Answer:
I_3/I_o = 0.1634
Explanation:
we have three polarizers, and can thus talk about three intervals of intensity of light:
I_o -- entering the first polarizer
I_1 -- leaving the first polarizer and entering the second
I_2 -- leaving the second polarizer and entering the third
I_3 -- leaving the third polarizer
From Malus' law, for the case of (linearly) polarized light falling on a (linear) polarizer, the output intensity Iout from the polarizer to the input intensity Iin, can be related as;
I_out = I_in•cos²(θ)
So we can relate the intensity of the light coming out of the second polarizer, I_2, to the intensity of the light coming into the second polarizer, I_1, by Malus' law:
I_2 = I_1cos²(23.9) = I_o/2(cos²(23.9))
Thus, in the same way, we can relate I_3 to I_2 as;
I_3 = (I_2)cos²(51.3) =I_o/2(cos²(23.9))cos²(51.3)
So, I_3/I_o = (1/2)cos²(23.9)•cos²(51.3)
= 0.5 x 0.8359 x 0.3909 = 0.1634