# Notation question about double slit interference pattern

I know that the conditions for constructive/destructive interference are given by the following equations:

$$\text{Constructive:} \space \space \space d\sin{\theta}=n\lambda \\ \text{Destructive: }\space \space \space d\sin{\theta}=(n+\frac{1}{2})\lambda$$

for $$n=0,1,2,3,...$$

the first minimum occurs at $$d\sin{\theta}=(\frac{1}{2})\lambda$$. Is this called the zeroth-order minimum or the first-order minimum? In other words, if a question asked me to find the distance between the first- and second- order minimum (given $$d, \lambda,L)$$, would I have to find the distance between

1. $$d\sin{\theta}=(\frac{1}{2})\lambda \space \space$$ and $$\space \space d\sin{\theta}=(1+\frac{1}{2})\lambda$$

or

1. $$d\sin{\theta}=(1+\frac{1}{2})\lambda \space \space$$ and $$\space \space d\sin{\theta}=(2+\frac{1}{2})\lambda$$
• The difference between adjacent minima or maxima is given by $d sin \theta_{diff} = \lambda$, independent of the terminology $n^{th}$ order minimum, which is new to me. – my2cts Jan 27 at 10:43
• @my2cts I have also found it here: web.mit.edu/viz/EM/visualizations/coursenotes/modules/… Page 6 of the pdf. – Nullspace Jan 27 at 11:22
• @my2cts So the distance between minima/maxima is always $\lambda$? – Nullspace Jan 27 at 11:23
• The change in optical path between consecutive maxima (or minima) is $\lambda$ . Not the distance between the fringes. – Vincent Fraticelli Jan 27 at 11:33
• @VincentFraticelli Yeah I confused that so I would first have to find $\theta$ and then $L \tan{\theta}=D$ would give me the distance between two constructive/destructive intereferences ($L$ is the distance between slit and screen). My question was if all the maxima/minima will have the same distance $D$ between them? – Nullspace Jan 27 at 12:17

If you use the approximations $$\sin (\theta )\approx \theta$$ , the two conditions are equivalent. This approximation is usual in the double slits experiment. If not, the fringes are paraboloids and are not equidistant.
4. The reason for (3) is that, for the double slit, $$d>>\lambda$$, so that $$\sin \theta \approx \theta$$, and the central fringes are very nearly equal spaced. This means that we don't really need to label orders.