# Multiple wavelengths of light through much larger single slit

littlepapa87
A text I am reading noted that as a multichromatic light source (consisting of many individual peaks) is passed through the my slit (order .25mm) the peaks will be spread over the neighboring wavelengths in a roughly Gaussian distribution. I am trying to understand why this is, and how I would go about finding the FWHM of the spread in wavelength space.

my application for this is Raman spectroscopy. For my work the light is scattering at 580nm 607nm along with a few others for vibrational transitions, and from 532-560ish for rotational transitions.

## Answers and Replies

RedX
A text I am reading noted that as a multichromatic light source (consisting of many individual peaks) is passed through the my slit (order .25mm) the peaks will be spread over the neighboring wavelengths in a roughly Gaussian distribution. I am trying to understand why this is, and how I would go about finding the FWHM of the spread in wavelength space.

my application for this is Raman spectroscopy. For my work the light is scattering at 580nm 607nm along with a few others for vibrational transitions, and from 532-560ish for rotational transitions.

I don't know much about real experiments, but isn't .25 mm a little too large to observe any interference with light? The ratio of the slit width to the light wavelength is like a 1000. Also I didn't realize vibrational and rotational modes are of the order of electronic transitions. I thought modes that had to do with the lattice are in the infrared or something.

Anyways, for a single slit, the formula for the intensity (when the screen is infinitely far away) is:

$$I=\frac{\lambda^2}{\pi^2L^2sin^2\theta} (sin(L sin\theta \pi / \lambda) )^2 = \left(\frac{sin(x)}{x}\right)^2$$

So L is the length of your slit (.25 mm), lambda is your wavelength, and theta is the angle from the slit to the point on the screen you want to calculate your interference. The intensity is normalized such that the value at theta=0 is 1. So if lambda has some distribution, you can find the distribution on the screen from this formula. I'm not exactly sure why this comes out to be Gaussian unless your distribution of light has a term inverse proportional to that sine function:

$$n(\lambda)=\left(\frac{1}{\lambda}\right)^2\left(\frac{1}{sin(L sin\theta \pi / \lambda)}\right)^2 *Gaussian(\lambda)$$

littlepapa87
We are using an incoming laser at 532nm so the shifts are actually quite small.

The slit is only being used to define our probe volume to eliminate unwanted light. Also I know that the Gaussian is only being used as an approximate function for the "smearing" across wavelengths that is being done when the light passes through the slit. My understanding was that it is more similar to the fuzzy edge of a shadow than to the interference pattern seen when light passes through a slit with a size similar to the wavelength.

Just for completeness sake, after passing through the slit the light is then passed through a spectrograph (collimating lens --> diffraction grating 1200lpmm --> focusing lens --> CCD) so that we may image the various peaks (transitions).

Does this make sense, or at this point should I go back an re-examine how the system is working?

It isn't critical, I am just curious how the apparatus is working so that on the side I might be able to "simulate" what I believe will happen.