Diffraction by a Single Slit: Locating the Minima — Question 5

Question 5

Oblique illumination makes the minimum angles asymmetric about the normal

For a slit under plane-wave incidence angle $\alpha$, use $a(\sin\theta-\sin\alpha)=m\lambda$ for nonzero integer m. Let $\sin\alpha=0.20$, $a=4.00\,\mu\mathrm m$, and $\lambda=600\,\mathrm{nm}$.

1. Calculate the m=-1 and m=+1 minimum angles.

2. Find all minimum orders in the open forward hemisphere and explain why the pattern is not symmetric in angle about the laboratory normal.

Question 5 - Solution

Oblique illumination makes the minimum angles asymmetric about the normal

(a) Method and calculation

$$\boxed{\sin\theta_m=0.20+0.15m}.$$$$\boxed{\theta_{-1}=2.86598^\circ,\quad\theta_{+1}=20.4873^\circ}.$$

(b) Interpretation and further calculation

$$-1<0.20+0.15m<1\Longrightarrow-8<m<16/3.$$$$\boxed{m=-7,\ldots,-1,1,\ldots,5:\ 12\ \text{minima}}.$$

Physical check

The symmetry is in the variable $\sin\theta-\sin\alpha$; the central phase-matched direction is $\theta=\alpha$, not zero.