A) 10%
B) 20%
C) 25%
D) 12.5%
Correct Answer: D
Solution :
Let \[{{R}_{0}}=\frac{{{V}^{2}}}{p}=\frac{{{\left( 30 \right)}^{2}}}{90}=10\Omega \] be the intensity of unpoIarised light, so the intensity of first Polaroid is \[I=\frac{V}{{{R}_{0}}}=\frac{30}{10}=3A\]On rotating through \[\text{I=}\frac{\text{V }\!\!'\!\!\text{ }}{\text{i}}\text{=}\frac{\text{120}}{\text{3}}\text{=40 }\!\!\Omega\!\!\text{ }\] the intensity of light from second Polaroid is \[R'=R+{{R}_{0}}\] \[\Rightarrow \] So, percentage of incident light transmitted through the system is 12.5%You need to login to perform this action.
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