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Question

(A) Let the angle between the direction of polarization and the $x-$axis be $\alpha$.According to Malus' Law,the transmitted intensity $I$ is given by

Vedclass · Accepted Answer

$203$

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(A) Let the angle between the direction of polarization and the $x-$axis be $\alpha$.According to Malus' Law,the transmitted intensity $I$ is given by $I = I_0 \cos^2(\alpha - 	heta)$,where $I_0$ is the incident intensity.For the intensities to be the same at different values of $	heta$,the values of $\cos^2(\alpha - 	heta)$ must be equal.This implies $(\alpha - 	heta) = \pm \phi$ or $(\alpha - 	heta) = 180^o \pm \phi$.Given $	heta_1 = 8^o, 	heta_2 = 38^o, 	heta_3 = 188^o, 	heta_4 = 218^o$.The average of these angles is $\alpha = \frac{8^o + 38^o + 188^o + 218^o}{4} = \frac{452^o}{4} = 113^o$.However,checking the symmetry: $(\alpha - 8^o) = -(\alpha - 38^o) \implies 2\alpha = 46^o \implies \alpha = 23^o$ or $23^o + 180^o = 203^o$.Thus,the angle is $203^o$.

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