The maximum numbers of possible interference maxima for slit separation equal to twice the wavelength in Young's double slit experiment is

The condition of interference maxima is $\begin{aligned} d \sin \theta & =n \lambda \\ \sin \theta & =\frac{n \lambda}{d}\end{aligned}$ Given, $\begin{gathered}d=2 \lambda \\ \sin \theta=\frac{n \lambda}{2 \lambda}=n / 2\end{gathered}$ The magnitude of $\sin \theta$ lies between 0 and 1 When $n=0, \sin \theta=0 \Rightarrow \theta=0$ When $n=1, \sin \theta=1 / 2 \Rightarrow \theta=30^{\circ}$ When $n=2, \sin \theta=1 \Rightarrow \theta=90^{\circ}$ Thus, there is central maximum $\left(\theta=0^{\circ}\right)$, on other side of it maxima lie at $\theta=30^{\circ}$ and $\theta=90^{\circ}$, so maximum number of possible interference maxima is 5.