The condition for interference maxima is given as, d‌sin‌θ=nλ‌quad...(I) Substituting the given values in above expression, we get 2λ‌sin‌θ=nλ 2‌sin‌θ=n The maximum value of sin‌θ comes out to be 1 The value of n comes out to be: n=2×1=2 Thus, total 5 integer is possible for equation (I) which are −2,−1,0,1,2