Can Diffraction Gratings Resolve Close Wavelengths in Visible Light Spectrum?

[cheff3r]

Homework Statement

A line corresponding to light of the wavelength 519.8 nm is observed at an angle 12 degrees in the second-order spectrum of a diffraction grating. The grating is 1 cm in width,
what is the line spacing of the grating?
how many per centimetre?
what is the lowest order in which the grating can resolve both this line at 519.8 nm and a near by line at 520 nm?
If these lines were to be resolved in the first-order diffraction pattern, what width would the grating need to be made?
If the grating is illuminated by light (400-700nm) what is the largest order for which complete spectrum of visible light can be observed?

Homework Equations

m*\lambda=d*
where m is the 'order'

The Attempt at a Solution

So I don't really now what I'm doing yet, but I want only help on the first couple parts then when I get them right possible help in the later sections
so for the first question
d= \frac{m*lambda}{sin (theta)} = \frac{2*519.8*10^-9}{sin 12} = 5*10^-6 m
so does this represent spacing between each diffraction? if this is the case then the next part is 5*10^-4 per centimetre??

Either way what the next part is not really related so I will attempt it as well (what is the lowest order in which the grating can resolve both this line at 519.8 nm and a near by line at 520 nm?)
so i used the formula twice
m*\lambda=d*sin \theta
once for each wave length (\lambda) and the plane was to solve it for different m's until it gave the same d, however I'm unsure what to do with the sine term can i drop it? (since we are not considering an angle)

Yeah so I am likely to be completely wrong with this (especially the last bit I tried) but I did a decent attempt the problem,
Has anyone one got a decent website for teaching this topic? (I can't find one)

 

[cheff3r]

Am I missing something like it must be a maximum intensity or something?

Can anyone point me towards a website on second-order spectrum stuff, i have looked can't find any decent information on the topic (only general ideas)

 

[Jolb]

In problems that use Bragg's law, depending on how the laser is shined on the grating (or crystal lattice), you might have a coefficient of 2 out front (c.f. x-ray diffractometry). Most questions that mention a diffraction grating use the equation with a coefficient of 2. Without a diagram of the problem I can't easily describe to you which to equation to use--if the light diffracts through a slit then there's no 2 but if it bounces through a layer or two of diffraction grating then multiply by 2.

Assuming there's no coefficient of 2, the way I learned Bragg's law is:
n\lambda = d\sin \theta_{n}
Lambda is the wavelength, theta is the angle from the midpoint (of the screen) of the peak in question, d is the size of the slit (spacing between gratings), and the peaks in question are numbered 1, 2, 3,... in order from the midpoint. (The peaks are symetrically located about the midpoint, i.e., there is an n=1 peak at +\theta_{1} and at -\theta_{1})

Therefore you can calculate the intra-layer spacing of the grating=d.

In your problem, they say the line is "second order", and then they say m is the "order", which to me means m=2, and so my equation would have n=2. Therefore,
d=\frac{n\lambda}{\sin \theta}=\frac{(2)(519.8\ \textup{nm})}{12 \ \textup{degrees}}
Which yields the same as you, about 5 microns between gratings. Therefore, there is 1 grating every 5 microns.

Now here's where you made a mistake.
\left (\frac{1 \ \textup{grating}}{5\ \textup{microns}} \right )\left ( \frac{10,000 \ \textup{microns}}{1 \ \textup{cm}} \right )=\frac{2000\ \textup{gratings}}{1 \ \textup{cm}}

However, I would double check to make sure you weren't supposed to use
n\lambda = 2d\sin \theta_{n}