# Optics: dispersive power of diffraction grating

1. Dec 28, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! Here is the problem I am trying to solve:

a) A source illuminates a grating in a spectroscopical element so that the principal maxima appear as thin bright bands (therefore the name "spectral lines"). Show that the angular width $\Delta \theta$ of such a line is inversely proportional to the width of the grating (in case of normal incidence).
b) Find an expression for the angular width per (small) wavelength's range, i.e. the dispersive power $\lim\limits_{\Delta \lambda \to 0} (\Delta \Theta/\Delta \lambda)$. Calculate the dispersive power for the 1st and 2nd order of a diffraction grating with a number of 700 line per cm (i.e. $\lambda \approx 500$nm).

2. Relevant equations

Grating equation: $a (\sin \theta_m - \underbrace{\sin \theta_e}_{=0}) = m \lambda$ where $a$ is the width of one slit and $m=0,1,2...$.

3. The attempt at a solution

a) We recall that the principal maxima occur for $\alpha = 0, \pm \pi, \pm 2\pi,...$. The minima occur at $\alpha= \pm \frac{\pi}{N}, \pm \frac{2 \pi}{N},...$ so I calculate $\Delta \alpha$ as the distance between the neighbour minima of a principal maxima, i.e. for example:

$\Delta \alpha = \frac{\pi}{N} - \big(- \frac{\pi}{N} \big) = \frac{2 \pi}{N}$

Recalling that $\alpha = \frac{k a}{2} \sin \theta$, we can obtain by differentiating $\Delta \alpha = \frac{k a}{2} \cos \theta_m \Delta \theta$ and thus:

$\frac{2 \pi}{N} = \frac{k a}{2} \cos \theta_m \Delta \theta \\ \implies \Delta \theta = \frac{2 \lambda}{N a \cos \theta_m} \\ \implies \Delta \theta \mbox{ is inversely proportional to}\frac{1}{N a}.$

So far so good (hopefully), but now I am stuck at b). I have been searching on internet and in the two books I have but I can't make sense of this limit as it is given in the problem. Neither could I find what this so-called dispersive power is. In the book of Hecht, there is something called the angular dispersion, which is defined as:

$\mathcal{D} = \frac{d \theta}{d \lambda}$.

Could that be the same thing? It looks slightly different though, the limit in the problem is not the definition of a derivative. Could you guys help me understand what I am supposed to do?

Julien.

2. Dec 28, 2016

The primary maxima occur at $m \lambda=d sin(\theta)$. There is another formula that gives the general form of the intensity as a function of angle for a grating with $N$ equally spaced lines. That formula reads $I(\theta)=I_o \frac{sin^2(\phi/2)}{sin^2(\phi/2)}$ where $\phi=\frac{2 \pi}{\lambda} d sin(\theta)$. This formula for the intensity is not too difficult to understand, and can be used to derive the width of the primary maxima lines. The primary maxima occur when the numerator and denominator are both zero, in which case the limit of the function at these angles is $I(\theta_{max})=N^2 I_o$. These primary maxima occur when (looking at the denominator), $\phi/2=m \pi=\frac{2 \pi}{\lambda}d sin(\theta)/2$. The result is $m \lambda=d sin(\theta)$ where $m$ is an integer. (Notice the numerator will also be zero at these points). $\\$ Now the numerator goes to zero (because of the $N$ ) many more times than the denominator. The width of each primary maximum is defined as the distance $\Delta \theta$ to the first adjacent zero of the numerator, (when the denominator is no longer zero.) We have $N \phi/2=N m \pi=N \pi d sin(\theta_{mo})$ and the next adjacent zero is $N \phi/2 =(N m +1) \pi =N \pi d sin(\theta_{m1})/\lambda$. Subtracting the $Nm \pi$ equation from $(N m+1) \pi$ equation gives us $\pi=N \pi d (sin(\theta_{m1})-sin(\theta_{mo}))/\lambda$ which gives $\Delta \theta= \lambda/(Nd)$. (using the approximation that $sin(\theta)=\theta$ ). $\\$ Now we also have $m \lambda=d sin(\theta)$ so that $m \Delta \lambda=d \Delta \theta$ (approximately). $\\$ Combining these gives $\frac{ \lambda }{\Delta \lambda}=N m$ which is the common expression for the resolving power of a spectrometer. Hopefully this helps to answer your question.

