p36 028

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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

.


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