Spectrometer CCD Homework: Calculate Spectral Range for m=4

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SUMMARY

The discussion centers on calculating the spectral range for a spectrometer equipped with a grating of 600 grooves/mm when observing in order m = 4, after changing to a larger CCD plate. The initial wavelength is 1228 nm at m = 5, with a grating spacing (d) of 1666.67 nm. The key equation used is mλ = d sin(θ), where the challenge lies in determining the angle θ for m = 4. The larger CCD plate allows for a broader wavelength coverage due to its increased size.

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  • Understanding of diffraction grating principles
  • Familiarity with the equation mλ = d sin(θ)
  • Knowledge of spectrometer components and their functions
  • Basic trigonometry for solving angles
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  • Calculate the angle θ for m = 4 using the equation mλ = d sin(θ)
  • Research the impact of CCD size on spectral range coverage
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Students and professionals in spectroscopy, optical engineering, and anyone involved in the design or analysis of spectrometers and CCD applications.

woaini
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Homework Statement


A spectroscopist has a spectrometer equipped with a grating with 600 grooves/mm. This grating puts a 1228 nm wavelength range on a spectroscopic plate (a CCD) in order m = 5. The spectroscopist now changes the spectroscopic plate CCD by one that is larger by a factor of 3. What spectral range will be accepted when the spectroscopist observes the lines in order m = 4 with the new, larger spectroscopic plate?

λ1=1228
d= 1666.67
m1=5
m2=4
λ2=?

Homework Equations



mλ=dsinθ

The Attempt at a Solution



sinθ=[(5)(1228)]/(1666.67)=3.68

However I can't solve for θ, and plug it into another question for m2=4. What step should I do next?

So what exactly does the electroscopic plate do?
 
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woaini said:
wavelength range on a spectroscopic plate (a CCD) in order m = 5. The spectroscopist now changes the spectroscopic plate CCD by one that is larger by a factor of 3.

i think the first set of data gives you an idea of angle at which the experiment is being done - my guess is that angle of diffraction gives you size of CCD detector.
if you use a larger size you may get a larger coverage of wave length /order of diffraction.
 

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