Barrier of Diffraction: Different Wavelengths, Same Color?

[Const@ntine]

I believe your value for a is incorrect.
570 grooves/mm = 570 grooves/mm × 1000 mm/m = 5.7×10⁵ grooves/m. Therefore a = 1/(5.7×10⁵) m/groove = 1.75×10^-6 m/groove.
Multiply by 10¹⁰ Angstrom/m to get a ≈ 18000 Angstroms/groove. This will increase your wavelengths by a factor of 10.

a is from the previous exercise I mentioned, which we did at the Lab, and checked it. Basically we did the same thing as here, but with Sodium Vapor, known wavelength (λ = 5893 Angstrom), for just one fringe (m=+-1) and using the formula for φ (tanφ = ...) and either (5) or (6) found a. The Professor checked it and said it was correct, so I have to use it here.

I don't think it's out of the question that all these values are incorrect, but since he told us to treat the exercise a certain way, I'll leave the results as is and ask him later. Just an FYI, he's a Post-Grad/Junior Professor so he might be glossing over some stuff to get it done quickly.

Also, can you show a sample calculation of how you got the values that you got? They disagree with mine.

On edit: I agree with the first two calculations even though it appears you have the wrong value for a. I disagree with the other two after "And again". You might wish to consider doing these on a spreadsheet with the same formula for all. Then either all of them will be correct or all wrong but easy to trouble shoot.

I just took the equation you came up with, and plugged in the angles, in the real values, not the absolute ones. For the 1st Order I put m = 1, and for the 2nd m = 2.

1st Order:
sin(70.9 - (-60,1)) + sin(-41,9 + (-60,1)) = 2*1*λ/(18300*10^-10 Angstrom) <=> λ = -2044 Angstrom

2nd Order:
sin(85,2 - (-54,7)) + sin(-26,9 + (-54,7)) = 2*2*λ/(18300*10^-10 Angstrom) <=> λ = -1579 Angstrom

Huh. Those are obviously different from before. I honestly don't know why. Maybe I added them in their absolute values before? Are these anywhere close to reality?

 

[kuruman]

Why are you using negative values for φ? I didn't. Like I wrote earlier, having the picture, on which the derivation of equations 5 & 6 was based, would help tremendously with the correct substitution of variables.