# Can this simple spectrometer be precise without collimation?

Consider the following spectrometer structure:

Here the slit can be changed in size, and the lens and detector can be moved closer to or farther from the diffraction grating. I was trying to get decent resolving power from such a spectrometer, but have failed so far, even with extremely tiny slit.

In fact, after I tried to draw the rays from a point source, as they extend after diffracting to the $1^\text{st}$ order, I found that they are actually don't converge in a single point, so it seems it's impossible to focus them to a single point. See this image (black are rays from source, red are first-order diffracted rays, green lines are the continuation of diffracted rays):

After adding a collimation lens between the slit and the grating I got much better focusing, so collimation seems to be the problem.

But really, is it possible, without adding a collimating lens between the slit and the grating, to get high spectral resolution (say, detect a nanometer difference in laser wavelength), without requiring that the slit-grating distance is very large?

• @ChrisMueller The lens actually is the objective of a webcam, and I've tried adjusting the focus in the whole range of its positions. But when the point corresponding to diffracted image became the smallest, it still appeared quite large, much larger than zeroth order point. – Ruslan Sep 22 '14 at 13:32
• Can you image an object which is the same distance from the webcam as the slit? If so, then you can probably increase the resolving power (from looking at your ray diagram) by adding an aperture between the slit and the diffraction grating. Better yet, add another lens between the slit and the diffraction grating to convert the point source to plane waves and forgo the aperture so you don't throw away more photons. – Chris Mueller Sep 22 '14 at 13:40
• @ChrisMueller yes, I was able to image the slit by rotating the webcam to it. As for adding aperture, it's basically a mild collimation of the beam. I've already succeeded in improving the resolution by adding a collimating lens. The question is whether it's somehow possible to do it without doing collimation, i.e. just by some simple processing (by objective or whatever) of outgoing rays from the diffraction grating. – Ruslan Sep 22 '14 at 13:46
• Many reflective grating spectrometers do their job without a lens at all. Just a slit, the grating and a projection surface (which is often curved to get a uniform scale). Can you try it without the lens? – dmckee --- ex-moderator kitten Sep 22 '14 at 14:34
• @dmckee well of course I can, but the result is obvious: almost uniform color in the captured frame. I guess the spectrometers you mention have a non-flat grating, which also focuses the light it reflects. Then indeed one would need no further lenses. – Ruslan Sep 22 '14 at 15:15

As you point out in your ray diagram, the point source no longer looks like a point source after the diffraction grating. You could get around this by collimating the input beam with another lens (which you did), with an aperture, or by moving the point source far away from the grating (effectively using the grating as an aperture). So, I think the answer to your question is no unless you are willing to better collimate the input source.

## Two Diffraction Gratings

You could, however, use a second diffraction grating to reproduce the original image. To see this, consider one of the rays in your diagram with initial angle at the grating of $\theta_1$. After the grating the ray in the first order diffraction will have angle $$\theta_2=\arcsin\left(\frac{\lambda}{d}-\sin(\theta_1)\right),$$ where $\lambda$ is the wavelength and $d$ is the grating's spacing. After the second grating the first order diffraction will have angle $$\theta_3=\arcsin\left(\frac{\lambda}{d}-\sin\left(\theta_2\right)\right)=\theta_1.$$ So, the angle of the ray is restored to its original value after the second diffraction grating. I.E. the image of the point source is restored.

The position of the ray's which form an image will be shifted by an amount determined by the spacing between the two gratings and the wavelength. To see this consider the position of the ray along the optical axis of the system. For this ray $\theta_1=0$ and $$\theta_2=\arcsin\left(\frac{\lambda}{d}\right)$$ As we just showed, the angle after the second grating will be $\theta_3=0$ again, but the ray will have translated by an amount given approximately by $$\delta x=L\ \theta_2=L\ \arcsin\left(\frac{\lambda}{d}\right),$$ where $L$ is the distance between the two gratings.

This will achieve exactly what you desire since the point sources of different wavelengths will be imaged to different locations on your CCD.

## Resolving Power

We can estimate the resolving power of such a spectrometer by considering two different closely spaced wavelengths: $\lambda_0$ and $\lambda_0+\delta\lambda$. In this case the separation on your CCD will be \begin{align} \mu&=\delta x_2-\delta x_1\\ &=L\left[\arcsin\left(\frac{\lambda_0+\delta\lambda}{d}\right)- \arcsin\left(\frac{\lambda_0}{d}\right)\right]\\ &\approx \frac{L\ \delta\lambda}{\sqrt{d^2-\lambda_0^2}}. \end{align} So the resolving power will be $$\delta\lambda=\frac{\mu}{L}\sqrt{d^2-\lambda_0^2},$$ where $\mu$ is the CCD pixel separation. Lets use a grating separation of $L=10\ \text{cm}$, a typical CCD pixel size is $\mu=20\ \mu\text{m}$, a CD has a diffraction grating spacing of $1.5\ \mu\text{m}$, and red light has a wavelength of $\lambda_0=600\ \text{nm}$. Putting all of these numbers in gives a resolving power at the CCD of $$\delta\lambda=0.25\ \text{nm}.$$ In reality, you will probably hit some imperfections in a simple home setup which will prevent you from reaching the full sensitivity (grating imperfections, relative grating misalignment, etc), but it seems like $1\ \text{nm}$ is very achievable.