Young's Interference Experiment — Question 5

Question 5

A displaced source translates the interference pattern

A point source is distance S before two slits and transverse distance h above their midpoint. The screen lies distance L after them. Work to first order in small angles with slit spacing d. Source-to-upper minus source-to-lower path is approximately $-dh/S$. Let $S=0.50\,\mathrm m$, $L=2.0\,\mathrm m$, and $h=0.75\,\mathrm{mm}$.

1. Derive the screen height at which the total two-path difference vanishes.

2. For $d=0.40\,\mathrm{mm}$ and $\lambda=600\,\mathrm{nm}$, calculate the shift in fringe spacings.

Question 5 - Solution

A displaced source translates the interference pattern

(a) Method and calculation

$$\Delta\ell=\left(-\frac{dh}{S}\right)+\left(-\frac{dy}{L}\right).$$$$\Delta\ell=0\Longrightarrow\boxed{y_0=-Lh/S=-3.00\,\mathrm{mm}}.$$

(b) Interpretation and further calculation

$$\Delta y=\frac{\lambda L}{d}=\boxed{3.00\,\mathrm{mm}}.$$$$\boxed{y_0/\Delta y=-dh/(S\lambda)=-1}.$$

Physical check

Raising the source moves the phase-matched fringe downward; this sign follows by adding the incoming and outgoing path differences.