JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 13)
A system of three polarizers P1, P2, P3 is set up such that the pass axis of P3 is crossed with respect to that of P1.
The pass axis of P2 is inclined at 60o to the pass axis of P3. When a beam of unpolarized light of intensity I0 is
incident on P1, the intensity of light transmitted by the three polarizers is I. The ratio ($${{{I_0}} \over I}$$) equals (nearly) :
10.67
5.33
16.00
1.80
Explanation
When unpolarized light of intensity I0 passes through P1, then intensity
I1 = $${{{I_0}} \over 2}$$
as we know, I = I0 cos2$$\theta $$
Given that angle between P2 & P3 = 60o and
P1 and P3 are crossed that means angle between P1 and P3 is 90o
so angle between P1 and P2 = 90° – 60° = 30°
I2 = $${{{I_0}} \over 2}{\cos ^2}30^\circ $$ = $${{3{I_0}} \over 8}$$
I3 = $${{3{I_0}} \over 8}{\cos ^2}60^\circ $$ = $${{3{I_0}} \over {32}}$$ = I
$$ \therefore $$ $${{{I_0}} \over I}$$ = $${{{I_0}} \over {{{3{I_0}} \over {32}}}}$$ = $${{32} \over 3}$$ = 10.67
I1 = $${{{I_0}} \over 2}$$
as we know, I = I0 cos2$$\theta $$
Given that angle between P2 & P3 = 60o and
P1 and P3 are crossed that means angle between P1 and P3 is 90o
so angle between P1 and P2 = 90° – 60° = 30°
I2 = $${{{I_0}} \over 2}{\cos ^2}30^\circ $$ = $${{3{I_0}} \over 8}$$
I3 = $${{3{I_0}} \over 8}{\cos ^2}60^\circ $$ = $${{3{I_0}} \over {32}}$$ = I
$$ \therefore $$ $${{{I_0}} \over I}$$ = $${{{I_0}} \over {{{3{I_0}} \over {32}}}}$$ = $${{32} \over 3}$$ = 10.67
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