An unpolarized light of intensity I_0 passes | vedclass

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(A) The intensity of unpolarized light after passing through a polarizer is $I_p = \frac{I_0}{2}$.Let $	heta$ be the angle of rotation produced by th

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The intensity of unpolarized light after passing through a polarizer is $I_p = \frac{I_0}{2}$.Let $	heta$ be the angle of rotation produced by the optically active solution.The light incident on the analyzer is polarized at an angle $	heta$ relative to the transmission axis of the analyzer (since the polarizer and analyzer are initially at $0^{\circ}$ relative to each other).According to Malus's Law, the intensity of light emerging from the analyzer is $I = I_p \cos^2(	heta) = \frac{I_0}{2} \cos^2(	heta)$.Given $I = \frac{3}{8} I_0$, we have $\frac{I_0}{2} \cos^2(	heta) = \frac{3}{8} I_0$.Simplifying this, $\cos^2(	heta) = \frac{3}{4}$, which gives $\cos(	heta) = \frac{\sqrt{3}}{2}$.Therefore, $	heta = 30^{\circ}$.

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