A) \[\left( \frac{{{I}_{0}}}{8} \right){{\sin }^{2}}2\theta \]
B) \[\left( \frac{{{I}_{0}}}{4} \right){{\sin }^{2}}2\theta \]
C) \[\left( \frac{{{I}_{0}}}{2} \right){{\cos }^{2}}\theta \]
D) \[{{I}_{0}}\,{{\cos }^{4}}\theta \]
Correct Answer: A
Solution :
The intensity of light transmitted through third Polaroid, \[I=\frac{{{I}_{0}}}{2}{{\cos }^{2}}\theta \] \[\therefore \] intensity of light transmitted through the last polaroid \[I'=\left( \frac{{{I}_{0}}}{2}{{\cos }^{2}}\theta \right).{{\cos }^{2}}(90-\theta )=\frac{{{I}_{0}}}{2}{{\cos }^{2}}\theta .{{\sin }^{2}}\theta \] \[=\frac{{{I}_{0}}\times 4{{\sin }^{2}}\theta .{{\cos }^{2}}\theta }{2\times 4}=\frac{{{I}_{0}}}{8}\times {{(2sin\theta .cos\theta )}^{2}}\] \[=\left( \frac{{{I}_{0}}}{8} \right){{\sin }^{2}}2\theta \]You need to login to perform this action.
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