Intensity in Single-Slit Diffraction, Qualitatively — Question 2

Question 2

Uniform phase gives the largest possible field for a fixed amplitude profile

Let a real nonnegative aperture amplitude be $A(u)$. Its field is $F(q)=\int A(u)e^{-iqu}du$. No phase mask is present. Define $F(0)=\int A(u)du$.

1. Use the triangle inequality to bound the field in every direction.

2. State the phase condition for equality and explain why a continuous opening of nonzero width generally has weaker off-axis lobes.

Question 2 - Solution

Uniform phase gives the largest possible field for a fixed amplitude profile

(a) Method and calculation

$$|F(q)|=\left|\int A(u)e^{-iqu}du\right|\leq\int|A(u)e^{-iqu}|du.$$$$\boxed{|F(q)|\leq F(0),\quad I(q)\leq I(0)}.$$

(b) Interpretation and further calculation

$$\text{Equality requires}\quad\boxed{\arg[e^{-iqu}]\ \text{constant wherever}\ A(u)>0}.$$$$\text{For an interval of nonzero width:}\quad q\ne0\Rightarrow\boxed{\text{nonconstant phase and }I(q)<I(0)}.$$

Physical check

The claim relies on a nonnegative, common-phase aperture field; an imposed phase mask can instead produce an axial null.