Respuesta :
To solve this problem it is necessary to apply the concepts related to the law of Malus.
The light intensity after passing through the first polarizer is given as
[tex]I_1 = \frac{I}{2}[/tex]
Where,
I = Intensity of unpolarized light
The expression for malus law is given as
[tex]I_2 = I_1 Cos^2 \theta[/tex]
Where
\theta = Angle between the first and second polarizers
[tex]I_1,I_2[/tex] = Intensities of the light after passing through the first and second polarizers respectively.
The intensity of light after passing first polarizer ( Perpendicular to the surface) is
[tex]I_1 = \frac{I}{2}[/tex]
[tex]I_1 = \frac{30}{2}[/tex]
[tex]I_1 = 15W/cm^2[/tex]
The angle between the first and second polarizers is given by
[tex]\theta = \theta_2 -\theta_1[/tex]
[tex]\theta = 162-14[/tex]
[tex]\theta =148\°[/tex]
The expression for Malus law is given by
[tex]I_2 = I_1 cos^2 \theta[/tex]
[tex]\theta = 148\°[/tex]
[tex]I_2 = (15)cos^2(148)[/tex]
[tex]I_2 = 10.78W/cm^2[/tex]
Therefore the intensity of the light after passing through both the polarizer is [tex]10.78W/cm^2[/tex]
The intensity of the light that emerges from the second polarizer should be considered as the [tex]10.78 W/cm^2.[/tex]
Calculation of the intensity;
Here we apply law of malus
The light intensity after passing should be
[tex]I_1 = I/2[/tex]
Here,
I = Intensity of unpolarized light
Now
[tex]= 30 /2[/tex]
= 15
Now
[tex]= 15 cos^2 (162-14)\\\\= 15 cos^2 (148)\\\\= 10.78 W/cm^2[/tex]
hence, The intensity of the light that emerges from the second polarizer should be considered as the [tex]10.78 W/cm^2.[/tex]
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