Why is the first minimum angle half what I calculated for 4-slit interference?

[Thefox14]

Homework Statement

A 4-slit interference pattern is created by 4 slits spaced 2μm apart. At what approximate angle θ does the intensity go to zero the first time, if the incoming light has a wavelength of 450 nm?
θ1st min =

Homework Equations

d*sinθ = m$\lambda$

The Attempt at a Solution

I thought this problem was as simple as using the above equation, setting dsinθ = $\lambda$/2 and solving for θ. When I did, I got 6.45 degrees; but the correct answer is supposed to be 3.22 degrees.

Why would the angle I got be 2x the angle I'm looking for?

 

[Redbelly98]

... dsinθ = $\lambda$/2 ...

That is for two-slit interference.

For four slits, you need to combine waves from all four slits and have sum to zero.

 

[Thefox14]

That is for two-slit interference.

For four slits, you need to combine waves from all four slits and have sum to zero.

Oh duh! haha thanks I got it now.

For anyone else who comes across this, though, here is what I did:

$I_{f} = I(\frac{sin(N\theta/2)}{sin(\theta/2)})^{2}$

So intensity will go to zero if we get the top part in the fraction to be 0. Then we know sin is zero in multiples of pi. So after that you get this equation:

$N\theta/2 = m*pi$

Simplify that and you end up with $sin\theta = \frac{m\lambda}{Nd}$

 

[iRaid]

Kind of off-topic, this had to be the worst part of class for me to be honest. Completely uninteresting and overall hard concept to grasp (for me).

@above that's a good idea because I hated it when I needed help asap and finally found a thread with my question but no help :p