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

For possible interference maxima on the screen, the condition is $d \sin \theta=n \lambda\ldots(i)$ Given : $d=$ slit - width $=2 \lambda$ $\begin{aligned} & \therefore & 2 \lambda \sin \theta & =n \lambda \\ \Rightarrow & & 2 \sin \theta & =n \end{aligned}$ The maximum value of $\sin \theta$ is 1 , hence, $n=2 \times 1=2$ Thus, Eq. (i) must be satisfied by 5 integer values $i e,-2,-1,0,1,2$. Hence, the maximum number of possible interference maxima is 5 .