A) 50%
B) 100%
C) 12.5%
D) 37.5%
Correct Answer: C
Solution :
Intensity of polarized light from first polarizer\[=\frac{100}{2}=50\] \[I={{I}_{0}}{{\cos }^{2}}\theta \] \[=50{{\cos }^{2}}60{}^\circ \] \[=\frac{50}{4}=12.5\]You need to login to perform this action.
You will be redirected in
3 sec