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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.

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closed as off-topic by David Z Dec 20 '13 at 10:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

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Hi 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. –  Brandon Enright Dec 3 '13 at 6:31
    
A minor note, I don't know if it also existed in your calculations, but 1228 nm is $1.228 \times 10^{-6}$ NOT $1.228 \times 10^{-9}$ –  Steven Goldade Dec 3 '13 at 8:40
1  
Seems 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 5th-order 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. –  Carl Witthoft Dec 3 '13 at 12:41

2 Answers 2

$\sin \theta = (m \lambda)/d = (5 \times 1228^{-6})/(1/600) = 3.684$

$d = (1/600)\times 3 = 0.005 \lambda = (d \sin \theta) /m = (0.005 \times 3.684)/4 = 0.004605\: \mathrm{mm} = 4605\: \mathrm{nm}$

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up vote -2 down vote accepted

So this was the answer given in the key:

enter image description here

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This 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? –  dmckee Dec 22 '13 at 17:48

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