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

The condition for possible interference maxima on the screen is2dsin⁡θ=nλ (2d= slit width ) given 2 d =2 λ 2λsin⁡θ=nλ or 2sin⁡θ=n n=2( maximum value of sin⁡θ=1)  Eq. (1) will be satisfied for integer values −2,−1,0,1, and  2, i.e., 5. So, the maximum number of possible interference maxima is 5