Double Slit Homework: Order Max at 730 and 558 nm

[woaini]

Homework Statement

A double slit apparatus is illuminated 730 and 558 nm wavelengths. They form a two-colored interference pattern on a measuring screen located at a distance of 2.28 m from the slits. The slit separation is 42 wavelengths of 730 nm light.
(a) What order (m) intensity maximum for λ = 730 nm light occurs near a distance of y = 0.205 m away from the central zeroth order?
(b) What order maximum for 558 nm light lies closest to this order?

λ1=730
λ2=558
L=2.28
m=?
d=42*730

Homework Equations

mλ=dsinθ
tanθ=y/L

The Attempt at a Solution

tanθ=.205/2.28 θ=5.14

a) m=(dsinθ)/λ = 3.76 (4?)

a) m=(dsinθ)/λ = 4.92 (5?)

So did I get the slit separation right? Are my orders suppose to be decimal numbers and do I round up or down?

 

[haruspex]

Your working for a) looks right, but it bothers me that it's not closer to a whole number. Doesn't strike me as particularly 'near'. As a check, I adjusted for the fact that d/D is really tan(theta), but it made hardly difference.
For b), although it's worded a little oddly, I think it's asking which order of 558nm max is nearest the 730nm max you just identified. Note that this not quite the same as asking which order of 558nm max is nearest 0.205m from the central axis. (But it still gives 5 as the answer.)

 

[woaini]

So if it's near the 4th order, I would do this:

arcsin((4*750)/(42*730)) = 5.61

L*tan(5.61)=0.224m
y2=0.224m

I'm lost here. :(

 

[haruspex]

So if it's near the 4th order, I would do this:

arcsin((4*750)/(42*730)) = 5.61

You mean 4*730, right? so it reduces to 4/42. But it's much simpler than this.
Write out the mλ=dsinθ equation for the two different wavelengths. Which variables are the same between the two equations?

 

[woaini]

The dsin remain constant right?

 

[haruspex]

The dsin remain constant right?

Yes, d and θ are constant. So what equation can you write for the rest?