28. Setting I = 2I
0
in Eq. 36-21 and solving for the smallest (in absolute value) two roots for φ/2,we find
φ = 2 cos
−1
1
√
2
=
±
π
2
rad .
Now,for small θ in radians,Eq. 36-22 becomes φ = 2πdθ/λ. This leads to two corresponding angle
values:
θ =
±
λ
4d
.
The difference between these two values is ∆θ =
λ
4d
−
−
λ
4d
=
λ
2d
.