Unpolarised light of intensity 32 , W m^{-2} | vedclass

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(D) Let the intensity of unpolarised light be $I_0 = 32 \, W m^{-2}$.After passing through the first polaroid,the intensity becomes $I_1 = \frac{I_0}{

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$30$

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(D) Let the intensity of unpolarised light be $I_0 = 32 \, W m^{-2}$.After passing through the first polaroid,the intensity becomes $I_1 = \frac{I_0}{2} = 16 \, W m^{-2}$.Let $	heta$ be the angle between the pass axis of the first and second polaroid.The intensity after the second polaroid is $I_2 = I_1 \cos^2 	heta = \frac{I_0}{2} \cos^2 	heta$.The angle between the second and third polaroid is $(90^{\circ} - 	heta)$ because the first and third polaroids are perpendicular.The intensity after the third polaroid is $I_3 = I_2 \cos^2(90^{\circ} - 	heta) = I_2 \sin^2 	heta$.Substituting $I_2$,we get $I_3 = \frac{I_0}{2} \cos^2 	heta \sin^2 	heta = \frac{I_0}{8} (2 \sin 	heta \cos 	heta)^2 = \frac{I_0}{8} \sin^2(2 	heta)$.Given $I_3 = 3 \, W m^{-2}$ and $I_0 = 32 \, W m^{-2}$,we have $3 = \frac{32}{8} \sin^2(2 	heta) = 4 \sin^2(2 	heta)$.$\sin^2(2 	heta) = \frac{3}{4} \implies \sin(2 	heta) = \frac{\sqrt{3}}{2}$.Thus,$2 	heta = 60^{\circ}$ or $120^{\circ}$,which gives $	heta = 30^{\circ}$ or $60^{\circ}$.

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