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Question

(C) Let the intensity of the incident light be $I_0$. According to Malus' law,the intensity of light after passing through a polariser is $I = I_{in}

Vedclass · Accepted Answer

$30 \%$ of the incident light

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(C) Let the intensity of the incident light be $I_0$. According to Malus' law,the intensity of light after passing through a polariser is $I = I_{in} \cos^2 	heta$,where $	heta$ is the angle between the incident light's polarisation plane and the polariser's axis.Since the first polariser's axis is parallel to the plane of polarisation of the incident light,the angle $	heta_1 = 0^{\circ}$. Thus,the intensity after the first polariser is $I_1 = I_0 \cos^2(0^{\circ}) = I_0$.For the subsequent polarisers,each is rotated by $30^{\circ}$ relative to the previous one. Therefore,the angle between the light's polarisation plane and the next polariser is $30^{\circ}$.Intensity after the second polariser: $I_2 = I_1 \cos^2(30^{\circ}) = I_0 (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} I_0$.Intensity after the third polariser: $I_3 = I_2 \cos^2(30^{\circ}) = (\frac{3}{4} I_0) 	imes \frac{3}{4} = (\frac{3}{4})^2 I_0$.Intensity after the fourth polariser: $I_4 = I_3 \cos^2(30^{\circ}) = (\frac{3}{4})^2 I_0 	imes \frac{3}{4} = (\frac{3}{4})^3 I_0$.Intensity after the fifth polariser: $I_5 = I_4 \cos^2(30^{\circ}) = (\frac{3}{4})^3 I_0 	imes \frac{3}{4} = (\frac{3}{4})^4 I_0$.Calculating the value: $I_5 = (0.75)^4 I_0 = 0.3164 I_0 \approx 31.6 \% I_0$.Comparing with the given options,the intensity is closest to $30 \%$ of the incident light.

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