Electromagnetic radiation is energy travelling as waves or photons. This is not the place or the time to go into this discussion [1] but Einstein and Infeld said it well:

Light, or the visible light spectrum range from \( \in [400 - 700] \) nm in the electromagnetic radiation spectrum. XX insert picture of Elmag spectrum here XX. Above, with higher wavelengths, you will find infrared radiation (also known as IR), and under you will find the ultraviolet radiation (also known as UV). We call it visible light due to the fact that our eyes can only "pick up" these wavelengths. For this project we will mostly focus on light \( \lambda \in [470,520] \) nm region.

You have probably heard that nothing can travel faster than light? But not that many (non-physicist) remembers what velocity light actually travels with. In most cases it's enough to say that light travels in \( \sim 3.0 \cdot 10^{8}\) m/s in vacuum or \( \sim 6.7 {\cdot 10^{8}} \)mph.

When talking about light there is a couple of expressions that is good to know the meaning of

• Monochromatic
• Coherent
• Wavelength
• Optical Path

\begin{align} \sin\theta = \end{align}

The light from a light-emitting diode, LED for short, can be viewed by the human eye as monochromatic. This means it appears to only light up as one color. From theory we know that a LED cannot be monochromatic, thus proving one of the biggest differences between a LED and a laser. More importantly, the light from a LED is not coherent, meaning that the light waves do not have the same frequency. A LED loses little energy to heat as it applies most of its energy to light up the diode. This is a very useful property.

To create a simple LED-circuit you only need a couple of components: a LED, a resistance, some wires, a circuit board and a voltage source. We used a blue LED lamp with the properties 20mA and 3.2V, a resistance of 1kΩ and a PHYWE power supply that allowed us to vary our voltage between 0-30V. The LED and the resistance were mounted to the circuit board using a soldering iron, before the wires were soldered to each side of the board (insert figure).

The given wavelength $\lambda$ for a lightsource can be found by XX INSERT FORMULA HERE XX, by knowing the grid spacing, $d$ and angle, $\theta$.

Continuing this part we will refer to light as waves. As the picture above shows, visible light range around ∈[400−700] nm also known as the wavelength, we also use the symbol \(\lambda\) of the light. Since it is easier to understand physics by looking at a specific example we will now use the rainbow as an example. Most people have had the pleasure to see this magnificent phenomenon in their life. To understand what happens to the light, we need to introduce the term refraction index. Which means in short terms that light will behave differently in different mediums if hit with an angle different than zero from the optical path. For air, the refraction index is simply one, but for water it is \(1.33\) both in vacuum. Using Snell’s law and the fact that the light travels from air to water we can see that the angle the light will be emitted(correct term??) by can be found

A laser is a device that emits monochromatic light amplificated by stimulated emission of electromagnetic radiation, hence the name (“Light Amplification by Stimulated Emission of Radiation”).

The first laser was built in 1960 by Theodore H. Maiman, based on theoretical work by Charles Hard Townes and Arthur Leonard Schawlow. What makes a laser different from other light sources is the fact that it emits its light coherently. This means that the stream of light will stay narrow over a long distance and can also be focused in a tight spot, like for example for laser cutting or a laser pointer.

A laser consists mainly of five parts: the gain medium, the laser pumping energy, the high reflector, an output coupler and a laser beam. The gain medium is a material which allows light to amplify. This is usually located in an optical cavity. What this means is that there is a mirror on either side of the medium, in order to make the light bounce back and forth. This amplifies the light, as it passes through the gain medium each time. One of the mirrors is usually not 100% reflective, since there will a small output (the laser beam). For anything to happen within the optical cavity we need a power source that allows the gain medium to amplify the light. This energy is supplied through a process called pumping and is usually either light or an electrical current.

When we use the term biolaser we refer to a biological component that posesses the same qualities as a laser. What we mean by this, is that the gain medium is now the biological component. Over to biology.

Bilde av set-up, med forklaring:

Voltage LED: our LED had a maximum at 20V, but as our sample light up at only 5V, we used this.

Led, filter (med spesifisering av begge filtre), speil (detaljer her og), oversikt over linser (ikke nødvendigvis med nøyaktig brennvidde, men hva de gjorde og hvor de sto), spektrometer og CCD-kamera.

What do we want to do? Observe specter to determine whether or not we have laserlight.

How are we expecting to observe the spectre emittet from our test? We were operating a CCD camera that did not include a lense, it only had the detector in place. In order to know if we will get an image we can see and work with (given that the rest of the setup works), we calculated how big the detected image would be.

We are working with wavelengths that go roughly from 505 nm-515 nm, so we wish to calculate the angle between the two waves when they pass through the slot of the spectrometre. We also know that 4000 bars occupy a space of 2.5 cm, so we can calculate d:

\begin{align} d = 2.5 \cdot 10^{-3} / 4000 = 6.25 \cdot 10^{-6} m \theta = sin^{-1} \left(\frac{505 \cdot 10^{-9}m}{6.25 \cdot 10^{-6}m}\right) = 0.081 rad \phi = sin^{-1} \left(\frac{515 \cdot 10^{-9} m}{6.25 \cdot 10^{-3}}\right) = 0.082 rad \end{align}

The distance from the grid to the tip of the spectrometre (where we would out our camera) is 0.3m.

So, the space between the two waves (and so, the size of our image) is:

begin{align} (0.082 - 0.081) \cdot 0.3m = 0.3mm \end{align}

To check if this works with our camera, we take the number of pixels pr 5 mm, and get:

\begin{align} \frac{0.005 m}{750 px} = 6.67 \cdot 10^{-6} = 7 \mu m \end{align}

Anything bigger than 7 \mu m should be easily observed, and our image should project at around 0.3 mm, so we are good to go!