Diffraction by a Double Slit — Question 8

Question 8

Fringe visibility can distinguish coherence loss from an envelope decrease

Equal parallel-polarized slits have an intensity envelope $I_s\operatorname{sinc}^2\beta$ and coherence modulus $|\gamma(D)|=e^{-(D/\ell_c)^2}$, where $D=d\sin\theta$. At nominal order m, $D=m\lambda$. A phase scan at order 10 gives visibility $e^{-1}$ for $\lambda=600\,\mathrm{nm}$.

1. Infer the coherence length and show why the envelope cancels from visibility.

2. If $d/a=12$, determine whether order 10 is inside the central envelope and predict visibility at order 5.

Question 8 - Solution

Fringe visibility can distinguish coherence loss from an envelope decrease

(a) Method and calculation

$$V=\frac{2I_s\operatorname{sinc}^2\beta|\gamma|}{2I_s\operatorname{sinc}^2\beta}=|\gamma|\quad\text{away from envelope zeros}.$$$$e^{-(10\lambda/\ell_c)^2}=e^{-1}\Longrightarrow\boxed{\ell_c=10\lambda=6.00\,\mu\mathrm m}.$$

(b) Interpretation and further calculation

$$|m|<d/a=12\Longrightarrow\boxed{m=10\ \text{is inside the central envelope}}.$$$$\boxed{V_5=e^{-(5/10)^2}=e^{-1/4}=0.778801}.$$

Physical check

Falling peak brightness and falling fringe contrast are different measurements; only the latter directly isolates coherence in this equal-beam model.