3. Dec 28, 2016

### JulienB

Hi @Charles Link and thank you for your answer. It's pretty clear and very helpful. However I still have one question: what is your interpretation of the limit expression given in the problem? One could write (I suppose):

$\lim\limits_{\Delta \lambda \to 0} (\Delta \theta/\Delta \lambda) = \lim\limits_{\Delta \lambda \to 0} \frac{\lambda}{N d \Delta \lambda} = \frac{1}{N d} \lim\limits_{\Delta \lambda \to 0} \frac{\lambda}{\Delta \lambda}$

so that the expression $(\lambda/\Delta \lambda)$ appears in the limit. But that is still not quite the same thing as just $(\lambda/\Delta \lambda)$. Can you clear up my misunderstanding?

Thank you very much again, I appreciate your help very much.

Julien.

4. Dec 28, 2016

$\Delta \lambda$ is not zero. The width of the diffraction pattern (i.e. the primary maximum and a couple of significant secondary maxima) from a perfectly monochomatic spectral line is not zero but rather the finite $\Delta \theta$ and $\Delta \lambda$. (The width is usually defined as the width of the primary peak from center of the peak (of the intensity function) to the first adjacent zero of the intensity function). $\\$ When you take $m \lambda =d sin(\theta)$ and take the derivative to get $\frac{d \lambda}{ d \theta}=d cos(\theta)/m =d/m$ (approximately), that is a mathematical process where you can let $d \theta$ and $d \lambda$ approach zero. The two things should not be confused. They are completely different.

Last edited: Dec 28, 2016
5. Dec 28, 2016

### JulienB

Yes I understand the difference, unfortunately the limit definition I mentioned in post #3 is the one given with the problem. I'm still searching that expression on the net and I'm still not finding anywhere something looking alike. If I give back the problem to my teacher without even mentioning the formula he gives, there is a problem either on his side (formulating) or on mine (interpreting).

Julien.

EDIT:
And surely $\lim\limits_{\Delta \lambda \to 0} (\Delta \theta/ \Delta \lambda) \neq \frac{d \theta}{d \lambda}$ or??

6. Dec 28, 2016

Your last equation is correct, but basically you can look at the monochromatic line as a blurred line that has a finite width. Another spectral line of a slightly different wavelength will blur into the first one (and be unresolvable from the first) unless the spacing in wavelength is greater than $\Delta \lambda_{minimum \, resolution}=\lambda/(Nm)$. $\\$ And note that these resolution equations are approximate. To easily resolve a doublet spectral line, the spacing $\Delta \lambda$ of the two lines should really be of the order of $\Delta \lambda \geq 2 \, \Delta \lambda_{minimum \, resolution}$.
$\\$ editing... I think the formula for the intensity is listed in many elementary textbooks that treat diffraction gratings and multi-slit interference. A reflective diffraction grating works with the same intensity equation as a transmissive (slits) grating where the lines of the grating act as Huygen's mirrors. (To be more precise, the phase (difference) $\phi$ is given by: $\$ $\phi=\frac{2 \pi}{\lambda}d(sin(\theta_i)+sin(\theta_r))$. I don't have a good text for you to read about diffraction grating spectroscopy. (Hecht and/or Hecht and Zajac may cover it somewhat thoroughly, but I don't have a copy of that book.)... I learned the subject many years ago and was a T.A. (teaching assistant) for an upper level undergraduate spectroscopy laboratory course.

Last edited: Dec 28, 2016
7. Dec 28, 2016

@JulienB Please see the edited version of post #6.

8. Dec 28, 2016

### JulienB

$m \lambda = a \sin \theta \implies \lambda = \frac{a \sin \theta}{m} \implies \frac{d \lambda}{d \theta} = \frac{a \cos \theta}{m} \\ (\Delta \theta)_{min} = \frac{\lambda}{N a \cos \theta_m} \implies \frac{(\Delta \theta)_{min}}{(\Delta \lambda)_{min}} = \frac{m}{a \cos \theta} = \frac{\lambda}{(\Delta \lambda)_{min}} \frac{1}{N a \cos \theta_m} \\ \implies \frac{\lambda}{(\Delta \lambda)_{min}} = N m$
That way I get the same result as you did in post #2.. But $(\lambda / \Delta \lambda)$ is the resolving power and not the dispersive power (which is what is asked in the problem) right? Then shouldn't the answer simply be:

$\frac{d \theta}{d \lambda} = \frac{m}{a \cos \theta}$

or approximated for small angles:

$\frac{d \theta}{d \lambda} = \frac{m}{a}$ ?

Thanks a lot again for your help.

Julien.

9. Dec 28, 2016

@JulienB Yes, what you have looks correct. And yes, I believe dispersive power is $\frac{d \theta}{d \lambda }$.

10. Dec 28, 2016

### JulienB

Very nice, thx a lot!

Julien.

11. Dec 28, 2016

And yes, part "a" asks for $\Delta \theta_{min}=\lambda/(Nd)=\lambda/w$ where $w$ is the width of the grating.