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

Step-by-step solution

For possible interference maxima on the screen, the condition is d =n (i) Given : d= slit - width =2 aligned & & 2 & =n & & 2 & =n aligned The maximum value of is 1 , hence, n=2 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 .