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

Step-by-step solution

The condition of interference maxima is aligned d & =n & = n d aligned Given, gathered d=2 = n 2 =n / 2 gathered The magnitude of lies between 0 and 1 When n=0, =0 =0 When n=1, =1 / 2 =30^ When n=2, =1 =90^ Thus, there is central maximum ( =0^ ) , on other side of it maxima lie at =30^ and =90^ , so maximum number of possible interference maxima is 5.