A) 16.00
B) 1.80
C) 5.33
D) 10.67
Correct Answer: D
Solution :
Since unpolarised light falls on \[{{P}_{1}}\Rightarrow \] intensity of light transmitted from \[{{P}_{1}}=\frac{{{I}_{0}}}{2}\] Pass axis of \[{{P}_{2}}\]will be at an angle of \[30{}^\text{o}\] with\[{{P}_{1}}\] \[\therefore \]Intensity of light transmitted from \[{{P}_{2}}=\frac{{{I}_{0}}}{2}{{\cos }^{2}}{{30}^{o}}=\frac{3{{I}_{0}}}{8}\] Pass axis of \[{{P}_{3}}\]is at an angle of \[60{}^\text{o}\] with \[{{P}_{2}}\] \[\therefore \]Intensity of light transmitted from \[{{P}_{3}}=\frac{3{{I}_{0}}}{8}{{\cos }^{2}}{{60}^{o}}=\frac{3{{I}_{0}}}{32}\] \[\therefore \]\[\left( \frac{{{I}_{0}}}{I} \right)=\frac{32}{3}=10.67\]You need to login to perform this action.
You will be redirected in
3 sec