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) \[{{I}_{0}}{{\cos }^{4}}\theta \]
Correct Answer: A
Solution :
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 need to login to perform this action.
You will be redirected in
3 sec