(C) According to Malus's law,the intensity of light transmitted through an analyser is given by $I_2 = I_1 \cos^2 \theta$.Here,$I_1 = I$ is the intens

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(C) According to Malus's law,the intensity of light transmitted through an analyser is given by $I_2 = I_1 \cos^2 \theta$.Here,$I_1 = I$ is the intensity of the incident linearly polarised light.The angle between the transmission axes of the polariser and the analyser is $\theta = 30^{\circ}$.Substituting these values into the formula:$I_2 = I \cos^2(30^{\circ})$Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have:$I_2 = I \left( \frac{\sqrt{3}}{2} \right)^2 = I \left( \frac{3}{4} \right) = \frac{3I}{4}$.

Class 12 Physics Wave Optics Polarisation of Light and Malus' Law

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After passing through a polariser,a linearly polarised light of intensity $I$ is incident on an analyser making an angle of $30^{\circ}$ with that of the polariser. The intensity of light emitted from the analyser will be

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A
$\frac{I}{2}$
B
$\frac{I}{3}$
C
$\frac{3I}{4}$
D
$\frac{2I}{3}$

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Wave Optics

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Polarisation of Light and Malus' Law

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After passing through a polariser,a linearly polarised light of intensity $I$ is incident on an analyser making an angle of $30^{\circ}$ with that of the polariser. The intensity of light emitted from the analyser will be

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$A$ beam of unpolarised light of intensity $I_0$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $45^{\circ}$ relative to that of $A$. The intensity of emergent light is:

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