Diffraction Grating and Dispersion

[lauraliz94]

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]

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...

 

[lauraliz94]

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

Then the equation reads...
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]

And here's me thinking you were supposed to derive this vital equation. You were nearly there!

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