In order to get lasing inside our bacteria we constructed an optical cavity inside and around the bacteria. An optical cavity is an arrangement of optical components which traps the light inside in a closed path (standing waves), where the light can resonate. For most normal lasers this is accomplished by two mirrors (figure 1A). In our project PHB granules are formed inside cells and by total internal reflection the light should be trapped inside them. Another way to form an optical cavity is by encapsulation of the bacteria in a material with a higher refractive index than the inside of the cell. This can be achieved by the enzyme silicatein, which polymerizes monomers such as silicic acid and tin dioxide, creating the desired reflective layer (figure 1B). In order to trap the light inside these cavities it is necessary that the wavelength of light ‘fits’ inside. Therefore standing waves have to be formed within the cavity. Here we will determine what the minimal required size of these cavities is to fit exactly one wavelength. This will not give us a definite answer whether we will get lasing, but a very strict lower bound of the required cavity size.

It is important that the waves follow a closed path forming a polygon so that constructive interference takes place. Constructive interference occurs when two waves travel in phase so that their amplitudes add up (figure 4). We need constructive interference so that every cycle in the resonater adds to the constructive interference. To make sure the waves travel in phase so that constructive interference can take place, the optical path length (OPL) is required to be an integer number of wavelengths (equation 1). In equation 1 n is the number of sides of length l, and m an integer number of wavelengths \(\lambda\).

$$OPL=n\cdot l = m\lambda$$

For intracellular lasers, oil droplets have been used before as an optical cavity (Humar et al., 2015). In our project we use PHB granules inside the E. coli instead of oil droplets as an optical cavity. This PHB granule will then be stained with the fluorescent dye Nile Blue with an emission maximum of 580 nm (Kitamura et al., 1994) to create the gain medium.

In order to get WGM resonance the light beam has to be reflected by total internal reflection every time it hits the surface. Therefore the light beam has to approach the surface at a minimal angle larger than the critical angle. From Snell’s law (equation 2) we can compute the critical angle (Hecht, 2001) where the incoming light is refracted to have an outgoing angle of exactly 90 degrees (figure 3C). In equation 2 and 3, \(n_1\) and \(n_2\) are the refractive indices of the materials at the interface.

$$n_1 sin(\theta_{i}) = n_2 sin(\theta_{f})$$ $$\theta_{c} = \arcsin\Big(\frac{n_2}{n_1}\sin\big(\frac{\pi}{2}\big)\Big)=\arcsin\Big(\frac{n_2}{n_1}\Big)$$

The outgoing light in total internal reflection should have an angle \(\theta_f>\theta_c\) to be reflected. When we look into geometrical optics the path of the light is as shown in figure 5. Since we want a closed optical path (polygon), we have an integer number of sides on the polygon. Using equation 4 from we can determine the number of sides \(n_{sides}\) of the polygon as in equation 5.

$$\phi = \pi - 2\theta_{f}$$ $$n_{sides} = \frac{2\pi}{\phi} = \frac{2\pi}{\pi-2\theta_{f}}$$

We can calculate the number of sides by using the reflected angle \(\theta_f\) in equation 4 and round the \(n_{sides}\) up to an integer. This will give us the smallest polygon possible for WGM. We can then calculate the smallest possible value of \(\theta_f\) to get total internal reflection with a closed optical path as in equation 6.

$$\theta_f = \frac{\pi}{2}-\frac{\pi}{round(n_{sides})} $$

Now that we know how many sides the polygon has, we can compute the minimal size for the granule sphere so that an integer number of wavelengths fit into the optical path length:

$$m\lambda = n_{sides}l$$ $$l = 2Rcos(\theta_{f})$$

We suggest that at least one wavelength should fit into \(l\) ( \(l=\lambda\) ) so that the waves can get trapped. Therefore the minimum size of the granule should be as in equation 9:

$$R = \frac{\lambda}{2\cos(\theta_{f})}$$

Results

For PHB granules the refractive index at a wavelength of 580 nm is \(n_{PHB} = 1.468\) (Huglin et al., 1991). The refractive index of cytosol is \(n_{cytoplasm} = 1.37\) (Liang et al., 2007). Putting these values into our model, we find that the minimal size of the PHB granules is \(1.7\mu m\) (figure 6). This is quite a bit larger than we expect the granules to be since the size of E. coli is only \(1-2 \mu m\) long and the granules will not fill the entire bacterium.

Since the size of the PHB granules that can be grown inside an E. coli bacterium was expected to be smaller than the required size for lasing, we investigated the option of making a biolaser out of the entire cell. Therefore we encapsulated the cell with a layer of biosilica and tin dioxide. The gain medium in this method is provided by fluorophores which we expressed in the cytosol of the cell. To determine the minimal size in this case the calculation was a bit more tricky than for the PGB granules, since we had to take the layer into account and the optical path has a star-like shape (figure 3A).

