• # question_answer Two polaroids are placed in the path of unpolarized beam of intensity ${{I}_{0}}$ such that no light is emitted from the second polaroid. If a third polaroid whose polarization axis makes an angle q with the polarization axis of first polaroid, is placed between these polaroids then the intensity of light emerging from the last polaroid will be           [UPSEAT 2005] A)            $\left( \frac{{{I}_{0}}}{8} \right){{\sin }^{2}}2\theta$         B)            $\left( \frac{{{I}_{0}}}{4} \right){{\sin }^{2}}2\theta$ C)            $\left( \frac{{{I}_{0}}}{2} \right){{\cos }^{4}}\theta$          D)            ${{I}_{0}}{{\cos }^{4}}\theta$

No light is emitted from the second polaroid, so ${{P}_{1}}$ and ${{P}_{2}}$ are perpendicular to each other Let the initial intensity of light is ${{I}_{0}}$. So Intensity of light after transmission from first polaroid = $\frac{{{I}_{0}}}{2}$.            Intensity of light emitted from ${{P}_{3}}$  ${{I}_{1}}=\frac{{{I}_{0}}}{2}{{\cos }^{2}}\theta$            Intensity of light transmitted from last polaroid i.e. from ${{P}_{2}}={{I}_{1}}{{\cos }^{2}}({{90}^{o}}-\theta )$= $\frac{{{I}_{0}}}{2}{{\cos }^{2}}\theta .{{\sin }^{2}}\theta$                 $=\frac{{{I}_{0}}}{8}$${{(2\sin \theta \cos \theta )}^{2}}$= $\frac{{{I}_{0}}}{8}{{\sin }^{2}}2\theta$. You will be redirected in 3 sec 