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(A) When unpolarised light of intensity $I_0$ passes through the first polariser,it becomes linearly polarised with intensity $I_1 = I_0 / 2$.Since th

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$\left(\frac{I_0}{8}ight) \sin^2 2	heta$

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(A) When unpolarised light of intensity $I_0$ passes through the first polariser,it becomes linearly polarised with intensity $I_1 = I_0 / 2$.Since the first and second polaroids were initially crossed (angle $90^{\circ}$),the third polaroid is placed at an angle $	heta$ with the first and $(90^{\circ} - 	heta)$ with the second.According to Malus' Law,the intensity after the second (middle) polariser is $I_2 = I_1 \cos^2 	heta = (I_0 / 2) \cos^2 	heta$.The intensity after the third (last) polariser is $I_3 = I_2 \cos^2(90^{\circ} - 	heta) = (I_0 / 2) \cos^2 	heta \sin^2 	heta$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have $\sin^2 2	heta = 4 \sin^2 	heta \cos^2 	heta$.Therefore,$I_3 = \frac{I_0}{2} \cdot \frac{\sin^2 2	heta}{4} = \frac{I_0}{8} \sin^2 2	heta$.

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