The critical angle can again be computed by equation 3, where \(n_1\) is the refractive index of the layer \(n_f\), and \(n_2\) is the refractive index of the buffer outside the cell \(n_b\). When the angle is slightly larger than the critical angle we have total internal reflection where the outgoing angle is equal to the incoming angle, therefore we may set \(\alpha = \theta_c\). In this model we will neglect the curvature of the surface which means that we can also set \(\theta_i=\alpha\). Using Snell’s Law (equation 2) again we can compute the outgoing angle \(\theta_0\) (equation 10) where \(n_f\) is the refractive index of the layer and \(n_c\) the refractive index of the cytosol.

$$\theta_0 = asin\Big(\frac{n_f}{n_c} sin(\alpha)\Big) $$

From \(\theta_0\) we can easily determine the angle \( \phi \) since all the angles in a triangle add up to \(\pi\) (equation 11).

$$\phi = \pi-2 \theta_0 = \pi-2\arcsin\Big(\frac{n_f}{n_c} sin(\alpha)\Big)$$

Here the path of the light is star-shaped and the \( OPL=k_{sides}\lambda\). Furthermore we will assume that the total path length per side is minimally \(\lambda\) and that one side is taken as the path from point A to D. The optical path length per side is given as \(OPL_{side}=2\cdot x+l\geq\lambda\). \(X\) and \(l\) can be determined as in equation 12. From equation 12 the minimal radius can be determined for the criteria where \(OPL_{sides} = \lambda\) (equation 13).

$$\left.\begin{matrix} OPL_{sides} =2x+l\geq \lambda \\ x=\frac{t}{\cos\alpha}\\ l =2R\cos\theta_o \end{matrix}\right\} OPL_{sides} = \lambda = \frac{2t}{\cos\alpha}+2R \cos\theta_o =\frac{2t}{\cos\alpha}+2R \sqrt{1-(\frac{n_f}{n_c}\sin\alpha)^2}\geq\lambda $$ $$ R_{min} = \frac{\lambda-\frac{2t}{\cos\alpha}}{2\cos\theta_o}$$

Using the minimal radius of R we can determine the angle \(\psi\) as in equation 14.

$$ \psi=2\arcsin(\frac{t \tan\alpha}{R_{min}})$$

Using \(\psi\) and \(\phi\) we can determine the minimal number of sides of the optical path (equation 15).

$$k_{sides} = \frac{2\pi}{\psi+\phi}$$

In equation 15, \(k_{sides}\) has to be an integer to have a closed optical path. Taken together the constraints as in equation 16 we get an equation for \(\phi\) and \(\psi\) as a function of \(\alpha\).

$$\left.\begin{matrix} \frac{2\pi}{k_{sides}}=\phi+\psi\\ \phi = \pi-2\arcsin(\frac{n_f}{n_c}\sin\alpha)\\ \psi = 2\arcsin(\frac{t\tan\alpha}{R_{min}})\\ OPL_{sides}=2x+l \rightarrow R_{min} = \frac{\lambda-\frac{2t}{\cos\alpha}}{2\sqrt{1-(\frac{n_f}{n_c}\sin\alpha)^2}} \end{matrix}\right\} \frac{2\pi}{k_{sides}}=\pi-2\arcsin(\frac{n_f}{n_c}\sin\alpha)+2\arcsin(\frac{2t\sin\alpha\sqrt{1-(\frac{n_f}{n_c}\sin\alpha)^2}}{\lambda\cos\alpha-2t})$$

Solving equation 16 for \(\alpha\) and using this in the equation for the radius results in the minimal radius required for WGM in a cell covered with a layer of biosilica or tin dioxide.

Results

Using the model described above we determined the minimal radius of a cell covered with polysilica and tin dioxide to create a cavity that allows for whispering gallery modes. The refractive index of polysilica is 1.47 (Liang et al., 2007). The refractive index of tin dioxide is 2 (Pradyot, 2003). For both cases we calculated the minimal radius assuming the thickness of the layer to be 50 nm and a wavelength of 509 nm, the emission wavelength of GFP. From our model we found a minimal diameter of \(1.2 \mu m\) and \(1.6 \mu m\) (figure 8) for polysilica and tin dioxide respectively. These sizes are comparable to the size of a bacterium, so whispering gallery modes should in principle be possible inside a covered cell. However, we made a number of simplifications in our model that will in practice probably prevent the system from lasing at this minimum scale. First, we did not take the polarization of the light into account. Second, the length scales we work with here are only about twice the wavelength, and the thickness of the boundary layer is much smaller than the wavelength. A more accurate description of what is happening within the cell would require us to work in the thin layer limit (known as Mie theory) as we do in Q4. There we calculate the quality factor for the cavity, which indicates how well the light can be trapped within the structure. Third, in this model we only determined whether light waves will fit into the cavity, in Q3 we will also determine whether lasing can take place with the size of the cell and the reflectivity of polysilica and tin dioxide.