Intensity in Double-Slit Interference — Question 8

Question 8

A constant background lowers visibility without moving the fringe positions

Equal coherent beams each contribute $I_0=4\,\mathrm{W/m^2}$. An incoherent uniform background contributes $B=6\,\mathrm{W/m^2}$, so $I=2I_0(1+\cos\delta)+B$.

1. Find the observed extrema and visibility.

2. If the background is unknown but the beam intensity is known, derive B from a measured visibility V; evaluate the largest B compatible with $V\geq0.80$.

Question 8 - Solution

A constant background lowers visibility without moving the fringe positions

(a) Method and calculation

$$\boxed{I_{\max}=4I_0+B=22\,\mathrm{W/m^2},\quad I_{\min}=B=6\,\mathrm{W/m^2}}.$$$$\boxed{V=\frac{4I_0}{4I_0+2B}=\frac{2I_0}{2I_0+B}=4/7}.$$

(b) Interpretation and further calculation

$$V(2I_0+B)=2I_0\Longrightarrow\boxed{B=2I_0(V^{-1}-1)}.$$$$\boxed{B_{\max}=8(1/0.80-1)=2.0\,\mathrm{W/m^2}}.$$

Physical check

A nonzero minimum does not by itself prove partial coherence: unequal amplitudes and additive background are separate possible causes.