To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.
Mathematically the expression is given as
[tex]I = I_0 cos^2\theta[/tex]
Where,
[tex]I_0[/tex]= Initial Intensity
I = Final Intensity after pass through the polarizer
[tex]\theta[/tex]= Angle between the polarizer and the light
Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as
[tex]\frac{I}{I_0} = \frac{1}{2}[/tex]
Using the Malus Law we have,
[tex]I = I_0 cos^2\theta[/tex]
[tex]cos^2\theta = \frac{I}{I_0}[/tex]
[tex]cos^2\theta = \frac{1}{2}[/tex]
[tex]\theta = cos^{-1}(\frac{1}{2})^2[/tex]
[tex]\theta = 75.52\°[/tex]
Angle with respect to maximum is [tex]90-75.52 = 14.48\°[/tex]
The angle that one of them should be placed so that the transmitted intensity is subsequently reduced by half is 45⁰.
The intensity of light according to Malus law is calculated as follows;
I = I₀cos²θ
where;
I₀ is the intensity of the unpolarized light
The intensity of the unpolarized light is maximum when θ = 0
I = I₀cos(0)²
I = I₀
When the intensity is halved,
[tex]\frac{I}{2} = I_0cos^2 \theta\\\\\frac{I_0}{2} = I_0cos^2 \theta\\\\\frac{I_0}{I_0} = 2cos ^2 \theta\\\\1 = 2cos^2 \theta\\\\\frac{1}{2} = cos ^2 \theta\\\\0.5 = cos ^2 \theta\\\\\sqrt{0.5} = cos\theta\\\\0.7071 = cos\theta\\\\\theta = cos^{-1} (0.7071)\\\\\theta = 45\ ^0[/tex]
Thus, the angle that one of them should be placed so that the transmitted intensity is subsequently reduced by half is 45⁰.
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