Answer :

Correct answer is : 45

$I_3=\frac{I_0}{2}\times <!--\; \times \; -->\frac{4{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}{4}$

$=\frac{I_0}{8}(\sin <!--\; \; -->2\mathrm{\theta <!--\; \theta \; -->}{)}^2$

For I₃ → max

sin2θ = 1

2θ = 90°

θ = 45°