An unpolarized light of intensity \( I_0 \) is passed through a polariser and is incident on an analyser making an angle of \( (\pi/6)\,\mathrm{rad} \) with that of the polariser. The intensity of the light transmitted from the analyzer is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 0.375\,I_0 \)
The polariser halves the intensity of unpolarised light. The analyser then reduces the polarised light by Malus's law, \( I = I_1\cos^{2}\theta \).
\( \displaystyle I_1 = \frac{I_0}{2} \)
\( \displaystyle I = I_1\cos^{2}\frac{\pi}{6} = \frac{I_0}{2} \times \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{I_0}{2} \times \frac{3}{4} = 0.375\,I_0 \)