Coincidence Time Measurement of Pulsed Lasers & "Useful" Applications

Members

Tan Wen Yi, Kim Mu Young

Abstract

Given two paths through a which a pulsed laser propagates, one might be interested in ensuring that the pulses coincide on the same spatial point at exactly the same time. One possible situation where such a coincidence may be useful is in the driving of Raman transitions via a pulsed laser. We intend to perform such a coincidence time measurement using commonly-available metals. By choosing a material with a work function on the order of the energies of two 355nm photons, the detection of a photocurrent through a metal would then indicate that both laser pulses arrived at the metal at approximately the same time. The project first aims to find such a metal and design a set-up to perform this experiment; after which, we intend to measure the timing resolution such a method is limited to. If time permits, we then intend to try to use this method to perhaps characterise the density/surface smoothness and other possible features of the metal object used.

Theory

Photoelectric Effect

The photoelectric effect is a simple yet revolutionary phenomena, being one of the first experiments to show the quantisation of energy. Every metal surface has a characteristic feature known as the work function ($\varphi$) which describes the amount of energy required for an electron to be ejected from its crystal structure into the continuum. If one illuminates such a metal surface using some light source, the energies of the electrons that are emitted are characterised by a simple equation:

$E_{electron}=hf-\varphi$

Where $h$ is the Planck's constant and $f$ is the frequency of light used.

There are three possible results based on the frequency of light.

$\{\begin{array}{c}hf<\varphi \to E_{electron}<0\to \text{No electrons emitted}\\ hf=\varphi \to E_{electron}=0\to \text{Surface electrons}\\ hf>\varphi \to E_{electron}>0\to \text{Electrons are emitted with kinetic energy}{E}_{electron}\end{array}$

A typical set-up used to test the model described above goes as the following (obtained from https://physicscatalyst.com/chemistry/photoelectric-effect.php, but will make our own later on):

A light source is used to illuminate a metal surface. If the frequency of the light is high enough, electrons are ejected from the surface. These ejected electrons are accelerated onto a detector via the application of an external potential. Once these electrons fall upon the detector, they travel through the wires, causing a current to be detected. As such, depending on the lack or presence of a current, one could tell if the frequency of light used is sufficiently high to overcome the work function of the metal.One additional note regarding this experiment is that the amount of current detected is not proportional to the frequency, but rather the intensity of light. While a higher frequency of light would provide the electrons with higher kinetic energies, this does not lead to a rise in current. However, a larger intensity, corresponding to a larger number of photons, would mean that more electrons are emitted, causing higher currents.

Pulsed Lasers

Pulsed lasers differ from the more commonly found continuous wave (cw) lasers in that they emit light in a discrete manner. The difference between the two can be understood from the following picture:

In addition to the usual parameters that characterises a laser; the wavelength and the power, pulsed lasers have two additional important numbers. The first is the pulse duration, which refers to how long each pulse lasts in the time domain. The second is the repetition rate, or how many pulses are emitted per second.

Experiment

Consider a situation where one has a pulsed laser split into two beam paths, which are then directed to coincide on a single metal plate. The material of the metal plate is chosen such that a single photon from one of the paths of the pulsed laser is insufficient to produce an electron ($hf<\varphi$). However, if both photons from each beam path is incident upon the metal plate, there is then sufficient energy for a photoelectron emission ($2hf>\varphi$). Since a pulsed laser is used, this will only happen when the pulses from each of the arms arrive at the metal plate at the same time. This would form the so-called "coincidence time measurement of pulsed lasers".

The main motivation and purpose of having such a set-up would be exactly when one requires the lengths of two or more beam paths to be exactly the same. Only when the lengths are approximately (depending on the pulse duration) equal to each other would a photoelectron be observed, allowing for precise measurements of distances.

Project Breakdown

Parameters of Key Devices

Laser - Coherent Paladin Compact

Details: 355nm, 15ps pulse duration, 120MHz repetition rate, 4W output power but about 400mW is actually available.

Photocathode - Copper sheet

Details: Work function of approximately 3.7eV

Vacuum - Random vacuum part lying around

Details: Have a pump to evacuate to around 10^(-3) Torr

High Voltage Source - High Voltage Amplifier

Details: Supposed max output of 800V, but only stable till about 500V or so.

Dynode - There are a few common materials for dynodes nowadays. The comparison chart is shown below (ALD: Atomic Layer Deposition. Basically impossible to use for us).

Steps Towards Success

This project has been broken down into the following steps:

  1) Design and Construction of DIY electron multiplier
  2) Design and Consturction of vaccum chamber (mainly an issue of getting things to fit inside nicely)
  3) Evacuation of vaccum chamber
  4) Start experiments.

But to ensure that the experiment would work, we first perform some simple simulations to estimate what would be the resources required to generate a high enough signal that may be detected.

Simulations

This project is generally heavily limited by the resources that we have on hand. As such, we initially perform some simple simulations to ensure that whatever resources we have are sufficient for the actual experiment to work.

Part 1: Photoemission

The energy associated to a wavelength of 355nm is given by

$E_{\lambda }=hf=h\ast c/\lambda \approx 3.493eV$

Compared to the work function of copper, ${\varphi }_{Cu}=4.7eV$, we can easily see that one photon is insufficent for the release of a photoelectron. We assumed here that there are no additional energies due to thermal excitations within the copper itself, the validity of which remains to be seen.

For two photons at 355nm, the resulting kinetic energy of the photoelectron is

$E_e=2hf-4.7\approx 2.285eV$

Part 2: Primary Dynode

We assume here that we are able to generate a stable 500V from the high voltage supply. In which case, due to the accleration of the photoelectron towards the positive charged dynode, the kinetic energy of the electron at the primary dynode is

$E_{e,dynode}=2.285+V\times e=502.285eV$

The value of which tells us that the kinetic energy at which the electrons are emitted do not actually matter much.

The number of secondary electrons emitted from the primary dynode depends on the energy of the incoming primary electron and the voltage the dynode is put at. For now, since we don't know what materials we are able to obtain, let's say that the number of secondary electrons produced is 2.

Part 3: Electron Multiplication

The number of electrons at the end of the multipler is simply given by the number of secondary electrons emitted at each dynode (which is approximately the same as the number of electrons emitted at the primary dynode) to the power of how many dynodes are available.

Typically, there is a 100V potential difference between the dynodes themselves. This means that with a maximum voltage of 500V, we should have 5 dynodes at 500V, 400V, 300V, 200V and 100V, and one additional dynode at ground to act as the anode.

With 2 secondary electrons per dynode, this would mean that we would have 2**5 = 32 electrons at the anode.

Part 4: Detection

We now need to see if the current generated by the multipled electrons is at a detectable scale.

The scaling goes as 1A = 6.242 \times 10^(18) electrons per second, meaning 32 electrons would give about 5*10**(-18) Amps.

Part 5: Scaling

In all of the above calculations, we have assumed that there is only one photoelectron emitted per pulse. However, the number of photoelectrons emitted is proportional to the intensity of the incident light. The exact scaling goes by

$N_e=N_{photon}=\frac{I\times A}{h\times f}$