16.00
1.80
5.33
10.67
Correct option is D. 10.67
Since unpolarised light falls on P_1 ⇒ intensity of light transmitted from P_1 = I_0/2.
Pass axis of P_3 is at an angle of 60° with P_2 ∴ Intensity of light transmitted from P_3 = (I_0/2) cos²60° = (I_0/2)(1/4) = I_0/8.
Intensity of light transmitted from P_2 = (I_0/2) cos²30° = (I_0/2)(3/4) = (3I_0)/8
Intensity of light transmitted from P_3 = (3I_0/8)cos²(60°) = (3I_0/8)(1/4) = (3I_0)/32
∴ (I_0/I) = 32/3 = 10.67