2 - \theta_1

Substituting values


\theta_d = 50^o - 0^o


=50^o

According to Malus Law the irradiance of light that would come out from the second polarizer is obtained by this mathematical expression


I(\theta_d) = I_1 cos^2(\theta_d)

Substituting values


I_2 = I(\theta_d) = 500 *cos (50)


=321.39 W/cm^2

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


\theta_d_1 = \theta_3 - \theta_1

Substituting the values


\theta_d_1 = 25^o -0^o


= 25^o

According to Malus Law the irradiance of light that would come out from the third polarizer is obtained by this mathematical expression


I(\theta_d_1) = I_1 cos^2(\theta_d_1)


I(\theta_d_1) = I_1 cos^2 (25)

Substituting values


= 500 cos^2(25)


I_3 = I(\theta_d_1) = 410.69 W/cm^2

The angle between the third polarizer and the second polarizer is mathematically evaluated as


\theta_d_2 = \theta_2 - \theta_3

Substituting the values


\theta_d_2 = 50^o -25^o [Note the third polarizer is placed at the


= 25^o 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


I(\theta_d_2) = I_3 cos^2(\theta_d_2)


I(\theta_d_1) = I_3 cos^2 (25)

Substituting values


= 410.69 cos^2(25)


I_4 = I(\theta_d_1) = 337.3 W/cm^2

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