Diffraction by a Double Slit — Question 9

Question 9

The integrated double-slit power is twice the single-slit power

In a paraxial one-dimensional model, $I_2(\beta)=4I_s\operatorname{sinc}^2\beta\cos^2(r\beta)$, with $r=d/a\geq1$ for nonoverlapping slits. Use $\int\operatorname{sinc}^2\beta\,d\beta=\pi$ over the real line and $\int\operatorname{sinc}^2\beta\cos(2r\beta)d\beta=\pi\max(1-|r|,0)$.

1. Integrate the double-slit intensity and compare with one slit.

2. Compare the axial intensities and explain the difference between peak and total-power ratios.

Question 9 - Solution

The integrated double-slit power is twice the single-slit power

(a) Method and calculation

$$\cos^2(r\beta)=\tfrac12[1+\cos(2r\beta)]\Longrightarrow\int I_2d\beta=2I_s[\pi+0].$$$$\int I_1d\beta=I_s\pi\Longrightarrow\boxed{P_2/P_1=2}.$$

(b) Interpretation and further calculation

$$\beta=0:\quad\boxed{I_2(0)/I_1(0)=4}.$$$$\boxed{\text{peak ratio }4\ne\text{integrated-power ratio }2}.$$

Physical check

Interference redistributes power among directions. The fourfold central peak is compensated by the intervening dark regions.