
After the first polarizer: \(I_1 = \frac{I_0}{2}\)
After the second polarizer: \(I_2 = I_1 \cos^2 60^\circ = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}\)
After the third polarizer: \(I_3 = I_2 \cos^2 45^\circ = \frac{I_0}{8} \times \frac{1}{2} = \frac{I_0}{16}\)
The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.