Diffraction Grating Spectrometry Question
So this was the answer given in the key:
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Mizuho
Updated on August 04, 2022Comments

Mizuho about 16 hours
Yes, this is a homework question, but I've already failed to solve it enough times that the online system hosting it isn't going to give me any marks, so I figure it's a good time to stop hitting the wall and actually ask. The question is as below:
A spectroscopist uses a spectrometer that has a grating with 600 grooves/mm. This grating can illuminate a CCD with a range of 1228 nm, in the 5th order (m = 5). One day, she buys a larger format CCD  one larger than her old one by a factor of 3. What spectral range of wavelengths can be measured when she now obscures the 4th order lines (m = 4) with her larger CCD?
Came out with answers like $1.228 \times 10^{6}$ and $4.17 \times 10^{7}$ so far because I have no idea what the process is. Hints would suffice.
Basically, everything I've tried so far revolves around a formula: $m\lambda=d\sin\theta$.
What did I do so far:
 $5(1.228\times10^{9})=\frac{0.001}{600}\times\sin\theta$ to get $\sin\theta$ and then put it back in to replace $\sin\theta$ as $\frac{4}{5}\sin\theta$. Then I went and searched up a diffraction grating to find out that the orders are not equidistant.
 $5(1.228\times10^{9})=\frac{0.001}{600}\times\sin 90^\circ$ where I assumed the angle is perpendicular.
 Other stuff I can't remember
I feel like I understood the wavelength portion of this incorrectly, but I have no idea why.

Brandon Enright over 8 yearsHi Mizuho, you need to describe what you've already tried and your basic thought process to have any hope of getting a good answer to your homework question.

Bugasu over 8 yearsA minor note, I don't know if it also existed in your calculations, but 1228 nm is $1.228 \times 10^{6}$ NOT $1.228 \times 10^{9}$

Carl Witthoft over 8 yearsSeems kind of odd: "grating can illuminate..." doesn't mean much to me. I suppose your teacher means the CCD is placed at a distance such that the 5thorder pattern fills N cm of space (the CCD width) with 1228nm spectral range, then at the same distance from the grating, place a CCD that is 3*N wide.

dmckee  exmoderator kitten over 8 yearsThis being someone else's work it should probably have a proper attribution. Saying it was "given in the key" is a good start, but is not really complete. Who should get the credit for the text?