# Find the Wavelength

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1. Sep 25, 2016

### allenlistar

1. The problem statement, all variables and given/known data
A student collects diffraction data using a lamp with known emission wavelengths of 425nm, 565nm, 600nm, and 700nm. These lines appeared on her spectroscope at 32mm, 59mm, 63mm, and 69mm(all measured from the same arbitrary 0mm position). With these data she is able to calibrate her spectroscope, and using this calibrated spectroscope she observes another lamp that has an emission line at 55mm. What is the wavelength of this emission line? (Use Excel to generate an equation of a line with a properly labeled graph)

2. Relevant equations

dsin(θ) = m (λ)

3. The attempt at a solution

In the equation above, I am provided with two out of four variables - I don't have the diffraction grating difference, nor do I have the angle at each wavelength.

What I thought is this: sin(theta) = x (spacing between bright fringes, i.e 32mm) / L (path length). If I substitute it in to the equation above, I would get dx / L = m (λ). I still am missing two variables. Even if I had tan(θ) = x(fringe spacing) / L, and I assumed sin(θ) ~ tan(θ) as the angle is small, I'm still utterly confused.

Now I attempt to address the last part of the problem in parentheses - plotting the equation - I had thought that in mλ = dsinθ, I would be able to find the slope to be some variable, but it seems I am still at a disadvantage without more information in the problem.

Any tips? I'd appreciate anything - better just tips rather than the whole solution if possible; I still want to try and arrive at the solution myself.

2. Sep 25, 2016

### Simon Bridge

Would it help if you knew the order "m" of the emission line(s)?
I think you need a sketch of how the spectroscope works: how do you get from $d\sin\theta = m\lambda$ to "mm from some arbitrary 0mm position"?

3. Sep 25, 2016

### allenlistar

I think I'm supposed to assume that m is 1 and each wavelength is related to each fringe spacing provided...unsure because the text above is all the professor provided.

I'll try looking up how the spectroscope works; but I believe the mm values given are x, or the fringe spacing - measured from the central maximum - perhaps the central maximum is the "0mm?"

4. Sep 25, 2016

### Simon Bridge

The text says that the 0mm position is "arbitrary" - therefore you cannot assume the position is from the central max.
Another thing to wonder about is if the spectroscope works so that the angle $\theta$ is small...

Bottom line: the data is useless unless you know how the spectroscope works ... ie. there could be a lens in it.
So you will have to check your notes to see what sort of thing is expected.

If you decide that the spectrscope data is consistent so that m is the same between reading, or that m=1 every time, then you have eliminated a variable. But the question was: does it help?

One of the ways to reduce the number of variables is to compare results between trials ... like if you look for ratios?