A polarized light of intensity I_{0} is passe | vedclass

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(D) According to Malus's Law,the intensity of the emergent light is given by:$I = I_{0} \cos^{2} 	heta$where $I$ is the intensity of the emergent pol

Vedclass · Accepted Answer

$I = I_{0} / 4$

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(D) According to Malus's Law,the intensity of the emergent light is given by:$I = I_{0} \cos^{2} 	heta$where $I$ is the intensity of the emergent polarized light,$I_{0}$ is the intensity of the incident polarized light,and $	heta$ is the angle between the pass axes of the two polarizers.Given that $	heta = 60^{\circ}$,we substitute this value into the formula:$I = I_{0} \cos^{2}(60^{\circ})$Since $\cos(60^{\circ}) = 1/2$,we have:$I = I_{0} 	imes (1/2)^{2}$$I = I_{0} 	imes (1/4)$$I = I_{0} / 4$Thus,the intensity of the emergent polarized light from the second polarizer is $I_{0} / 4$.

$A$ polarized light of intensity $I_{0}$ is passed through another polarizer whose pass axis makes an angle of $60^{\circ}$ with the pass axis of the former. What is the intensity of emergent polarized light from the second polarizer?

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