# Diffraction Grating and Dispersion

1. The problem statement
A diffraction grating has a slit separation of 2094.0 nm. What is the dispersion of the 2nd order lines at an angle of 30 degrees (in degrees per micrometer)?

## Homework Equations

dsinΘ=mλ
y=(mDλ)/d

3. Attempt at a solution
I began by finding λ using the first equation, so...
2λ=.002094(sin(.30))
λ=.00000548205
Note: I changed d from nm to mm before I began any calculations

We haven't gone over this in class yet, so I'm confused on where to go from here. I believe it has something to do with the second equation listed above, but I don't know what variable represents the dispersion or if these are really even the equations I need. I would like some guidance as for what steps to take next or what equations I should be using- I think just a push in the right direction could help immensely.

BvU
Homework Helper
Hello LL, welcome to PF :)

Check your calculations ! sin(.30) is not the sine of 30 degrees ! I don't see how you can end up with 5.5 nm wavelength (assuming you report it in mm -- why do you do that if they want an answer in degrees per micrometer ?).

And the angular dispersion is nothing else than ##d\theta\over d\lambda##.
So you're better off with the first eqn than with the second (that is -- if I am not mistaken -- a distance y off-axis on a screen at distance D).

Now, what is ##d\theta\over d\lambda## at ##\ \ \theta = \pi/6\ \ ## if ## \ d\sin\theta = 2\lambda\ ## ?

It's more math than physics...

Hi BvU!

Alright, I was able to work on the problem in class- I was missing a vital equation that made the whole process really easy.

The equation should be D=m/(d)cosΘ

So if...
D=dispersion
m=2
d=2.094 micrometers
Θ=30 degrees

D=(2)/(2.094cos(30))
D=1.10 deg/micrometer

I'm sorry about the confusion, I really appreciated your help! You were right, it was just figuring out what math is the right math!

BvU
The "math" (physicists have their own ideas about how to deal with differentials :) ) is real easy: $$\ d\sin\theta = 2\lambda\quad\Rightarrow\ d \cos\theta \; d\theta = 2\; d\lambda\quad \Rightarrow\ {d\theta\over d\lambda} = {2\over d\cos\theta}$$