How Many Interference Maxima Appear Within the Central Diffraction Peak?

[akmphy]

Homework Statement

Light with a wavelength of 530 nm is incident on 2 slits that are spaced 1.0mm apart. If each slit has a width of .1mm, how many interference maxima lie within the central diffraction peak?

Homework Equations

sin( degree) = m(lambda)/d

The Attempt at a Solution

I have no idea what the question is asking, but I started with:

arc sin ( 1 * 530e-9)/ .001
I got .03 degrees for the half width of the central diffraction.
Double this degree, .06, for the entire width of the central diffraction.

I then did:
[sin(.06) * .001]/530e-9 = m
m = 1.97

So wrong! The answer is 21.
Please help me understand this.
Thanks.

 

[ehild]

The halfwidth of the diffraction pattern is about lambda/width of the slit, so it is 0.304°.
The maxima of the interference pattern of the two slits are apart by lambda/(distance between the slits).

ehild