Young's Interference Experiment — Question 3

Question 3

Oblique illumination shifts the zero-order direction

A plane wave reaches two narrow slits at incidence angle $\alpha$ from the normal. Define the phase difference at outgoing angle $\theta$ by $\delta=(2\pi d/\lambda)(\sin\theta-\sin\alpha)$. Let $d=20\,\mu\mathrm m$, $\lambda=500\,\mathrm{nm}$, and $\sin\alpha=0.10$.

1. Find the zero-order outgoing direction and the two adjacent bright directions.

2. Determine the bright-fringe order that lies on the normal to the slit plane.

Question 3 - Solution

Oblique illumination shifts the zero-order direction

(a) Method and calculation

$$\delta=2\pi m\Longrightarrow\boxed{\sin\theta_m=0.10+0.025m}.$$$$\boxed{\theta_0=5.7392^\circ,\quad\theta_{-1}=4.3012^\circ,\quad\theta_{+1}=7.1808^\circ}.$$

(b) Interpretation and further calculation

$$\theta=0:\quad m=-\frac{d\sin\alpha}{\lambda}=-\frac{20(0.10)}{0.50}.$$$$\boxed{m=-4}.$$

Physical check

The central phase-matched direction follows the incident wave, while the laboratory screen center can carry a nonzero interference order.