Diffraction Gratings — Question 3

Question 3

Increasing slit count inserts more zeros between principal maxima

A six-slit grating has array factor $I/I_1=[\sin(3\delta)/\sin(\delta/2)]^2$. Consider one phase interval $0<\delta<2\pi$ between neighboring principal maxima.

1. Locate every zero in this open interval.

2. State the number of secondary maxima between the principal peaks and explain why the endpoints are not zeros.

Question 3 - Solution

Increasing slit count inserts more zeros between principal maxima

(a) Method and calculation

$$\sin(3\delta)=0\Longrightarrow\delta=p\pi/3.$$$$\boxed{\delta=\pi/3,2\pi/3,\pi,4\pi/3,5\pi/3:\ 5\ \text{zeros}}.$$

(b) Interpretation and further calculation

$$\boxed{N-2=4\ \text{secondary maxima between these zeros}}.$$$$\delta=0,2\pi:\quad\lim\frac{\sin(3\delta)}{\sin(\delta/2)}=\pm6\Longrightarrow\boxed{I=36I_1}.$$

Physical check

Numerator zeros that coincide with denominator zeros must be evaluated by a limit; counting them as dark lines would remove the principal orders.