JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 17)
A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be
zero, is :
71.6o
90o
18.4o
45o
Explanation
I = I0 cos2 $$\theta $$
$$ \Rightarrow $$ $${{{I_0}} \over {10}}$$ = I0 cos2 $$\theta $$
$$ \Rightarrow $$ cos $$\theta $$ = $${1 \over {\sqrt {10} }}$$
$$ \Rightarrow $$ $$\theta $$ = 71.6o
$$ \therefore $$ $$\phi $$ = 90 - 71.6 = 18.4o
$$ \Rightarrow $$ $${{{I_0}} \over {10}}$$ = I0 cos2 $$\theta $$
$$ \Rightarrow $$ cos $$\theta $$ = $${1 \over {\sqrt {10} }}$$
$$ \Rightarrow $$ $$\theta $$ = 71.6o
$$ \therefore $$ $$\phi $$ = 90 - 71.6 = 18.4o
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