Intensity in Single-Slit Diffraction, Qualitatively — Question 7

Question 7

Blocking half a slit can fill a former dark minimum

A uniform slit of width a has central intensity $I_c$. At its first minimum, $q=2\pi/a$. One half of the slit is then covered, keeping the field amplitude in the remaining half unchanged. Use the phasor arc for the surviving half.

1. Find the surviving-half arc angle and its resultant field relative to the old central field.

2. Calculate the intensity at the former minimum and compare it with the new on-axis intensity.

Question 7 - Solution

Blocking half a slit can fill a former dark minimum

(a) Method and calculation

$$\Delta\phi_{\mathrm{half}}=q(a/2)=\pi;\quad E_{c,\mathrm{half}}=E_c/2.$$$$\boxed{|E_{\mathrm{half}}|=(E_c/2)\frac{\sin(\pi/2)}{\pi/2}=E_c/\pi}.$$

(b) Interpretation and further calculation

$$\boxed{I_{\mathrm{former\ zero}}=I_c/\pi^2=0.101321I_c}.$$$$I_{c,\mathrm{new}}=I_c/4\Longrightarrow\boxed{I_{\mathrm{former\ zero}}/I_{c,\mathrm{new}}=4/\pi^2}.$$

Physical check

The old zero depended on cancellation between both halves; removing one half removes that cancellation.