Simple Spectrometer and Spectroscopy Question

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SUMMARY

The discussion centers on calculating wavelengths using a 9650 line/cm spectroscope based on measured angles of +26°38', +41°02', -26°18', and -40°27'. The formula used is λ = (d/m)sin(θ). The participant, Zach, expresses confusion about the presence of only two distinct spectrum lines and the implications of measurement error. Ultimately, Zach resolved the issue by averaging the angles, acknowledging potential measurement errors in the textbook "Physics for Scientists and Engineers" by Giancoli.

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Homework Statement


Two first-order spectrum lines are measured by a 9650line/cm spectroscope at angles, on each side of the center, of +26*38', +41*02' and -26*18', -40*27'. Calculate the wavelengths based on these data.

Homework Equations


[tex]\lambda[/tex]= (d/m)sin([tex]\theta[/tex])

The Attempt at a Solution



i know how to solve similar problems... nothing complicated. but I'm confused how there are only two different spectrum lines in this problem that can be seen with different angles on either side... i thought the incident ray would be oriented at 0 and would then be diffracted by the slits an equal amount up and down. IE if a ray diffracts so you see it at 30*, how could it not be -30*, in the downward direction? or would i for some reason use something like [tex]\Delta\theta[/tex] and use the difference between two rays?

Thank you!

-Zach

edit...
Or is possible that there is some kind of error that the book (Physics for Scientists and Engineers, Giancoli 4th) would want to you take into account for and just count the similar opposite numbers as statistically equivalent?

or could it be a continuous spectrum from +26*38' to +41*02' and then another from -26*18' to -40*27'?

just brainstorming here... really at a loss, any input is greatly appreciated
 
Last edited:
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Usually with these measurements, you find the diffraction position on both sides.
This gives you (by addition) double the angle for (in this case) the 1st order.
Taking twice the angle in this way and dividing by two gives a more accurate value.
 
Yes that is what I ended up doing. I just turned it in. I ended up just chalking it up to measurement error, and averaged the two for the angle. That was just really out of character for the textbook. First problem in 35 Chapters I've seen that would have some kind of practical measurement error involved.

After handing it in, even my professor had no good answer for it and figured the same as you Stone.

Thanks for the input!
 

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