Diffraction by a Double Slit — Question 6

Question 6

A half-cycle phase offset changes the number of central-envelope bright orders

Equal slits have $d/a=2.50$. A uniform phase plate adds relative phase pi, so the intensity is $I=4I_s\operatorname{sinc}^2\beta\sin^2\alpha$, with $\alpha=\pi d\sin\theta/\lambda$ and $\beta=(a/d)\alpha$.

1. Find the nominal bright-order condition and classify the axis.

2. Count the nominal bright directions strictly inside the central envelope and determine whether the envelope-edge orders survive.

Question 6 - Solution

A half-cycle phase offset changes the number of central-envelope bright orders

(a) Method and calculation

$$\sin^2\alpha=1\Longrightarrow\boxed{\alpha=(m+1/2)\pi};\quad\alpha=0\Longrightarrow\boxed{I=0}.$$$$|\beta|<\pi\Longrightarrow|m+1/2|<2.50.$$

(b) Interpretation and further calculation

$$\boxed{m+1/2=-1.5,-0.5,0.5,1.5:\ 4\ \text{directions}}.$$$$m+1/2=\pm2.5\Longrightarrow\beta=\pm\pi\Longrightarrow\boxed{\text{both edge orders are missing}}.$$

Physical check

A phase plate shifts the interference factor but leaves the individual slit envelope fixed.