Diffraction Gratings — Question 9

Question 9

One incorrectly phased slit reduces the principal grating peak coherently

At a principal order, nine slits of a ten-slit array contribute field $E_1$ in phase. A defect makes the tenth contribute $E_1e^{i\epsilon}$ with unchanged amplitude. Let $I_1$ be the intensity from one slit.

1. Derive the peak intensity as a function of the phase error.

2. Compare the peak for $\epsilon=\pi$ with the peak after the defective slit is blocked; normalize both to the perfect ten-slit peak.

Question 9 - Solution

One incorrectly phased slit reduces the principal grating peak coherently

(a) Method and calculation

$$E/E_1=9+e^{i\epsilon}\Longrightarrow\boxed{I/I_1=82+18\cos\epsilon}.$$$$\epsilon=0:\quad I_{\mathrm{perfect}}=100I_1.$$

(b) Interpretation and further calculation

$$\epsilon=\pi:\quad\boxed{I/I_{\mathrm{perfect}}=64/100=0.64}.$$$$\text{Defective slit blocked:}\quad\boxed{I/I_{\mathrm{perfect}}=81/100=0.81}.$$

Physical check

Removing a destructive contribution can raise the principal peak while reducing total transmitted power.