Three linear polarisers are placed coaxially. | vedclass

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(C) When an already polarised beam passes through a polariser,the intensity $I$ of light obtained is given by Malus' law: $I = I_0 \cos^2 	heta$,wher

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$30$

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(C) When an already polarised beam passes through a polariser,the intensity $I$ of light obtained is given by Malus' law: $I = I_0 \cos^2 	heta$,where $I_0$ is the intensity of incident polarised light and $	heta$ is the angle between the plane of polarisation and the transmission axis of the polariser.Given that each polariser absorbs $10 \%$ of the light,the transmission factor is $k = 0.9$.$1$. For the first polariser $(P_1)$: The incident light is vertical. The transmission axis is at $30^{\circ}$ to the vertical. So,$	heta_1 = 30^{\circ}$.$I_1 = k \cdot I_0 \cdot \cos^2(30^{\circ}) = 0.9 \cdot 100 \cdot (\sqrt{3}/2)^2 = 0.9 \cdot 100 \cdot 0.75 = 67.5 \, W/m^2$.$2$. For the second polariser $(P_2)$: The light incident on $P_2$ is polarised at $30^{\circ}$. The transmission axis of $P_2$ is at $60^{\circ}$ to the vertical. The angle between the incident light and the axis is $	heta_2 = 60^{\circ} - 30^{\circ} = 30^{\circ}$.$I_2 = k \cdot I_1 \cdot \cos^2(30^{\circ}) = 0.9 \cdot 67.5 \cdot 0.75 = 45.5625 \, W/m^2$.$3$. For the third polariser $(P_3)$: The light incident on $P_3$ is polarised at $60^{\circ}$. The transmission axis of $P_3$ is at $90^{\circ}$ to the vertical. The angle between the incident light and the axis is $	heta_3 = 90^{\circ} - 60^{\circ} = 30^{\circ}$.$I_3 = k \cdot I_2 \cdot \cos^2(30^{\circ}) = 0.9 \cdot 45.5625 \cdot 0.75 \approx 30.75 \, W/m^2$.The final intensity is approximately $30 \, W/m^2$.

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