A) 50 %
B) 100 %
C) 12.5 %
D) 37.5 %
Correct Answer: C
Solution :
[c] Intensity of polarized light form first polarizer \[=\frac{100}{2}=50%\] \[I=50{{\cos }^{2}}60{}^\circ =\frac{50}{2}=12.5%\]
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