In a Young's double slit experiment, if the slit separation is twice the wavelength of light used, then the maximum number of interference maxima is

For possible interference maxima on the screen the condition is \(\begin{aligned} d \sin \theta & =n \lambda \quad \ldots (i) \\ \text{Here, } d & =2 \lambda \text { (Given) } \\ 2 \lambda \sin \theta & =n \lambda \\ 2 \sin \theta & =n \end{aligned}\) The maximum value of \(\sin \theta\) is 1 . Hence, \(n=2 \times 1=2\) Thus the 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 .