# Diffraction Grating Spectrometry Question [closed]

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:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

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
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

$\sin \theta = (m \lambda)/d = (5 \times 1228^{-6})/(1/600) = 3.684$
$d = (1/600)\times 3 = 0.005 \lambda = (d \sin \theta) /m = (0.005 \times 3.684)/4 = 0.004605\: \mathrm{mm} = 4605\: \mathrm{nm}$