Question

A polaroid is placed at $45^{\circ }$ to an incoming light of intensity $I_0$ . Now, the intensity of light passing through the polaroid after polarisation would be,

• A. $I_0$
• B. $\frac{I_0}{2}$
• C. $\frac{I_0}{4}$
• D. zero

Answer 1

Answer: (B)

According to Malus's law
It states that the intensity of the plane-polarized light passing via an analyzer correlates directly to the angle of the square of cosine in between polariser plane and the analyzer's transmission axes.
The expression of the intensity of light passing through polaroid after polarisation is stated as $I=I_0{\cos }^2\theta$ , here $I_0$ represents the intensity of incoming light and $\theta$ is the angle of incoming light on the polaroid.
Given here $\theta =45^{\circ }$ .
Thus, the intensity of light passing through polaroid after polarisation is $I=I_0{\left(\cos \right)}^2\left({\left(45\right)}^{{}^{\circ }}\right)$
$I=I_0{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{I_0}{2}$ .
After polarisation, the intensity reduces to half.