Diffraction and the Wave Theory of Light — Question 4

Question 4

Complementary screens have equal off-axis diffraction intensities

In scalar diffraction, complementary aperture transmissions satisfy $t_A(u)+t_B(u)=1$. The resulting fields obey $E_A+E_B=E_{\mathrm{free}}$. At an off-axis far-field point take $E_{\mathrm{free}}=0$; on axis it is generally nonzero.

1. Derive the off-axis field and intensity relation between the complementary screens.

2. On axis, let $E_A=0.30E_{\mathrm{free}}$ with the same phase. Find the complementary intensity and explain why intensities cannot be added to obtain the unobstructed result.

Question 4 - Solution

Complementary screens have equal off-axis diffraction intensities

(a) Method and calculation

$$E_{\mathrm{free}}=0\Longrightarrow\boxed{E_B=-E_A,\quad I_B=I_A}.$$$$\boxed{\arg E_B-\arg E_A=\pi\pmod{2\pi}}\quad(E_A\ne0).$$

(b) Interpretation and further calculation

$$E_B=E_{\mathrm{free}}-E_A=0.70E_{\mathrm{free}}\Longrightarrow\boxed{I_B=0.49I_{\mathrm{free}}}.$$$$I_A+I_B=0.58I_{\mathrm{free}};\quad2\Re(E_AE_B^*)/|E_{\mathrm{free}}|^2=0.42\Longrightarrow\boxed{0.58+0.42=1}.$$

Physical check

The complementary-field relation is coherent and linear in amplitude. Its equal-intensity consequence excludes the unobstructed forward beam.