Calculating Min. D for .6nm Wavelength Difference & 1cm Diffraction Grating

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To resolve two wavelengths differing by 0.6nm using a 1cm diffraction grating with 105 slits, a distance D to the detector must be calculated. The derived equation indicates D should be 8.3mm, but this seems low given the small wavelength difference. The relationship between the number of slits and resolving power is emphasized, suggesting that more slits can reduce the necessary distance D. There is a concern that the initial equation may not accurately reflect this dependency. Overall, further verification of the calculations and understanding of the principles involved is needed.
Jupiter
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Consider two wavelengths that differ by .6nm and a 1cm diffraction grating that has 105 slits. How great must D (dist. to detector) be in order to resolve these wavelengths from a source containing them at better than ΔY=0.1mm?

From the intensity equation, I find that a maximum will occur whenever Yd/D=n2λ. So I need (solving for D)
D=\frac{d\Delta Y}{2\Delta \lambda}=8.3\textrm{mm}. My work seems fine, but my answer seems off the mark. .6nm is a small wavelength difference. I'd expect a large D.
Verify anyone?
(d is the dist. between slit, which I take to be 1/105cm)
 
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I am surprised no one can help me. Has no one ever studied this before?
 
Your equation seems fishy since it shows no dependence on the number of slits. The more slits in the grating, the greater the resolving power. (Also, the distance D need not be all that big if there are lots of slits.) Check this out:
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/gratres.html
 

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