Question
Two polarizers are so oriented that the maximum amount of light is transmitted. To what fraction of its maximum value is the intensity of the transmitted light reduced when the second polarizer is rotated through $(a) 300 , (b) 600$ ?
✓
Answer
Data : $\theta=30^{\circ}, 60^{\circ}$,
$I _2= I _2 \cos ^2 \theta$
where $I_1$ is the intensity of the incident light and $I_2$ is that of the transmitted light.
$(i)$ For $\theta=30^{\circ}$
$I_2=I_1\left(\cos 30^{\circ}\right)^2=I_1\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4} I_1$
$I_2=0.75 I _1$
$=75 \% \text { of } I _1$
$(ii)$ For $\theta=60^{\circ}$
$I_2=I_1\left(\cos 60^{\circ}\right)^2$
$=I_1\left(\frac{1}{2}\right)^2=\frac{1}{4} I_1$
$=0.25 I _1=25 \%$ of $I _1$
$I _2= I _2 \cos ^2 \theta$
where $I_1$ is the intensity of the incident light and $I_2$ is that of the transmitted light.
$(i)$ For $\theta=30^{\circ}$
$I_2=I_1\left(\cos 30^{\circ}\right)^2=I_1\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4} I_1$
$I_2=0.75 I _1$
$=75 \% \text { of } I _1$
$(ii)$ For $\theta=60^{\circ}$
$I_2=I_1\left(\cos 60^{\circ}\right)^2$
$=I_1\left(\frac{1}{2}\right)^2=\frac{1}{4} I_1$
$=0.25 I _1=25 \%$ of $I _1$
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