Two polaroids have their polarizing direction | vedclass

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(A) According to Malus' Law,the intensity of transmitted light is given by $I = I_0 \cos^2 	heta$,where $I_0$ is the maximum intensity and $	heta$ i

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(A) According to Malus' Law,the intensity of transmitted light is given by $I = I_0 \cos^2 	heta$,where $I_0$ is the maximum intensity and $	heta$ is the angle between the polarizing directions of the two polaroids.We are given that the intensity drops to half,so $I = \frac{I_0}{2}$.Substituting this into the equation: $\frac{I_0}{2} = I_0 \cos^2 	heta$.$\cos^2 	heta = \frac{1}{2} \implies \cos 	heta = \frac{1}{\sqrt{2}}$.This gives $	heta = 45^o$.Since the polaroids were initially parallel $(	heta = 0^o)$,the angle through which either polaroid must be turned is $45^o$ relative to the initial position. However,the question asks for the angle of rotation such that the intensity drops by half. If we rotate one polaroid by $45^o$,the angle between them becomes $45^o$. The options provided suggest the answer is $135^o$,which corresponds to $180^o - 45^o$ (the supplementary angle). Thus,the angle is $45^o$ or $135^o$.

Two polaroids have their polarizing directions parallel so that the intensity of a transmitted light is maximum. The angle through which either polaroid must be turned if the intensity is to drop by one-half is.....$^o$

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