An unpolarised light of intensity 64 text{ Wm | vedclass

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Question

(A) Let the intensity of the unpolarised light be $I_0 = 64 	ext{ Wm}^{-2}$.After passing through the first polarizer, the intensity becomes $I_1 = \

Vedclass · Accepted Answer

$\frac{1}{2} \sin^{-1} \left( \frac{\sqrt{3}}{2} \right)$

Vedclass · Answer

(A) Let the intensity of the unpolarised light be $I_0 = 64 	ext{ Wm}^{-2}$.After passing through the first polarizer, the intensity becomes $I_1 = \frac{I_0}{2} = 32 	ext{ Wm}^{-2}$.Let $	heta$ be the angle between the transmission axes of the first and second polarizers.The intensity after the second polarizer is $I_2 = I_1 \cos^2 	heta = 32 \cos^2 	heta$.The angle between the first and third polarizer is $90^\circ$ (crossed).Therefore, the angle between the second and third polarizer is $(90^\circ - 	heta)$.The intensity after the third polarizer is $I_3 = I_2 \cos^2(90^\circ - 	heta) = I_2 \sin^2 	heta$.Substituting $I_2$, we get $I_3 = (32 \cos^2 	heta) \sin^2 	heta = 32 (\sin 	heta \cos 	heta)^2 = 32 \left( \frac{\sin 2	heta}{2} ight)^2 = 8 \sin^2 2	heta$.Given $I_3 = 6 	ext{ Wm}^{-2}$, so $8 \sin^2 2	heta = 6$, which means $\sin^2 2	heta = \frac{6}{8} = \frac{3}{4}$.Thus, $\sin 2	heta = \frac{\sqrt{3}}{2}$, which implies $2	heta = \sin^{-1} \left( \frac{\sqrt{3}}{2} ight)$.Therefore, $	heta = \frac{1}{2} \sin^{-1} \left( \frac{\sqrt{3}}{2} ight)$.

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