# A Problem About Multi Slit Interference

To my understanding,in multi slit interference, 2 equations are correct. $$n*\lambda=d*sin(\theta)$$ and $$S=(\lambda*D)/d$$ In the problem below, however, the 2 equations gave different solutions. I hope to understand why that is.

The problem is below:

Using the first equation $$n*\lambda=d*sin(\theta)$$, the answer I got was $$1.51$$. Specifically:

• When we solve the equation for $$\theta$$ with $$700$$ nm, we get $$22$$ degrees. When we solve with $$400$$ nm, we get $$12$$ degrees. We then take the tangent of the 2 and multiply both by $$8.13$$, then subtract them. This is how we get $$1.51$$ m.

Using the second equation $$S=(\lambda*D)/d$$, the answer I got was $$1.31$$. Specifically:

• I subbed the given values and $$400$$ nm into the equation, then did the same for $$700$$ nm. Then I multiplied the 2 results by $$2$$ (to get the distance from center to second principle maxima), then subtracted both.

So why would the 2 equations give different solutions?

I suspect that your second formula has $$S$$ as the fringe separation on a screen that is $$D$$ away from slits which are $$d$$ apart.
That being so the second formula can be derived from the first formula on the assumption that the angle $$\theta$$ is small such that $$\tan \theta \approx \sin \theta \approx\theta$$.
So you have something like $$n \lambda = d \sin \theta_{\rm n} \Rightarrow n \lambda \approx d \,\dfrac {s_{\rm n}}{D}$$ and $$(n+1) \lambda = d \sin \theta_{\rm n+1} \Rightarrow (n+1) \lambda \approx d \,\dfrac {s_{\rm n+1}}{D}$$.
Subtraction and rearrangement gives $$s_{\rm n+1} - s_{\rm n} = S \approx \dfrac{\lambda D}{d}$$